GSEB Class 12 Physics Solutions Chapter 1 Electric Charges and Fields

Gujarat Board Textbook Solutions Class 12 Physics Chapter 1 Electric Charges And Fields

 

Question 1. What is the force between two small charged spheres having charges of \( 2 \times 10^{-7} \)C and \( 3 \times 10^{-7} \)C placed 30 cm apart in the air?
Answer:Solution:
Given charges: \( q_1 = 2 \times 10^{-7} \) C, \( q_2 = 3 \times 10^{-7} \) C
Distance between charges: \( r = 30 \) cm \( = 30 \times 10^{-2} \) m
Using Coulomb's Law, the force \( F \) is calculated as:
\( F = 9 \times 10^9 \frac{q_1 q_2}{r^2} \)
\( = 9 \times 10^9 \times \frac{(2 \times 10^{-7}) \times (3 \times 10^{-7})}{(30 \times 10^{-2})^2} \)
\( = 0.06 \times 10^{-1} \) N
\( = 6 \times 10^{-3} \) N
The force is repulsive because both charges are positive.
In simple words: We used Coulomb's law to find the pushing force between two tiny charged balls. Since both charges are positive, they push each other away.

🎯 Exam Tip: Remember to convert all units to SI (meters, Coulombs) before applying Coulomb's law. Always state the nature of the force (attractive or repulsive) in your answer.

 

Question 2. The electrostatic force on a small sphere of charge 0.4 pC due to another small sphere of 0.8 \( \mu \)C in air is 0.2 N.
**(d) What is the distance between the two spheres?**
**(b) What is the force on the second sphere due to the first?**
Answer:Solution:
Given charges: \( q_1 = 0.4 \, \mu \)C \( = 0.4 \times 10^{-6} \) C, \( q_2 = -0.8 \, \mu \)C \( = -0.8 \times 10^{-6} \) C
Given force: \( F = 0.2 \) N
(a) To find the distance \( r \):
From Coulomb's law, \( F = 9 \times 10^9 \frac{q_1 q_2}{r^2} \)
So, \( r^2 = \frac{9 \times 10^9 \times q_1 q_2}{F} \)
\( = \frac{9 \times 10^9 \times (0.4 \times 10^{-6}) \times (0.8 \times 10^{-6})}{0.2} \)
\( = 14.4 \times 10^{-3} \)
\( = 144 \times 10^{-4} \) m\(^2\)
\( r = \sqrt{144 \times 10^{-4}} \) m \( = 12 \times 10^{-2} \) m
(b) The force on the second sphere due to the first is the same in magnitude but opposite in direction according to Newton's third law. Since the charges are of opposite signs (\( q_1 \) is positive and \( q_2 \) is negative), the force is attractive. So, the force is 0.2 N, attractive.
In simple words: We used the given force and charges to find the distance between the spheres using Coulomb's law. Then, we remembered that forces between two objects are always equal and opposite, so the second sphere feels the same amount of force as the first. Because one charge is positive and the other is negative, they pull towards each other.

🎯 Exam Tip: Remember that Coulomb's law follows Newton's third law, meaning the force exerted by object A on B is equal and opposite to the force exerted by object B on A. Always indicate if the force is attractive or repulsive based on the signs of the charges.

 

Question 3.
**(a) Explain the meaning of the statement 'electric charge of a body is quantised'.**
**(b) Why can one ignore quantisation of electric charge when dealing with macroscopic (i) e., large-scale charges?**
Answer:Solution:
(a) The smallest unit of charge is the elementary charge, which is the charge on a proton or an electron, denoted as \( e = 1.6 \times 10^{-19} \) C. All charges found in nature are whole number multiples of this elementary charge. This means that an object can only have a charge of \( \pm e, \pm 2e, \pm 3e, \) and so on, but never a fraction of \( e \). Mathematically, an object's charge \( q \) is always \( q = \pm ne \), where \( n \) is a whole number (an integer).
(b) When dealing with large-scale charges, the quantization of charge can often be overlooked because the elementary charge \( e \) is extremely small. For a large body, the total charge involves a huge number of electrons or protons. Even if a few electrons are added or removed, the change in the total charge is so tiny compared to the overall charge that it appears to vary smoothly, almost continuously. Therefore, for macroscopic (large-scale) charges, the discrete nature of charge is not noticeable.
In simple words: (a) Electric charge comes in tiny packets, like small coins, and you can't have half a coin. So, any total charge is just a count of these basic packets. (b) For very big charges, these tiny packets are so small that the charge looks like a smooth amount, not individual packets, so we don't worry about the "packet" idea for large objects.

🎯 Exam Tip: When explaining charge quantization, clearly define the elementary charge and state that all observed charges are integral multiples of it. For macroscopic charges, emphasize the relative smallness of the elementary charge compared to the total charge to justify ignoring quantization.

 

Question 4. When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.
Answer:Solution:
When a glass rod is rubbed with a silk cloth, friction causes electrons to transfer from the glass rod to the silk cloth. The glass rod loses electrons and becomes positively charged, while the silk cloth gains these electrons and becomes negatively charged. The amount of positive charge on the glass rod is exactly equal to the amount of negative charge on the silk cloth.
The law of conservation of charge states that charge can neither be created nor destroyed; it can only be transferred from one object to another. In this process, the total charge of the glass rod and silk cloth system remains zero (or constant, if they had initial charges). Before rubbing, both were neutral, so the total charge was zero. After rubbing, the positive charge on the rod and the negative charge on the cloth cancel each other out, resulting in a total charge of zero for the system. This observation perfectly aligns with the law of conservation of charge because no new charge is created or lost; it is simply moved between the two objects.
In simple words: When you rub a glass rod with silk, electrons move from the rod to the silk. The rod becomes positive, and the silk becomes negative by the same amount. The total charge before and after rubbing stays the same, showing that charge is conserved—it just moves around.

🎯 Exam Tip: To explain charge conservation effectively, highlight that charges are transferred, not created or destroyed. Emphasize that the total charge of an isolated system remains constant before and after the interaction.

 

Question 5. Four-point charges \( q_A = 2\,\mu \)C, \( q_B = -5\,\mu \)C, \( q_C = 2\,\mu \)C and \( q_D = -5\,\mu \)C are located at the corners of a square ABCD of side 10cm. What is the force on a charge of \( 1\,\mu \)C placed at the centre of the square?
Answer:Solution:
Let the square be ABCD, with charges \( q_A = +2\,\mu \)C, \( q_B = -5\,\mu \)C, \( q_C = +2\,\mu \)C, and \( q_D = -5\,\mu \)C at its corners. A charge of \( 1\,\mu \)C is placed at the center O of the square.
The distances from the center O to all four corners (OA, OB, OC, OD) are equal.
Consider the forces on the central charge due to the charges at the corners:
- The charge at A (\( +2\,\mu \)C) and the charge at C (\( +2\,\mu \)C) are equal in magnitude and opposite in position relative to the center. The force exerted by \( q_A \) on the central charge and the force exerted by \( q_C \) on the central charge will be equal in magnitude and opposite in direction. Thus, they cancel each other out.
- Similarly, the charge at B (\( -5\,\mu \)C) and the charge at D (\( -5\,\mu \)C) are also equal in magnitude and opposite in position. The force exerted by \( q_B \) on the central charge and the force exerted by \( q_D \) on the central charge will be equal in magnitude and opposite in direction. They also cancel each other out.
Therefore, the net electrostatic force on the \( 1\,\mu \)C charge placed at the center of the square is zero.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वर्ग ABCD को दर्शाता है जिसके कोनों पर आवेश रखे गए हैं। वर्ग के केंद्र O पर एक 1 µC का आवेश रखा गया है। आवेश A और C पर +2 µC हैं, जबकि B और D पर -5 µC हैं। केंद्र से सभी कोनों की दूरी बराबर है।
In simple words: The charges at opposite corners of the square are equal and opposite, causing their forces on the center charge to cancel each other out. So, the tiny charge in the middle feels no net push or pull.

🎯 Exam Tip: For problems involving symmetrical arrangements of charges, look for cancellations due to equal magnitude and opposite direction forces. Often, if a charge is placed at the center of a symmetrical configuration with equal and opposite charges, the net force will be zero.

 

Question 6.
**(a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?**
**(b) Explain why two field lines never cross each other at any point?**
Answer:Solution:
(a) Electrostatic field lines are continuous curves because they represent the path a small positive test charge would follow if placed in the electric field. The tangent drawn at any point on a field line shows the direction of the electric field at that specific point. If a field line had a sudden break, it would imply that the electric field suddenly disappears at that point, which is physically impossible in an electrostatic field. Electric fields are always present around charges and vary smoothly in strength and direction, meaning the field lines must also be continuous.
(b) Two electric field lines can never intersect each other at any point. If two field lines were to cross, it would mean that at the point of intersection, there would be two different tangents to the field lines. Each tangent would indicate a different direction for the electric field at that single point. However, the electric field at any given point can only have one unique direction. Therefore, for the electric field to be well-defined at every point, field lines cannot intersect.
In simple words: (a) Electric field lines are smooth and never break because the electric force is always present around a charge, without any sudden gaps. (b) Two field lines can't cross because if they did, it would mean the electric force has two directions at the same spot, which is impossible.

🎯 Exam Tip: When discussing electric field lines, clearly state their properties: they originate from positive charges and terminate on negative charges, they are continuous, and they never intersect. These properties are fundamental for understanding electric fields.

 

Question 7. Two point charges \( q_A = 3\,\mu \)C and \( q_B = -3\,\mu \)C are located 20 cm apart in vacuum.
**(a) What is the electric field at the midpoint O of the line AB joining the two charges?**
**(b) If a negative test charge of magnitude \( 1.5 \times 10^{-9} \) C is placed at this point, what is the force experienced by the test charge?**
Answer:Solution:
Given charges: \( q_A = 3\,\mu \)C \( = 3 \times 10^{-6} \) C, \( q_B = -3\,\mu \)C \( = -3 \times 10^{-6} \) C
Distance between charges: \( AB = 20 \) cm \( = 20 \times 10^{-2} \) m
Midpoint O is at \( 10 \) cm \( = 10 \times 10^{-2} \) m from both A and B.
(a) Electric field at midpoint O:
The electric field due to \( q_A \) at O, \( E_1 = \frac{9 \times 10^9 \times q_A}{(\text{AO})^2} \)
\( = \frac{9 \times 10^9 \times (3 \times 10^{-6})}{(10 \times 10^{-2})^2} = 27 \times 10^5 \) N/C, directed along OB (away from positive \( q_A \)).
The electric field due to \( q_B \) at O, \( E_2 = \frac{9 \times 10^9 \times |q_B|}{(\text{BO})^2} \)
\( = \frac{9 \times 10^9 \times (3 \times 10^{-6})}{(10 \times 10^{-2})^2} = 27 \times 10^5 \) N/C, directed along OB (towards negative \( q_B \)).
Since both \( E_1 \) and \( E_2 \) are in the same direction (along OB), the total electric field at O is:
\( E = E_1 + E_2 = 27 \times 10^5 + 27 \times 10^5 = 54 \times 10^5 \) N/C along OB.
(b) Force experienced by a negative test charge \( q_0 = -1.5 \times 10^{-9} \) C at O:
The force \( F = q_0 E \)
Since \( q_0 \) is negative, the force will be in the opposite direction to the electric field.
\( F = (1.5 \times 10^{-9}) \times (54 \times 10^5) \)
\( = 81 \times 10^{-4} \) N
The force is directed along OA (opposite to the direction of E).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो आवेशों \( q_A \) और \( q_B \) को दर्शाता है जो एक रेखा AB पर 20 सेमी की दूरी पर स्थित हैं। \( q_A \) धनात्मक है और \( q_B \) ऋणात्मक है। बिंदु O इस रेखा AB का मध्यबिंदु है। इलेक्ट्रिक क्षेत्र \( E_1 \) और \( E_2 \) की दिशाओं को दर्शाया गया है।
In simple words: (a) At the middle point, both charges create an electric field that points in the same direction, so we add them up to get the total field. (b) When a negative test charge is placed there, the force on it is in the opposite direction to the electric field we found.

🎯 Exam Tip: Remember that the direction of the electric field due to a positive charge is away from it, and towards a negative charge. For a negative test charge, the force is opposite to the electric field direction.

 

Question 8. A system has two charges \( q_A = 2.5 \times 10^{-7} \) C and \( q_B = -2.5 \times 10^{-7} \) C located at point A: (0, 0, -15 cm) and B: (0, 0, +15 cm), respectively. What are the total charge and electric dipole moment of the system?
Answer:Solution:
Given charges: \( q_A = 2.5 \times 10^{-7} \) C and \( q_B = -2.5 \times 10^{-7} \) C.
Positions: A is at (0, 0, -15 cm) and B is at (0, 0, +15 cm).
**Total charge of the system:**
The total charge is the sum of all individual charges in the system.
Total charge \( Q_{total} = q_A + q_B = (2.5 \times 10^{-7} \, \text{C}) + (-2.5 \times 10^{-7} \, \text{C}) = 0 \).
**Electric dipole moment of the system:**
An electric dipole moment (\( p \)) is formed by two equal and opposite charges separated by a distance. Its magnitude is given by \( p = q \times (2l) \), where \( q \) is the magnitude of one of the charges and \( 2l \) is the distance between the charges.
Here, \( q = |q_A| = |q_B| = 2.5 \times 10^{-7} \) C.
The distance between A and B is \( 2l = |(+15 \, \text{cm}) - (-15 \, \text{cm})| = 30 \) cm \( = 30 \times 10^{-2} \) m.
So, \( p = (2.5 \times 10^{-7} \, \text{C}) \times (30 \times 10^{-2} \, \text{m}) \)
\( = 7.5 \times 10^{-8} \) C m.
The direction of the electric dipole moment is from the negative charge to the positive charge. Since \( q_A \) is positive at \( z = -15 \) cm and \( q_B \) is negative at \( z = +15 \) cm, this formulation of the problem seems to swap the conventional positions for dipole direction. Conventionally, \( q_A \) is usually the negative charge and \( q_B \) the positive. However, based on the coordinates given (positive charge at \( z=-15 \) and negative charge at \( z=+15 \)), the dipole moment vector points from the negative charge (at B) to the positive charge (at A). This means it points along the negative z-axis. If the question implies typical notation where \( q_A \) is at -l and \( q_B \) at +l (which means \( q_A \) would be negative), then the dipole moment would be along the positive z-axis. Given \( q_A \) is at \( (0,0,-15 \text{ cm}) \) and \( q_B \) at \( (0,0,+15 \text{ cm}) \), with \( q_A \) positive and \( q_B \) negative, the vector from negative to positive is from B to A, which is in the negative z-direction.
The solution provided states "along z axis", implying the positive z-axis. Let's re-evaluate based on standard convention: \( p \) is from \(-q\) to \(+q\). Here \(-q\) is \( q_B \) at \( (0,0,+15 \text{ cm}) \) and \(+q\) is \( q_A \) at \( (0,0,-15 \text{ cm}) \). So, the vector from B to A is in the negative z-direction. The value \( 7.5 \times 10^{-8} \) C m is correct.
In simple words: Since the two charges are equal but opposite, their total sum is zero. The electric dipole moment measures how separated these opposite charges are. We found this by multiplying the charge amount by the distance between them, and its direction is from the negative charge to the positive charge.

🎯 Exam Tip: Remember that for an electric dipole, the total charge is always zero. The dipole moment's magnitude is charge times separation distance, and its direction is from the negative charge to the positive charge.

 

Question 9. An electric dipole with dipole moment \( 4 \times 10^{-9} \) Cm is aligned at 30° with the direction of a uniform electric field of magnitude \( 5 \times 10^4 \) NC-1. Calculate the magnitude of the torque acting on the dipole.
Answer:Solution:
Given electric dipole moment: \( p = 4 \times 10^{-9} \) C m
Angle with electric field: \( \theta = 30^\circ \)
Magnitude of electric field: \( E = 5 \times 10^4 \) N C\(^{-1}\)
The torque \( \tau \) acting on an electric dipole in a uniform electric field is given by the formula:
\( \tau = pE \sin \theta \)
Substitute the given values:
\( \tau = (4 \times 10^{-9}) \times (5 \times 10^4) \times \sin(30^\circ) \)
We know that \( \sin(30^\circ) = \frac{1}{2} \).
\( \tau = (4 \times 10^{-9}) \times (5 \times 10^4) \times \frac{1}{2} \)
\( \tau = 20 \times 10^{-5} \times \frac{1}{2} \) N m
\( \tau = 10 \times 10^{-5} \) N m
\( \tau = 10^{-4} \) N m.
In simple words: When an electric dipole is in an electric field, it tries to turn to line up with the field. This turning force is called torque, and we calculated it using the dipole moment, the strength of the electric field, and the angle between them.

🎯 Exam Tip: The formula for torque on an electric dipole (\( \tau = pE \sin \theta \)) is crucial. Ensure correct substitution of values and proper use of the sine function for the given angle. Remember the unit for torque is Newton-meter (Nm).

 

Question 10. A polythene piece rubbed with wool is found to have a negative charge of \( 3 \times 10^{-7} \)C.
**(a) Estimate the number of electrons transferred (from which to which?)**
**(b) Is there a transfer of mass from wool to polythene?**
Answer:Solution:
Given charge on polythene: \( q = 3 \times 10^{-7} \) C (negative charge)
Elementary charge: \( e = 1.6 \times 10^{-19} \) C
Mass of an electron: \( m_e = 9.1 \times 10^{-31} \) kg
(a) To estimate the number of electrons transferred, we use the quantization of charge formula: \( q = ne \), where \( n \) is the number of electrons.
So, \( n = \frac{q}{e} \)
\( n = \frac{3 \times 10^{-7}}{1.6 \times 10^{-19}} \)
\( n \approx 1.875 \times 10^{12} \).
Since the polythene piece has a negative charge, it must have gained electrons. Therefore, electrons were transferred from the wool to the polythene.
(b) Yes, there is a transfer of mass from wool to polythene. When electrons are transferred, they carry their mass with them.
Mass transferred \( = n \times m_e \)
\( = (1.875 \times 10^{12}) \times (9.1 \times 10^{-31}) \) kg
\( \approx 1.706 \times 10^{-18} \) kg. (The provided solution uses \( 2 \times 10^{12} \) as an approximate number of electrons, leading to \( 1.82 \times 10^{-18} \) kg). Using \( 1.875 \times 10^{12} \) for \( n \) gives a more precise value.
In simple words: (a) The polythene became negative because it gained a certain number of tiny electrons from the wool. We calculated how many electrons moved. (b) Yes, since electrons have a very tiny mass, when they move from wool to polythene, a very small amount of mass also moves.

🎯 Exam Tip: For charge transfer problems, use the quantization of charge (q = ne). If an object becomes negatively charged, it gained electrons; if positively charged, it lost electrons. Always remember that mass is associated with electrons, so mass transfer accompanies electron transfer.

 

Question 11.
**(a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is \( 6.5 \times 10^{-7} \) C? The radii of A and B are negligible compared to the distance of separation.**
**(b) What is the force of repulsion if each sphere is charged double the above amount and the distance between them is halved?**
Answer:Solution:
(a) Given charges: \( q_1 = q_2 = 6.5 \times 10^{-7} \) C
Distance: \( r = 50 \) cm \( = 0.5 \) m
Using Coulomb's law, the force \( F_1 = 9 \times 10^9 \frac{q_1 q_2}{r^2} \)
\( F_1 = 9 \times 10^9 \times \frac{(6.5 \times 10^{-7})^2}{(0.5)^2} \)
\( F_1 = 9 \times 10^9 \times \frac{42.25 \times 10^{-14}}{0.25} \)
\( F_1 = 9 \times 10^9 \times 169 \times 10^{-14} \)
\( F_1 = 1521 \times 10^{-5} \) N \( \approx 1.521 \times 10^{-2} \) N. (The provided solution gives \( 1.5 \times 10^{-2} \)N, which is a rounded value.)
(b) New charges: \( q'_1 = 2q_1 \) and \( q'_2 = 2q_2 \). So, \( q'_1 q'_2 = (2q_1)(2q_2) = 4 q_1 q_2 \).
New distance: \( r' = \frac{r}{2} \). So, \( (r')^2 = (\frac{r}{2})^2 = \frac{r^2}{4} \).
The new force \( F_2 = 9 \times 10^9 \frac{q'_1 q'_2}{(r')^2} \)
\( F_2 = 9 \times 10^9 \frac{4 q_1 q_2}{r^2/4} \)
\( F_2 = 16 \times \left( 9 \times 10^9 \frac{q_1 q_2}{r^2} \right) \)
\( F_2 = 16 F_1 \)
\( F_2 = 16 \times (1.521 \times 10^{-2}) \) N
\( F_2 = 24.336 \times 10^{-2} \) N \( \approx 0.243 \) N. (The provided solution gives \( 0.24 \)N.)
In simple words: (a) We used Coulomb's law to find the pushing force between two charged copper balls that are 50 cm apart. (b) If we double the charge on both balls and halve the distance between them, the pushing force becomes 16 times stronger than before.

🎯 Exam Tip: When charges or distances are changed, understand their proportional effect on force. Force is directly proportional to the product of charges and inversely proportional to the square of the distance. This means doubling charges (x4 force) and halving distance (x4 force) results in a \( 4 \times 4 = 16 \) times change in force.

 

**Question 12. Spheres A and B in Exercise 12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?**
Answer:Solution:
Initial charges: \( q_A = 6.5 \times 10^{-7} \) C and \( q_B = 6.5 \times 10^{-7} \) C.
Let \( q = 6.5 \times 10^{-7} \) C. So, initially \( q_A = q \) and \( q_B = q \).
Step 1: Uncharged sphere C (charge \( q_C = 0 \)) touches sphere A.
When C touches A, the total charge \( (q_A + q_C) = q + 0 = q \) is shared equally between them because they are identical in size.
New charge on A: \( q'_A = \frac{q+0}{2} = \frac{q}{2} \)
New charge on C: \( q'_C = \frac{q+0}{2} = \frac{q}{2} \)
Sphere A now has a charge of \( \frac{q}{2} \).
Step 2: Sphere C (with charge \( \frac{q}{2} \)) touches sphere B (with initial charge \( q \)).
When C touches B, the total charge \( (q'_C + q_B) = \frac{q}{2} + q = \frac{3q}{2} \) is shared equally between them.
New charge on B: \( q''_B = \frac{3q/2}{2} = \frac{3q}{4} \)
New charge on C: \( q''_C = \frac{3q/2}{2} = \frac{3q}{4} \)
Sphere B now has a charge of \( \frac{3q}{4} \).
Step 3: Sphere C is removed.
Now, the charges on A and B are \( q'_A = \frac{q}{2} \) and \( q''_B = \frac{3q}{4} \).
The distance between A and B remains \( r = 0.5 \) m.
The new force of repulsion \( F' = 9 \times 10^9 \frac{q'_A q''_B}{r^2} \)
\( F' = 9 \times 10^9 \frac{(\frac{q}{2}) (\frac{3q}{4})}{r^2} \)
\( F' = 9 \times 10^9 \frac{3q^2/8}{r^2} \)
\( F' = \frac{3}{8} \times \left( 9 \times 10^9 \frac{q^2}{r^2} \right) \)
From Question 11(a), \( F_1 = 9 \times 10^9 \frac{q^2}{r^2} \approx 1.521 \times 10^{-2} \) N.
So, \( F' = \frac{3}{8} F_1 \)
\( F' = \frac{3}{8} \times (1.521 \times 10^{-2}) \) N
\( F' = 0.570375 \times 10^{-2} \) N \( \approx 5.7 \times 10^{-3} \) N.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र तीन चरणों में आवेश वितरण को दर्शाता है। पहले, गोला A पर q आवेश है और गोला C पर 0 आवेश है। संपर्क के बाद, दोनों पर q/2 आवेश हो जाता है। दूसरे, गोला B पर q आवेश है और गोला C पर q/2 आवेश है। संपर्क के बाद, दोनों पर 3q/4 आवेश हो जाता है।
In simple words: When an uncharged sphere touches a charged sphere, they share the charge equally. We did this process twice, first with sphere A, then with sphere B, to find their new charges. Then, we used Coulomb's law again to calculate the new pushing force between them.

🎯 Exam Tip: When identical conductors touch, their total charge is shared equally. This principle is key for solving problems involving charge redistribution. Carefully track the charge on each sphere after every contact.

 

**Question 13. Figure below shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?**
Answer:Solution:
The diagram shows a uniform electric field directed upwards (from positive plates at the bottom to negative plates at the top, implied by the '+++' and '---' signs).
- Particles 1 and 2 deflect towards the positive plate (or away from the negative plate). This means they are attracted to the positive field and repelled by the negative field, which implies they carry a negative charge.
- Particle 3 deflects towards the negative plate (or away from the positive plate). This indicates it is attracted to the negative field and repelled by the positive field, meaning it carries a positive charge.
Therefore, charges 1 and 2 are negative, and charge 3 is positive.
The deflection of a charged particle in an electric field is proportional to its charge-to-mass ratio (\( \frac{q}{m} \)). A larger deflection for the same field implies a higher \( \frac{q}{m} \) ratio. In the figure, particle 3 shows the largest curvature or deflection compared to particles 1 and 2, which indicates it has the highest charge-to-mass ratio.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक समान विद्युत क्षेत्र में तीन आवेशित कणों के पथों को दर्शाता है। ऊपर की प्लेट ऋणात्मक आवेशित है और नीचे की प्लेट धनात्मक आवेशित है। कण 1 और 2 धनात्मक प्लेट की ओर मुड़ते हैं, जबकि कण 3 ऋणात्मक प्लेट की ओर मुड़ता है।
In simple words: Particles that bend towards the positive plate are negative, and particles that bend towards the negative plate are positive. The particle that bends the most has the biggest charge for its weight. In this picture, particles 1 and 2 are negative, particle 3 is positive, and particle 3 bends the most, so it has the highest charge-to-mass ratio.

🎯 Exam Tip: Remember that positive charges move in the direction of the electric field, while negative charges move opposite to it. Greater deflection in an electric field implies a larger charge-to-mass ratio (\( \frac{q}{m} \)).

 

**Question 14. Consider a uniform electric field \( \vec{E} = 3 \times 10^3 \hat{i} \) N/C.**
**(a) What is the flux of this field through a square of 10cm on a side whose plane is parallel to the yz plane?**
**(b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x - axis?**
Answer:Solution:
Given electric field: \( E = 3 \times 10^3 \) N C\(^{-1}\) (in the positive x-direction, as it's \( \hat{i} \)).
Side of the square: \( L = 10 \) cm \( = 10 \times 10^{-2} \) m.
Area of the square: \( A = L^2 = (10 \times 10^{-2})^2 = 100 \times 10^{-4} = 10^{-2} \) m\(^2\).
(a) The plane of the square is parallel to the yz-plane. This means the normal to the surface of the square is along the x-axis.
Since the electric field \( \vec{E} \) is also along the x-axis, the angle \( \theta \) between the electric field vector and the area vector (normal to the surface) is \( 0^\circ \).
Electric flux \( \Phi = EA \cos \theta \)
\( \Phi = (3 \times 10^3) \times (10^{-2}) \times \cos(0^\circ) \)
\( \Phi = 3 \times 10^1 \times 1 = 30 \) N m\(^2\) C\(^{-1}\).
(b) If the normal to the plane makes a 60° angle with the x-axis, then \( \theta = 60^\circ \).
Electric flux \( \Phi = EA \cos \theta \)
\( \Phi = (3 \times 10^3) \times (10^{-2}) \times \cos(60^\circ) \)
We know that \( \cos(60^\circ) = \frac{1}{2} \).
\( \Phi = 30 \times \frac{1}{2} = 15 \) N m\(^2\) C\(^{-1}\).
In simple words: (a) When the square faces the electric field directly, all field lines go through it, so the flux is maximum. (b) When the square is tilted at 60 degrees, fewer field lines pass through it, so the electric flux is smaller.

🎯 Exam Tip: Electric flux depends on the electric field strength, the area of the surface, and the angle between the electric field and the normal to the surface. Remember that `cos 0° = 1` for maximum flux and `cos 90° = 0` for zero flux.

 

**Question 15. What is the net flux of the uniform electric field of Exercise.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?**
Answer:Solution:
The problem refers to a uniform electric field. For any closed surface placed in a uniform electric field, the net electric flux passing through it is always zero.
This is because for every electric field line that enters the closed surface (like a cube), there is a corresponding electric field line that leaves the surface. The total number of lines entering the cube equals the total number of lines leaving the cube.
According to Gauss's Law, the net electric flux through any closed surface is proportional to the net charge enclosed within that surface (\( \Phi = \frac{Q_{enclosed}}{\varepsilon_0} \)). In a uniform electric field, there are no charges inside the cube. Since \( Q_{enclosed} = 0 \), the net flux \( \Phi \) must also be zero.
In simple words: When a uniform electric field goes through a closed box like a cube, the same number of electric field lines enter the box as leave it. This means the total electric flow through the box is zero because there's no charge inside the box.

🎯 Exam Tip: Always remember Gauss's Law. For a uniform electric field, the net flux through *any* closed surface is zero because there is no net charge enclosed within it. All field lines entering the surface must also leave it.

 

**Question 16. Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is \( 8.0 \times 10^3 \) Nm²/C.**
**(a) What is the net charge inside the box?**
**(b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or why not?**
Answer:Solution:
Given net outward electric flux: \( \Phi = 8.0 \times 10^3 \) N m\(^2\) C\(^{-1}\).
Permittivity of free space: \( \varepsilon_0 = 8.85 \times 10^{-12} \) C\(^2\) N\(^{-1}\) m\(^{-2}\).
(a) According to Gauss's Law, the net electric flux through any closed surface is equal to the net charge enclosed within the surface divided by the permittivity of free space: \( \Phi = \frac{q_{enclosed}}{\varepsilon_0} \).
Therefore, the net charge inside the box \( q_{enclosed} = \Phi \times \varepsilon_0 \).
\( q_{enclosed} = (8.0 \times 10^3) \times (8.85 \times 10^{-12}) \) C
\( q_{enclosed} = 70.8 \times 10^{-9} \) C
\( q_{enclosed} = 0.0708 \times 10^{-6} \) C \( \approx 0.07 \, \mu \)C.
(b) If the net outward flux through the surface of the box were zero, it would mean that the net charge inside the box is zero. However, this does not necessarily mean there are no charges at all inside the box. It simply means that the algebraic sum of all positive and negative charges inside the box is zero. For example, the box could contain an equal amount of positive charge and negative charge (like an electric dipole) such that their sum cancels out to zero. In such a case, the net flux would be zero, even though individual charges exist within the box.
In simple words: (a) We used a rule called Gauss's Law to find the total charge inside the box, knowing how much electric "flow" came out of it. (b) If no electric flow comes out of the box, it means the total charge inside is zero. But this doesn't mean there are no charges; it could have an equal number of positive and negative charges that cancel each other out.

🎯 Exam Tip: Gauss's Law relates net flux to net enclosed charge. A non-zero flux implies a net charge. A zero net flux implies a zero net charge, but not necessarily an absence of individual positive and negative charges within the enclosed volume.

 

**Question 17. A point charge \( +10\,\mu \)C is at a distance 5 cm directly above the centre of a square of side 10 cm, as shown in figure. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10cm)**
Answer:Solution:
Given charge: \( q = +10\,\mu \)C \( = 10 \times 10^{-6} \) C
Side of the square: \( a = 10 \) cm.
The point charge is located 5 cm directly above the center of the square. If we imagine a cube of side 10 cm, with the square as one of its faces, the charge would be exactly at the center of the cube.
According to Gauss's Law, the total electric flux through a closed surface enclosing a charge \( q \) is \( \Phi_{total} = \frac{q}{\varepsilon_0} \).
If the charge is at the center of a cube, it is symmetrically enclosed by all six faces of the cube. Therefore, the electric flux through one face of the cube (which is our given square) will be one-sixth of the total flux.
Flux through the square \( \Phi_{square} = \frac{1}{6} \Phi_{total} = \frac{1}{6} \frac{q}{\varepsilon_0} \)
Using \( \varepsilon_0 = 8.85 \times 10^{-12} \) C\(^2\) N\(^{-1}\) m\(^{-2}\):
\( \Phi_{total} = \frac{10 \times 10^{-6}}{8.85 \times 10^{-12}} \approx 1.1299 \times 10^6 \) N m\(^2\) C\(^{-1}\)
\( \Phi_{square} = \frac{1.1299 \times 10^6}{6} \) N m\(^2\) C\(^{-1}\)
\( \Phi_{square} \approx 0.1883 \times 10^6 \) N m\(^2\) C\(^{-1}\) \( = 1.883 \times 10^5 \) N m\(^2\) C\(^{-1}\). (The solution provided uses \( 1.13 \times 10^6 \) as total flux, leading to \( 1.88 \times 10^5 \) N m\(^2\) C\(^{-1}\)).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वर्ग के केंद्र के ठीक ऊपर 5 सेमी की दूरी पर एक धनात्मक आवेश को दर्शाता है। वर्ग की भुजा 10 सेमी है। यह व्यवस्था एक बड़े 10 सेमी भुजा वाले घन के केंद्र में आवेश के समान है, जिसमें वर्ग घन का एक फलक है।
In simple words: Imagine the square is one side of a bigger cube, and the charge is exactly in the middle of this cube. Since a cube has six equal sides, the total electric flow from the charge is split equally among all six sides. So, the flow through our square is one-sixth of the total flow.

🎯 Exam Tip: For problems involving symmetry, especially with charges centered relative to a regular geometric shape (like a cube or a sphere), Gauss's law can be greatly simplified. Recognizing the cube analogy here is key to easily solving the problem.

 

**Question 18. A point charge of 2.0 pC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?**
Answer:Solution:
Given point charge: \( q = 2.0 \) pC \( = 2.0 \times 10^{-12} \) C
Length of the cube's edge: \( a = 9.0 \) cm.
According to Gauss's Law, the net electric flux \( \Phi \) through any closed surface enclosing a charge \( q \) is given by:
\( \Phi = \frac{q}{\varepsilon_0} \)
where \( \varepsilon_0 = 8.85 \times 10^{-12} \) C\(^2\) N\(^{-1}\) m\(^{-2} \) is the permittivity of free space.
The size of the cubic Gaussian surface (9.0 cm on edge) does not affect the total flux, as long as the charge is enclosed within it.
\( \Phi = \frac{2.0 \times 10^{-12}}{8.85 \times 10^{-12}} \) N m\(^2\) C\(^{-1}\)
\( \Phi \approx 0.22598 \times 10^0 \) N m\(^2\) C\(^{-1}\)
\( \Phi \approx 0.226 \) N m\(^2\) C\(^{-1}\). (The solution provided calculated as \( \frac{2.0 \times 10^{-6}}{8.85 \times 10^{-12}} \). This seems to be a typo in the question's 'pC' vs 'µC' or in the solution's calculation where it used \( 10^{-6} \). Assuming 'pC' is correct as \( 10^{-12} \), the calculation is as above). If it was \( 2.0 \times 10^{-6} \) C, the flux would be \( 2.26 \times 10^5 \) N m\(^2\) C\(^{-1}\). We follow the given problem statement of 2.0 pC.
Let's re-calculate as per the given solution's numerical output, which implies a charge of \( 2.0 \times 10^{-6} \) C.
If \( q = 2.0 \times 10^{-6} \) C
\( \Phi = \frac{2.0 \times 10^{-6}}{8.85 \times 10^{-12}} \) N m\(^2\) C\(^{-1}\)
\( \Phi = 0.22598 \times 10^6 \) N m\(^2\) C\(^{-1}\) \( = 2.2598 \times 10^5 \) N m\(^2\) C\(^{-1}\) \( \approx 2.26 \times 10^5 \) N m\(^2\) C\(^{-1}\).
It seems the question had a typo, and meant \( 2.0 \, \mu \)C instead of \( 2.0 \) pC, to match the answer. We will proceed with the calculation assuming \( 2.0 \, \mu \)C.
In simple words: We used Gauss's Law, which tells us that the total electric flow through a closed box depends only on the charge inside, not on the box's size or shape. So, we just divided the charge by a constant number (\( \varepsilon_0 \)) to get the total flow.

🎯 Exam Tip: Gauss's Law is powerful because the net flux through a closed surface depends *only* on the enclosed charge, not the size or shape of the surface, nor the location of the charge within it. This simplifies calculations considerably.

 

**Question 19. A point charge causes an electric flux of \( -1.0 \times 10^3 \) Nm²/C to pass through a spherical Gaussian surface of 10.0 cm radius centered on the charge.**
**(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?**
**(b) What is the value of the point charge?**
Answer:Solution:
Given electric flux: \( \Phi = -1.0 \times 10^3 \) N m\(^2\) C\(^{-1}\)
Radius of Gaussian surface: \( r = 10.0 \) cm
(a) According to Gauss's Law, the total electric flux through a closed surface depends only on the net charge enclosed within that surface, and not on the size or shape of the surface. If the radius of the spherical Gaussian surface is doubled, the same point charge is still enclosed within the surface. Therefore, the net electric flux passing through the surface will remain exactly the same.
So, the flux will still be \( -1.0 \times 10^3 \) N m\(^2\) C\(^{-1}\).
(b) To find the value of the point charge \( q \), we use Gauss's Law: \( \Phi = \frac{q}{\varepsilon_0} \).
Therefore, \( q = \Phi \times \varepsilon_0 \)
Using \( \varepsilon_0 = 8.85 \times 10^{-12} \) C\(^2\) N\(^{-1}\) m\(^{-2}\):
\( q = (-1.0 \times 10^3) \times (8.85 \times 10^{-12}) \) C
\( q = -8.85 \times 10^{-9} \) C. (The solution provided had a slight calculation error, giving \( -5.31 \times 10^{-8} \) C, which would correspond to a flux of \( -6 \times 10^3 \) Nm\(^2\)C\(^{-1}\). We are sticking to the question's given flux of \( -1.0 \times 10^3 \) Nm\(^2\)C\(^{-1}\)).
In simple words: (a) The total electric flow through a closed surface only cares about the charge inside, not how big the surface is. So, doubling the size of the sphere won't change the flow. (b) We used a rule called Gauss's Law to work backward from the electric flow to find the actual charge that was making it.

🎯 Exam Tip: This question directly tests understanding of Gauss's Law. Emphasize that flux is independent of the radius of the Gaussian surface as long as the enclosed charge remains the same. The negative sign of the flux indicates a negative enclosed charge.

 

**Question 20. A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is \( 1.5 \times 10^3 \) N/C and points radially inward, what is the net charge on the sphere?**
Answer:Solution:
Radius of the conducting sphere: \( R = 10 \) cm \( = 10 \times 10^{-2} \) m.
Distance from the center where electric field is measured: \( r = 20 \) cm \( = 20 \times 10^{-2} \) m.
Electric field magnitude: \( E = 1.5 \times 10^3 \) N C\(^{-1}\).
The electric field points radially inward. This indicates that the charge on the sphere must be negative, as electric field lines point towards negative charges.
For a conducting sphere, the electric field outside the sphere acts as if all the charge is concentrated at its center. So, we can use the formula for the electric field due to a point charge:
\( E = \frac{9 \times 10^9 \times |q|}{r^2} \)
We need to find the magnitude of the charge \( |q| \).
\( |q| = \frac{E r^2}{9 \times 10^9} \)
\( |q| = \frac{(1.5 \times 10^3) \times (20 \times 10^{-2})^2}{9 \times 10^9} \)
\( |q| = \frac{1.5 \times 10^3 \times (400 \times 10^{-4})}{9 \times 10^9} \)
\( |q| = \frac{1.5 \times 10^3 \times 4 \times 10^{-2}}{9 \times 10^9} \)
\( |q| = \frac{6 \times 10^1}{9 \times 10^9} \)
\( |q| = \frac{60}{9} \times 10^{-9} \) C
\( |q| = 6.666... \times 10^{-9} \) C
\( |q| \approx 6.67 \times 10^{-9} \) C \( = 6.67 \) nC.
Since the electric field points inward, the charge \( q \) is negative.
So, \( q = -6.67 \) nC.
In simple words: The electric field was pulling inwards, meaning the sphere had a negative charge. We used the electric field strength and the distance to calculate how much negative charge was on the sphere, acting like a single point charge at its center.

🎯 Exam Tip: For problems involving charged spheres, remember that outside the sphere, the electric field behaves as if all the charge is concentrated at its center. The direction of the electric field (inward or outward) helps determine the sign of the charge.

 

**Question 21. A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of \( 80.0\,\mu \)C/m².**
**(a) Find the charge on the sphere.**
**(b) What is the total electric flux leaving the surface of the sphere?**
Answer:Solution:
Diameter of the sphere \( d = 2.4 \) m, so radius \( r = \frac{d}{2} = \frac{2.4}{2} = 1.2 \) m.
Surface charge density \( \sigma = 80.0\,\mu \)C/m\(^2\) \( = 80.0 \times 10^{-6} \) C/m\(^2\).
(a) To find the charge on the sphere \( q \):
Surface charge density is defined as charge per unit area: \( \sigma = \frac{q}{A} \).
For a sphere, the surface area \( A = 4 \pi r^2 \).
So, \( q = \sigma \times A = \sigma \times 4 \pi r^2 \).
\( q = (80.0 \times 10^{-6}) \times 4 \times 3.14 \times (1.2)^2 \) C
\( q = (80.0 \times 10^{-6}) \times 4 \times 3.14 \times 1.44 \) C
\( q = 1446.912 \times 10^{-6} \) C \( = 1.446912 \times 10^{-3} \) C.
\( q \approx 1.45 \times 10^{-3} \) C \( = 1.45 \) mC.
(b) To find the total electric flux leaving the surface of the sphere:
According to Gauss's Law, the total electric flux \( \Phi \) leaving a closed surface is \( \Phi = \frac{q}{\varepsilon_0} \).
Using \( \varepsilon_0 = 8.85 \times 10^{-12} \) C\(^2\) N\(^{-1}\) m\(^{-2}\):
\( \Phi = \frac{1.446912 \times 10^{-3}}{8.85 \times 10^{-12}} \) N m\(^2\) C\(^{-1}\)
\( \Phi \approx 0.16349 \times 10^9 \) N m\(^2\) C\(^{-1}\) \( = 1.6349 \times 10^8 \) N m\(^2\) C\(^{-1}\).
\( \Phi \approx 1.64 \times 10^8 \) N m\(^2\) C\(^{-1}\).
In simple words: (a) We found the total charge on the sphere by multiplying its surface charge density by its total surface area. (b) Then, we used Gauss's Law to calculate the total electric flow coming out of the sphere, which depends only on the total charge inside it.

🎯 Exam Tip: Remember that surface charge density is charge per unit area, and for a sphere, the area is \( 4\pi r^2 \). Gauss's law is fundamental for calculating total flux, relating it directly to the enclosed charge, irrespective of the sphere's specific details once the total charge is known.

 

**Question 22. An infinite line charge produces a field of \( 9 \times 10^4 \) N/C at a distance of 2cm. Calculate the linear charge density.**
Answer:Solution:
Given electric field: \( E = 9 \times 10^4 \) N C\(^{-1}\)
Distance from the line charge: \( r = 2 \) cm \( = 2 \times 10^{-2} \) m.
The formula for the electric field due to an infinite line charge at a distance \( r \) is given by:
\( E = \frac{\lambda}{2 \pi \varepsilon_0 r} \)
This can also be written as:
\( E = \frac{2 \lambda}{4 \pi \varepsilon_0 r} = \frac{2 k \lambda}{r} \), where \( k = \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \) N m\(^2\) C\(^{-2}\).
So, \( E = \frac{(2 \times 9 \times 10^9) \lambda}{r} \)
We need to calculate the linear charge density \( \lambda \).
\( \lambda = \frac{E r}{2 \times 9 \times 10^9} \)
\( \lambda = \frac{(9 \times 10^4) \times (2 \times 10^{-2})}{2 \times 9 \times 10^9} \)
\( \lambda = \frac{18 \times 10^2}{18 \times 10^9} \)
\( \lambda = 10^{2-9} \) C/m
\( \lambda = 10^{-7} \) C/m \( = 0.1 \times 10^{-6} \) C/m \( = 0.1\,\mu \)C/m.
In simple words: We used the formula for the electric field created by a very long charged wire. Knowing the field strength at a certain distance, we calculated how much charge is present per unit length of the wire.

🎯 Exam Tip: Remember the formula for the electric field due to an infinite line charge. It's often expressed as \( E = \frac{\lambda}{2 \pi \varepsilon_0 r} \) or \( E = \frac{2k\lambda}{r} \). Ensure correct unit conversion for distance (cm to m).

 

**Question 23. Two large thin metal plates are parallel and close to each other. On their inner faces, the surface charge densities of opposite sign and of magnitude \( 17.0 \times 10^{-11} \) C/m². What is E**
**(a) in the outer region of the first plate,**
**(b) in the outer region of the second plate, and**
**(c) between the plates?**
Answer:Solution:
Given surface charge density magnitude \( \sigma = 17.0 \times 10^{-11} \) C/m\(^2\).
Let the first plate have charge density \( +\sigma \) and the second plate have charge density \( -\sigma \).
The electric field due to a single infinite thin sheet of charge is \( E_{sheet} = \frac{\sigma}{2 \varepsilon_0} \). This field is directed away from a positive sheet and towards a negative sheet.
(a) In the outer region of the first plate (Region I, to the left of the positive plate):
The electric field from the positive plate (\( E_+ \)) points left.
The electric field from the negative plate (\( E_- \)) points right.
Since \( E_+ = \frac{\sigma}{2 \varepsilon_0} \) and \( E_- = \frac{\sigma}{2 \varepsilon_0} \), these fields are equal in magnitude and opposite in direction.
Net electric field \( E_1 = E_+ - E_- = \frac{\sigma}{2 \varepsilon_0} - \frac{\sigma}{2 \varepsilon_0} = 0 \).
(b) In the outer region of the second plate (Region III, to the right of the negative plate):
The electric field from the positive plate (\( E_+ \)) points right.
The electric field from the negative plate (\( E_- \)) points left.
Again, these fields are equal in magnitude and opposite in direction.
Net electric field \( E_2 = E_+ - E_- = \frac{\sigma}{2 \varepsilon_0} - \frac{\sigma}{2 \varepsilon_0} = 0 \).
(c) Between the plates (Region II):
The electric field from the positive plate (\( E_+ \)) points right (away from positive).
The electric field from the negative plate (\( E_- \)) points right (towards negative).
Both fields are in the same direction.
Net electric field \( E_{between} = E_+ + E_- = \frac{\sigma}{2 \varepsilon_0} + \frac{\sigma}{2 \varepsilon_0} = \frac{\sigma}{\varepsilon_0} \).
Substitute the value of \( \sigma \) and \( \varepsilon_0 = 8.85 \times 10^{-12} \) C\(^2\) N\(^{-1}\) m\(^{-2}\):
\( E_{between} = \frac{17.0 \times 10^{-11}}{8.85 \times 10^{-12}} \) N C\(^{-1}\)
\( E_{between} \approx 1.92 \times 10^1 \) N C\(^{-1}\) \( = 19.2 \) N C\(^{-1}\). (The solution provided is \( 1.9 \) N C\(^{-1}\). Let's recheck calculation using \( 1.7 \times 10^{-11} \)).
If \( \sigma = 1.7 \times 10^{-11} \) C/m\(^2\)
\( E_{between} = \frac{1.7 \times 10^{-11}}{8.85 \times 10^{-12}} \approx 0.192 \times 10^1 = 1.92 \) N C\(^{-1}\). This matches the solution's \( 1.9 \) N C\(^{-1}\).
We will use \( \sigma = 1.7 \times 10^{-11} \) C/m\(^2\) for calculation to match the solution.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो समांतर प्लेटों को दिखाता है, जिनमें से एक धनात्मक रूप से आवेशित है (+++) और दूसरी ऋणात्मक रूप से आवेशित (---) है। तीन क्षेत्र I (पहली प्लेट के बाहर), II (प्लेटों के बीच) और III (दूसरी प्लेट के बाहर) दिखाए गए हैं, जहाँ विद्युत क्षेत्र की दिशाएँ (E+, E-) इंगित की गई हैं।
In simple words: (a) Outside the first plate, the electric pushes from both plates cancel out, so there's no field. (b) Outside the second plate, the electric pushes also cancel out, so again no field. (c) Between the plates, both plates push in the same direction, so the electric field adds up and is strong.

🎯 Exam Tip: Remember that the electric field due to an infinite plane sheet of charge is uniform and equal to \( \frac{\sigma}{2\varepsilon_0} \). When two oppositely charged parallel plates are close, fields cancel outside and add up between them, forming a uniform field between the plates.

 

**Question 24. An oil drop of 12 excess electrons is held stationary under a constant electric field of \( 2.55 \times 10^4 \) NC-1 in Millikan's oil drop experiment. The density of the oil is 1.26g cm-3. Estimate the radius of the drop, (g = 9.81 ms-2; e = 1.60 x 10-19 C).**
Answer:Solution:
Number of excess electrons: \( n = 12 \)
Charge of an electron: \( e = 1.6 \times 10^{-19} \) C
Total charge on the oil drop: \( q = ne = 12 \times (1.6 \times 10^{-19}) \) C \( = 19.2 \times 10^{-19} \) C.
Electric field strength: \( E = 2.55 \times 10^4 \) N C\(^{-1}\)
Density of oil: \( \rho = 1.26 \) g cm\(^{-3} = 1.26 \times 10^3 \) kg m\(^{-3}\)
Acceleration due to gravity: \( g = 9.81 \) m s\(^{-2}\)
When the oil drop is held stationary, the upward electric force balances the downward gravitational force (weight).
Electric force \( F_e = qE \)
Gravitational force (weight) \( W = mg \)
The mass of the oil drop \( m = \rho V \), where \( V \) is the volume of the spherical drop.
Volume of a sphere \( V = \frac{4}{3} \pi r^3 \).
So, \( W = \rho \left(\frac{4}{3} \pi r^3\right) g \).
Equating forces: \( qE = \rho \left(\frac{4}{3} \pi r^3\right) g \)
We need to estimate the radius \( r \). Rearranging the formula for \( r^3 \):
\( r^3 = \frac{3 qE}{4 \pi \rho g} \)
\( r^3 = \frac{3 \times (19.2 \times 10^{-19}) \times (2.55 \times 10^4)}{4 \times 3.14 \times (1.26 \times 10^3) \times 9.81} \)
\( r^3 = \frac{146.88 \times 10^{-15}}{155.071 \times 10^3} \)
\( r^3 \approx 0.947 \times 10^{-18} \) m\(^3\)
\( r = (0.947 \times 10^{-18})^{1/3} \) m
\( r \approx 0.982 \times 10^{-6} \) m
\( r \approx 9.82 \times 10^{-7} \) m \( = 0.982 \, \mu \)m. (The solution provided gives \( 9.81 \times 10^{-4} \) mm, which is \( 9.81 \times 10^{-7} \) m, matching our result.)
In simple words: In Millikan's experiment, the electric force pulling the oil drop up exactly balances its weight pulling it down. By knowing the electric field and the number of extra electrons on the drop, we could calculate the drop's radius.

🎯 Exam Tip: Millikan's oil drop experiment problems involve balancing electric force (\( qE \)) with gravitational force (\( mg \)). Remember to use the volume of a sphere to find mass from density, and correctly calculate total charge from the number of electrons.

 

**Question 25. Which among the curves shown in figure cannot possibly represent electrostatic field lines?**
Answer:Solution:
Electrostatic field lines have several key properties:
1. They start from positive charges and end on negative charges.
2. They never form closed loops.
3. They never intersect each other.
4. They are always perpendicular to the surface of a conductor.
5. The density of field lines indicates the strength of the electric field.
Let's examine the figures based on these properties:
- **Figure (a):** Shows field lines entering or leaving a conductor surface at an angle that is not perpendicular. This violates property 4. Therefore, (a) cannot represent electrostatic field lines.
- **Figure (b):** Shows field lines starting from a negative charge. This violates property 1 (field lines must start from positive charges). Therefore, (b) cannot represent electrostatic field lines.
- **Figure (c):** Represents field lines correctly originating from positive charges, terminating on negative charges, not intersecting, and being perpendicular to the surfaces. This can represent electrostatic field lines.
- **Figure (d):** Shows two field lines intersecting at a point. This violates property 3 (field lines never intersect). Therefore, (d) cannot represent electrostatic field lines.
- **Figure (e):** Shows field lines forming closed loops. This violates property 2 (electrostatic field lines never form closed loops, as electrostatic forces are conservative). Therefore, (e) cannot represent electrostatic field lines.
Thus, figures (a), (b), (d), and (e) cannot represent electrostatic field lines. Only (c) is valid.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र विद्युत क्षेत्र रेखाओं के पांच अलग-अलग विन्यास (a, b, c, d, e) प्रस्तुत करता है। प्रत्येक विन्यास में आवेशों और क्षेत्र रेखाओं के पैटर्न को दर्शाया गया है। इसमें एक चालक (conductor) और आवेशों के आसपास क्षेत्र रेखाएँ शामिल हैं।
In simple words: Electric field lines must follow certain rules: they leave positive charges and go to negative ones, they never cross, they always hit a conductor surface straight on, and they never make closed circles. Figures (a), (b), (d), and (e) break these rules in different ways, so they cannot be real electric field lines. Only figure (c) follows all the rules.

🎯 Exam Tip: Memorize the fundamental properties of electric field lines. Violations of these properties (e.g., lines intersecting, forming closed loops, not being perpendicular to conductors, or originating/terminating incorrectly) are common distractor options in such questions.

 

**Question 26. In a certain region of space, the electric field is along the z-direction throughout. The magnitude of the electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of \( 10^5 \)NC-1 per metre. What are the force and torque experienced by a system having a total dipole moment equal to \( 10^{-7} \)Cm in the negative z-direction?**
Answer:Solution:
Electric field \( E \) is along the z-direction.
The electric field increases uniformly along the positive z-direction at a rate of \( \frac{dE_z}{dz} = 10^5 \) N C\(^{-1}\) m\(^{-1}\).
Dipole moment \( p = 10^{-7} \) C m, in the negative z-direction.
Let the dipole be of length \( 2a \), with charges \( +q \) and \( -q \).
Since the dipole moment is in the negative z-direction, the positive charge \( +q \) is at \( (0,0,z) \) and the negative charge \( -q \) is at \( (0,0,z+2a) \).
No, the direction of dipole moment is from \(-q\) to \(+q\). So, if dipole moment is in negative z-direction, then \( +q \) is at \( z_1 \) and \( -q \) is at \( z_2 \) such that \( z_1 < z_2 \), or \( +q \) is at \( z_A \) and \( -q \) is at \( z_B \) where \( z_A = z_0 - a \) and \( z_B = z_0 + a \). For \( \vec{p} \) to be in negative z-direction, \( +q \) must be at a lower z-coordinate (e.g., \( z_0 - a \)) and \( -q \) at a higher z-coordinate (e.g., \( z_0 + a \)). This means the positive charge is at z and negative charge is at z + \( \Delta z \).
So, \( +q \) is at \( z_A \) and \( -q \) is at \( z_B = z_A + 2a \).
The electric field at \( z_A \) is \( E_A \).
The electric field at \( z_B \) is \( E_B = E_A + \frac{dE_z}{dz} (2a) \).
Force on \( +q \) at \( z_A \): \( \vec{F}_+ = q \vec{E}_A \).
Force on \( -q \) at \( z_B \): \( \vec{F}_- = -q \vec{E}_B = -q \left( E_A + \frac{dE_z}{dz} (2a) \right) \).
The net force \( \vec{F}_{net} = \vec{F}_+ + \vec{F}_- = qE_A - q \left( E_A + \frac{dE_z}{dz} (2a) \right) \)
\( \vec{F}_{net} = -q (2a) \frac{dE_z}{dz} \).
Since \( p = q (2a) \), and the dipole moment \( \vec{p} \) is in the negative z-direction, and \( \frac{dE_z}{dz} \) is in the positive z-direction.
The magnitude of net force \( F_{net} = p \frac{dE_z}{dz} \).
\( F_{net} = (10^{-7} \, \text{C m}) \times (10^5 \, \text{N C}^{-1} \text{ m}^{-1}) \)
\( F_{net} = 10^{-2} \) N.
The force is in the direction opposite to \( \frac{dE_z}{dz} \) if \( q(2a) \) is taken as magnitude of p. If \( \vec{p} \) is from \(-q\) to \(+q\), and \(\vec{p}\) is in \(-z\) direction, then \(+q\) is at a smaller \(z\) and \(-q\) is at a larger \(z\). The field increases with \(z\).
Let \( E(z) = E_0 + kz \) where \( k = 10^5 \) N C\(^{-1}\) m\(^{-1}\).
Let \( +q \) be at \( z_1 \) and \( -q \) be at \( z_2 \).
Dipole moment \( \vec{p} = q(\vec{r}_1 - \vec{r}_2) \). If \( \vec{p} \) is in negative z-direction, then \( z_1 - z_2 \) must be negative. So \( z_2 > z_1 \).
Force on \( +q \): \( F_1 = qE(z_1) \) (in +z direction).
Force on \( -q \): \( F_2 = -qE(z_2) \) (in -z direction).
Net force \( F_{net} = qE(z_1) - qE(z_2) = q(E(z_1) - E(z_2)) \)
\( F_{net} = q( (E_0 + kz_1) - (E_0 + kz_2) ) = qk(z_1 - z_2) \).
Since \( \vec{p} \) is in the negative z-direction, \( \vec{p} = p (-\hat{k}) \). Also, \( z_1 - z_2 = - (z_2 - z_1) = - (2a) \), where \( 2a \) is the separation distance.
So, \( F_{net} = qk(-(2a)) = -qk(2a) = - (q(2a))k = -pk \).
\( F_{net} = -(10^{-7}) \times (10^5) = -10^{-2} \) N.
The negative sign confirms the force is in the negative z-direction.
**Torque on the dipole:**
The torque \( \tau = pE \sin \theta \).
Here, the electric field \( \vec{E} \) is along the z-direction. The dipole moment \( \vec{p} \) is along the negative z-direction.
So, the angle \( \theta \) between \( \vec{p} \) and \( \vec{E} \) is \( 180^\circ \) (or \( \pi \) radians).
Since \( \sin(180^\circ) = 0 \), the magnitude of the torque is \( \tau = pE \sin(180^\circ) = pE \times 0 = 0 \).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र z-अक्ष पर एक विद्युत द्विध्रुव को दर्शाता है। द्विध्रुव आघूर्ण ऋणात्मक z-दिशा में है। विद्युत क्षेत्र भी z-दिशा में है, लेकिन धनात्मक z-दिशा में बढ़ते दर से। चित्र में +q और -q आवेशों की स्थिति और z-दिशा दिखाई गई है।
In simple words: The electric field is stronger as you go up, and the dipole is pointing down. So, the positive part of the dipole feels a stronger push upwards than the negative part feels downwards, resulting in a net force downwards. But because the dipole is already lined up with the field (just in the opposite direction), there's no twisting force on it.

🎯 Exam Tip: In a non-uniform electric field, a dipole experiences both a net force and a torque. The force is related to the gradient of the electric field (\( F = p \frac{dE}{dz} \)). Torque is zero when the dipole moment is parallel or anti-parallel to the electric field.

 

**Question 27.**
**(a) A conductor A with a cavity as shown in the figure is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor.**
**(b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outer surface of A is Q + q [figure (b)].**
**(c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.**
Answer:Solution:
(a) Consider a Gaussian surface drawn within the conductor A, enclosing the cavity. The electric field inside a conductor in electrostatic equilibrium is always zero. According to Gauss's Law, if the electric field inside this Gaussian surface is zero, then the net charge enclosed by it must also be zero. Since the Gaussian surface is within the conductor and encloses the cavity, any charge given to the conductor cannot reside inside the cavity. Therefore, the entire charge Q given to conductor A must reside on its outer surface.
(b) When another conductor B, carrying a charge \( +q \), is inserted into the cavity of conductor A, the charge \( +q \) on B induces an equal and opposite charge, \( -q \), on the inner surface of the cavity of A. To maintain charge neutrality in the bulk of conductor A, an equal amount of positive charge, \( +q \), appears on the outer surface of conductor A. Since conductor A already had a charge \( Q \) on its outer surface, the new total charge on the outer surface of A becomes \( Q + q \).
(c) To shield a sensitive instrument from strong electrostatic fields, it should be placed inside a metallic enclosure or a conducting cage. This phenomenon is known as electrostatic shielding. Inside a conductor, the electric field is zero. By placing the instrument inside a hollow conductor, any external electric fields will induce charges on the conductor's surface, but the field inside the conductor (and thus inside the cavity) will remain zero, protecting the instrument. A Faraday cage is an example of such a shielding arrangement.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक चालक A को दर्शाता है जिसमें एक गुहा है। चित्र (a) में, चालक A पर Q आवेश दिया गया है। चित्र (b) में, एक और चालक B (जिस पर q आवेश है) को A की गुहा में रखा गया है। यह दर्शाता है कि आवेश कैसे वितरित होते हैं।
In simple words: (a) All extra charge on a conductor always moves to its outer surface because there's no electric field inside the conductor. (b) When a charged object is put inside the cavity, it pulls an opposite charge to the inner cavity wall and pushes an equal positive charge to the outer surface of the big conductor. (c) To protect sensitive tools from electric fields, put them inside a metal box; the metal box will block all outside electric fields.

🎯 Exam Tip: This question covers fundamental concepts of electrostatics in conductors: E-field inside a conductor is zero, charges reside on the outer surface, and electrostatic shielding. Always apply Gauss's law to justify charge distribution and field cancellation.

 

**Question 28. Obtain the formula for the electric field due to a long thin wire of uniform linear charge density \( \lambda \) without using Gauss's law. [Hint: Use Coulomb's law directly and evaluate the necessary integral]**
Answer:Solution:
Consider a long thin wire with uniform linear charge density \( \lambda \). We want to find the electric field at a point P at a perpendicular distance \( r \) from the wire.
Let the wire lie along the y-axis, and point P be on the x-axis at \( (r, 0) \).
Consider a small element \( dy \) of the wire at a distance \( y \) from the origin. The charge on this element is \( dq = \lambda dy \).
The distance from this element \( dq \) to point P is \( \sqrt{r^2 + y^2} \).
The electric field \( dE \) produced by \( dq \) at P is:
\( dE = \frac{1}{4 \pi \varepsilon_0} \frac{dq}{ (r^2 + y^2) } = \frac{1}{4 \pi \varepsilon_0} \frac{\lambda dy}{(r^2 + y^2)} \).
This electric field \( dE \) has two components: \( dE_x \) (perpendicular to the wire) and \( dE_y \) (parallel to the wire).
By symmetry, the \( dE_y \) components due to elements on opposite sides of the origin will cancel each other out. So, the net electric field will only be in the x-direction.
\( dE_x = dE \cos \theta \), where \( \theta \) is the angle between \( dE \) and the x-axis.
From the geometry, \( \cos \theta = \frac{r}{\sqrt{r^2 + y^2}} \).
So, \( dE_x = \frac{1}{4 \pi \varepsilon_0} \frac{\lambda dy}{(r^2 + y^2)} \frac{r}{\sqrt{r^2 + y^2}} = \frac{\lambda r}{4 \pi \varepsilon_0} \frac{dy}{(r^2 + y^2)^{3/2}} \).
To integrate, let \( y = r \tan \theta' \), then \( dy = r \sec^2 \theta' d\theta' \).
And \( r^2 + y^2 = r^2 (1 + \tan^2 \theta') = r^2 \sec^2 \theta' \).
As \( y \) goes from \( -\infty \) to \( +\infty \) (for an infinitely long wire), \( \theta' \) goes from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \).
\( E_x = \int_{-\pi/2}^{\pi/2} \frac{\lambda r}{4 \pi \varepsilon_0} \frac{r \sec^2 \theta' d\theta'}{(r^2 \sec^2 \theta')^{3/2}} \)
\( E_x = \int_{-\pi/2}^{\pi/2} \frac{\lambda r^2 \sec^2 \theta' d\theta'}{4 \pi \varepsilon_0 r^3 \sec^3 \theta'} \)
\( E_x = \frac{\lambda}{4 \pi \varepsilon_0 r} \int_{-\pi/2}^{\pi/2} \frac{1}{\sec \theta'} d\theta' = \frac{\lambda}{4 \pi \varepsilon_0 r} \int_{-\pi/2}^{\pi/2} \cos \theta' d\theta' \)
\( E_x = \frac{\lambda}{4 \pi \varepsilon_0 r} [\sin \theta']_{-\pi/2}^{\pi/2} \)
\( E_x = \frac{\lambda}{4 \pi \varepsilon_0 r} [\sin(\pi/2) - \sin(-\pi/2)] \)
\( E_x = \frac{\lambda}{4 \pi \varepsilon_0 r} [1 - (-1)] = \frac{\lambda}{4 \pi \varepsilon_0 r} [2] \)
\( E = \frac{\lambda}{2 \pi \varepsilon_0 r} \).
This is the formula for the electric field due to an infinitely long thin wire.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक अनंत लंबाई के सीधे आवेशित तार को दर्शाता है, जिस पर रैखिक आवेश घनत्व \( \lambda \) है। बिंदु O मूल बिंदु है, और P वह बिंदु है जहाँ विद्युत क्षेत्र की गणना की जा रही है, जो तार से लंबवत दूरी r पर है। चित्र में तार के एक छोटे से खंड \( dq \) को दिखाया गया है, और कोण \( \theta \) तथा \( \theta' \) को भी परिभाषित किया गया है, जिसका उपयोग क्षेत्र के घटकों को हल करने के लिए किया जाता है।
In simple words: We imagine the wire made of many tiny charged pieces. Each piece makes a small electric field. By adding up (integrating) the sideways parts of all these tiny fields from the whole wire, we found the total electric field at a point near the wire. The vertical parts of the fields cancel out because of symmetry.

🎯 Exam Tip: Deriving the electric field using integration requires careful setup of coordinates, defining a differential charge element, breaking the field into components, and using symmetry to simplify the integration limits. Mastering the trigonometric substitutions (like \( y = r \tan \theta' \)) is key for successful evaluation of such integrals.

 

**Question 29. It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so-called 'up quark (denoted by u) of charge \( +\frac{2}{3}e \), and the 'down' quark (denoted by d) of charge \( -\frac{1}{3}e \) have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.**
Answer:Solution:
Given quark charges:
Up quark (u): charge \( +\frac{2}{3}e \)
Down quark (d): charge \( -\frac{1}{3}e \)
Both protons and neutrons consist of three quarks.
**For a proton:**
A proton has a net charge of \( +e \). We need to find a combination of three quarks (u and d) that adds up to \( +e \).
Let's try combinations:
- If it were 3 up quarks: \( 3 \times (+\frac{2}{3}e) = +2e \) (Incorrect)
- If it were 3 down quarks: \( 3 \times (-\frac{1}{3}e) = -e \) (Incorrect)
- Consider 2 up quarks and 1 down quark (uud):
Charge = \( (+\frac{2}{3}e) + (+\frac{2}{3}e) + (-\frac{1}{3}e) = \frac{4}{3}e - \frac{1}{3}e = \frac{3}{3}e = +e \).
This matches the charge of a proton. So, the composition of a proton is uud.
**For a neutron:**
A neutron has a net charge of \( 0 \). We need to find a combination of three quarks (u and d) that adds up to \( 0 \).
- Consider 1 up quark and 2 down quarks (udd):
Charge = \( (+\frac{2}{3}e) + (-\frac{1}{3}e) + (-\frac{1}{3}e) = \frac{2}{3}e - \frac{2}{3}e = 0 \).
This matches the charge of a neutron. So, the composition of a neutron is udd.
Therefore, a possible quark composition for a proton is uud, and for a neutron is udd.
In simple words: Protons have a total charge of +1, and neutrons have zero charge. By combining "up" quarks (which have a charge of \(+\frac{2}{3}\)) and "down" quarks (which have a charge of \(-\frac{1}{3}\)), we found that a proton is made of two "up" quarks and one "down" quark (uud), and a neutron is made of one "up" quark and two "down" quarks (udd).

🎯 Exam Tip: Remember the charges of up and down quarks. For these types of problems, combine the quark charges to match the known total charge of the proton (\(+e\)) and neutron (\(0\)), keeping in mind that each consists of three quarks.

GSEB Class 12 Physics Electric Charges And Fields Additional Important Questions And Answers

Question 1. A glass rod rubbed with silk is brought close to tw>o uncharged spheres in contact with each other inducing charges on them as shown in the figure. What happens when
(a) the spheres are slightly separated.
(b) the glass rod is subsequently removed.
(c) the spheres are separated far apart.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो अनावेशित गोलों A और B को एक-दूसरे के संपर्क में दिखाता है। एक कांच की छड़, जिसे रेशम से रगड़ा गया है (और इस प्रकार धनात्मक रूप से आवेशित है), गोलों के पास लाई जाती है। इससे गोलों पर आवेश प्रेरित होते हैं: गोला A पर ऋणात्मक आवेश और गोला B पर धनात्मक आवेश प्रेरित होता है।
Answer:
(a) When the spheres are slightly moved apart, the induced charges largely remain in their respective positions on each sphere due to mutual attraction.
(b) After the glass rod is taken away, the charges redistribute. The negative charges on sphere A and positive charges on sphere B are attracted to each other, so they move to the surfaces of the spheres that are closest to each other.
(c) When the spheres are moved far apart from each other, the charges on each sphere spread out uniformly over their entire surfaces because there is no longer any significant electrostatic interaction between the two spheres.In simple words: This question explains how charges move and rearrange on objects when another charged object is brought near, and then removed or separated. It shows how induction works and how charges distribute themselves.

🎯 Exam Tip: Understand the concept of charging by induction and how charges redistribute on conductors. Focus on the final charge distribution in different separation stages.

Question 2. Although ordinary rubber is an insulator, the special rubber tyres of aircraft are made slightly conducting. Explain why.

Answer: Aircraft moving through the air build up static charge due to friction with the atmosphere. If this charge isn't removed, it can build up to dangerous levels, potentially causing sparks or explosions, especially from sharp metal parts. By making the tires slightly conductive, this built-up charge can flow through the tires and safely discharge into the ground when the aircraft lands or takes off.In simple words: Airplanes get charged up from rubbing against the air. If the tires can conduct electricity a little, this charge goes into the ground when the plane lands, stopping dangerous sparks.

🎯 Exam Tip: This question tests your understanding of static electricity, charge accumulation, and safety measures in real-world applications. Relate the properties of materials (insulators vs. conductors) to practical safety designs.

Question 3. Vehicles carrying inflammable material usually have metallic ropes touching the ground during motion. Why?

Answer: Vehicles transporting flammable substances become electrically charged due to friction while driving. This build-up of static electricity could create sparks, which might ignite the flammable materials. To prevent this, metallic ropes or chains are attached to the vehicle, allowing the accumulated charge to continuously flow safely into the earth.In simple words: Trucks carrying flammable goods use metal chains that drag on the ground. This helps remove static electricity that builds up from friction, preventing sparks that could cause a fire.

🎯 Exam Tip: Relate the principles of static charge and discharge to safety applications, especially when dealing with hazardous materials. Emphasize the importance of earthing/grounding.

Question 4. Is coulomb a very big unit of charge?

Answer: Yes, the coulomb is a very large unit of electric charge. For instance, the repulsive force between two charges, each measuring one coulomb, when placed one meter apart in air or a vacuum, is an immense `\( 9 \times 10^9 \)` Newtons.In simple words: Yes, a coulomb is a huge amount of charge. Two 1-coulomb charges push each other away with a massive force if placed a meter apart.

🎯 Exam Tip: Remember the magnitude of Coulomb's constant (`\( 9 \times 10^9 \text{ Nm}^2/\text{C}^2 \)`), which demonstrates why a single coulomb is a very large charge unit in practical scenarios.

Question 5. The electric field can be studied in terms of electric field intensity.
(a) What is an electric field?
(b) Define electric field strength. Give its unit and dimensional formula
(c) Who introduced the concept of electric field?
(d) Is the electric field a scalar quantity or a vector quantity?

Answer:
(a) An electric field is the space around an electric charge where other charges or charged objects experience an electric force. It shows the region where the influence of a charge can be felt.
(b) Electric field strength, also called electric field intensity, is the force experienced by a unit positive test charge placed at a specific point in the field. Its unit is Newtons per Coulomb (N/C), and its dimensional formula is `\( [\text{M}^1\text{L}^1\text{T}^{-3}\text{A}^{-1}] \)`.
(c) Michael Faraday was the scientist who first introduced the idea of an electric field.
(d) The electric field is a vector quantity, meaning it has both magnitude and direction.In simple words: An electric field is the area around a charge where other charges feel a push or pull. Electric field strength tells you how strong this push or pull is at any spot. Michael Faraday came up with this idea. It has a direction, so it's a vector.

🎯 Exam Tip: Clearly distinguish between electric field and electric field strength, remembering their definitions, units, and whether they are scalar or vector quantities for examination purposes.

Question 6. Show mathematically that the electric field strength due to a short electric dipole at a distance 'r' along its axis is twice that at the same distance along its equatorial axis.

Answer: We need to show that the electric field strength along the axial line of a short electric dipole is twice the magnitude of the electric field strength along its equatorial line, at the same distance 'r' from the center.
The electric field along the axial line (`\( \text{E}_{\text{axial}} \)`), when `\( r \)` is much larger than the dipole length, is given by:
`\( \text{E}_{\text{axial}} = \frac{1}{4\pi\varepsilon_0} \frac{2\text{p}}{\text{r}^3} \)`
The electric field along the equatorial line (`\( \text{E}_{\text{equatorial}} \)`), under the same condition (`\( r \gg \text{a} \)`), is given by:
`\( \text{E}_{\text{equatorial}} = \frac{1}{4\pi\varepsilon_0} \frac{-\text{p}}{\text{r}^3} \)`
Comparing these two expressions, we can observe that:
`\( \text{E}_{\text{axial}} = 2 \times (-\text{E}_{\text{equatorial}}) \)`
Considering only the magnitudes, we find that:
`\( |\text{E}_{\text{axial}}| = 2 \times |\text{E}_{\text{equatorial}}| \)`
Thus, the electric field strength along the axial line of a short electric dipole is indeed twice the magnitude of the electric field strength along its equatorial line at the same distance 'r'.In simple words: For a small electric dipole, the electric force field you feel straight out from its ends (axial line) is twice as strong as the field you feel sideways from its middle (equatorial line), if you are the same distance away.

🎯 Exam Tip: Remember the formulas for electric field strength for a short dipole on both axial and equatorial lines, and how their magnitudes compare. The negative sign for the equatorial field indicates its direction relative to the dipole moment.

Question 7. What is the total force acting on the dipole?

Answer: The total force acting on an electric dipole when it is placed in a uniform electric field is zero. This is because the forces on the positive and negative charges of the dipole are equal in magnitude and opposite in direction, canceling each other out.In simple words: If an electric dipole is in a steady electric field, the total push or pull on it is zero because the forces on its two ends balance each other.

🎯 Exam Tip: Understand that a uniform electric field results in zero net force on a dipole, but can produce a torque, which is a key concept in dipole behavior.

Question 8. What is the effect of keeping the dipole in the field?

Answer: When an electric dipole is placed in an electric field, its behavior depends on its orientation. If it's aligned parallel to the field, it remains in a stable position (stable equilibrium). If it's aligned opposite to the field (antiparallel), it's in an unstable position (unstable equilibrium). If the dipole is at any angle to the field, it will experience a turning force (torque), causing it to rotate until it aligns with the field.In simple words: If a dipole is put in an electric field, it will try to line up with the field. If it's already lined up, it stays still (stable). If it's upside down, it's wobbly (unstable). If it's sideways, it spins until it lines up.

🎯 Exam Tip: Differentiate between stable and unstable equilibrium for a dipole in an electric field, and recognize that an inclined dipole experiences torque, which is crucial for analyzing its rotational motion.

Question 9. What happens if the field is not uniform?

Answer: If an electric dipole is placed in a non-uniform electric field, it will experience both a net force (translational motion) and a torque (rotational motion). The forces on the positive and negative charges will no longer be equal in magnitude and opposite in direction, leading to a net force that causes it to move, in addition to the turning effect.In simple words: If the electric field isn't even everywhere, an electric dipole will not only spin (rotate) but also move from its place (translate), because the forces on its ends won't perfectly cancel out.

🎯 Exam Tip: Contrast the behavior of a dipole in uniform vs. non-uniform electric fields, noting the presence of both translational force and torque in the latter case.

Question 10. The following four insulating spheres contain magnet, dipole, proton, and neutron. Pick the odd one out of the following based on the net flux.

ℹ️ चित्र व्याख्या (Diagram Explanation): यहाँ चार गोले दिखाए गए हैं। गोला (i) में 'N' और 'S' लेबल वाला एक चुंबक है। गोला (ii) में एक ऋणात्मक और एक धनात्मक आवेश वाला विद्युत द्विध्रुव है। गोला (iii) में एक धनात्मक आवेशित प्रोटॉन है। गोला (iv) में एक उदासीन न्यूट्रॉन है।
Answer: We are given four insulating spheres, each containing a different item: (i) a magnet, (ii) an electric dipole, (iii) a proton, and (iv) a neutron. We need to identify which one has a non-zero net electric flux.
(i) A magnet produces magnetic fields, not electric flux.
(ii) An electric dipole has equal positive and negative charges, so its net charge is zero, resulting in zero net electric flux.
(iii) A proton has a positive charge, so it will produce a non-zero net outward electric flux.
(iv) A neutron has no net charge, so it produces zero net electric flux.
Therefore, the sphere containing the proton (iii) is the odd one out because it has a non-zero net electric flux.In simple words: We have spheres with a magnet, an electric dipole (plus and minus charges that cancel), a proton (positive charge), and a neutron (no charge). Only the proton sphere has a total electric field flowing out of it (net flux) because it has a charge. So, the proton is the odd one out.

🎯 Exam Tip: Apply Gauss's Law, which states that net electric flux through a closed surface depends only on the net charge enclosed within it. Remember that magnets and dipoles have zero net electric charge (or produce magnetic flux, not electric flux).

Question 11. How does the electric field (E) vary with distance (r) in the following cases?
(i) Uniformly charged spherical shell
(ii) Uniformly charged solid sphere.
(iii) Uniformly charged plane sheet of large size.
Also, plot a graph between E and r in each case.

Answer:
(i) For a uniformly charged spherical shell:
- Outside the shell, the electric field `\( \text{E} \)` is inversely proportional to the square of the distance `\( \text{r} \)` from the center, i.e., `\( \text{E} \propto \frac{1}{\text{r}^2} \)`.
- On the surface of the shell, the electric field is also proportional to `\( \frac{1}{\text{r}^2} \)`, where `\( \text{r} \)` is the shell's radius.
- Inside the shell, the electric field `\( \text{E} \)` is zero.
(ii) For a uniformly charged solid sphere:
- Outside the sphere, the electric field `\( \text{E} \)` is inversely proportional to the square of the distance `\( \text{r} \)` from the center, i.e., `\( \text{E} \propto \frac{1}{\text{r}^2} \)`.
- On the surface of the sphere, the electric field is proportional to `\( \frac{1}{\text{r}^2} \)`.
- Inside the sphere, the electric field `\( \text{E} \)` is directly proportional to the distance `\( \text{r} \)` from the center, i.e., `\( \text{E} \propto \text{r} \)`.
(iii) For a uniformly charged plane sheet of large size:
- The electric field `\( \text{E} \)` is constant and equal to `\( \frac{\sigma}{2 \varepsilon_{0}} \)`, meaning it does not depend on the distance `\( \text{r} \)` from the sheet.In simple words:
- (i) For a hollow charged ball: field is zero inside, and outside it gets weaker as `\( \frac{1}{\text{r}^2} \)`.
- (ii) For a solid charged ball: field grows stronger as `\( \text{r} \)` inside, and outside it gets weaker as `\( \frac{1}{\text{r}^2} \)`.
- (iii) For a big flat charged sheet: the field is always the same strength, no matter how far you are from it.

🎯 Exam Tip: Memorize the variations of electric field with distance for these common charge distributions (spherical shell, solid sphere, infinite sheet) as they are frequently tested using Gauss's Law. Be prepared to sketch the `\( \text{E} \)` vs. `\( \text{r} \)` graphs.

Question 12. Give the expression for the time period of oscillation of an electric dipole in a uniform electric field.

Answer: The formula for the time period of oscillation (`\( \text{T} \)`) of an electric dipole when placed in a uniform electric field is given by:
`\( \text{T} = 2\pi\sqrt{\frac{\text{I}}{\text{pE}}} \)`
In this formula:
- `\( \text{I} \)` represents the moment of inertia of the electric dipole.
- `\( \text{p} \)` stands for the electric dipole moment.
- `\( \text{E} \)` denotes the strength of the uniform electric field.In simple words: The time it takes for an electric dipole to swing back and forth in a steady electric field is given by a formula that uses `\( 2\pi \)` times the square root of its moment of inertia divided by its dipole moment multiplied by the field strength.

🎯 Exam Tip: Understand the physical significance of each variable in the time period formula for an oscillating dipole and recall this formula accurately for calculations and theoretical questions.

Question 13. A pendulum bob of mass 40 mg and carrying a charge of `\( 2 \times 10^{-5} \text{C} \)` is at rest in a horizontal uniform electric field of `\( 2 \times 10^4 \text{NC}^{-1} \)`. Find the tension in the string of the pendulum and the angle it makes with the vertical.

ℹ️ चित्र व्याख्या (Diagram Explanation): चित्र एक आवेशित पेंडुलम बॉब दिखाता है जो एक क्षैतिज विद्युत क्षेत्र में लटका हुआ है। इस पर तीन बल कार्य कर रहे हैं: नीचे की ओर गुरुत्वाकर्षण बल (mg), डोरी के साथ तनाव बल (T), और क्षैतिज विद्युत क्षेत्र के कारण बल (qE)। बॉब संतुलन में है, और डोरी ऊर्ध्वाधर से एक कोण `\( \theta \)` बनाती है। तनाव `\( \text{T} \)` को `\( \text{T} \cos \theta \)` (ऊर्ध्वाधर) और `\( \text{T} \sin \theta \)` (क्षैतिज) घटकों में विभाजित किया गया है।
Answer: A small pendulum bob, with a mass of `\( 40 \text{ mg} \)` (which is `\( 40 \times 10^{-6} \text{ kg} \)` or `\( 40 \times 10^{-3} \text{ g} \)` based on the given solution's calculation approach) and a charge of `\( 2 \times 10^{-5} \text{ C} \)`, is held motionless in a horizontal electric field of `\( 2 \times 10^4 \text{ N/C} \)`. We need to calculate the tension in the pendulum's string and the angle it forms with the vertical direction.
Forces acting on the bob for equilibrium:
(i) Gravitational force (weight), `\( \text{mg} \)`, pulling it downwards.
(ii) Tension, `\( \text{T} \)`, in the string, pulling it upwards along the string.
(iii) Electrostatic force, `\( \text{F} = \text{qE} \)`, pushing it horizontally due to the electric field.
For the bob to be in equilibrium (at rest), the forces must balance. We resolve tension `\( \text{T} \)` into vertical and horizontal components.
The vertical component of tension balances the weight:
`\( \text{T} \cos \theta = \text{mg} \)` (Equation 1)
The horizontal component of tension balances the electrostatic force:
`\( \text{T} \sin \theta = \text{qE} \)` (Equation 2)
Dividing Equation 2 by Equation 1 gives:
`\( \frac{\text{T} \sin \theta}{\text{T} \cos \theta} = \frac{\text{qE}}{\text{mg}} \)`
`\( \implies \tan \theta = \frac{\text{qE}}{\text{mg}} \)`
Substitute the given values:
`\( \text{q} = 2 \times 10^{-5} \text{ C} \)`
`\( \text{E} = 2 \times 10^4 \text{ N/C} \)`
`\( \text{m} = 40 \text{ mg} = 40 \times 10^{-6} \text{ kg} \)` (or `\( 40 \times 10^{-3} \text{ g} \)` in the solution's calculation logic)
`\( \text{g} = 9.8 \text{ m/s}^2 \)`
Using `\( \text{m} = 40 \times 10^{-3} \text{ kg} \)` as implied by the solution:
`\( \tan \theta = \frac{(2 \times 10^{-5}) \times (2 \times 10^4)}{(40 \times 10^{-3}) \times 9.8} = \frac{4 \times 10^{-1}}{0.392} = 1.0204 \)`
`\( \implies \theta = \arctan(1.0204) \approx 45.57^\circ \)`
Now, calculate tension `\( \text{T} \)` using Equation 1:
`\( \text{T} = \frac{\text{mg}}{\cos \theta} = \frac{(40 \times 10^{-3} \text{ kg}) \times 9.8}{\cos(45.57^\circ)} = \frac{0.392}{0.7001} \approx 0.5599 \text{ N} \)`
Therefore, `\( \text{T} = 0.5599 \text{ N} \)` (which can be written as `\( 5.60 \times 10^{-1} \text{ N} \)`).In simple words: A charged ball hanging like a pendulum is pulled down by gravity and pushed sideways by an electric field. The string holds it in place. We calculate the angle the string makes and how strong it pulls by balancing these forces. The angle is about 45.6 degrees, and the string's pull (tension) is about 0.56 Newtons.

🎯 Exam Tip: To solve equilibrium problems involving multiple forces, draw a clear free-body diagram, resolve forces into components, and apply the conditions for equilibrium (`\( \Sigma \text{F}_x = 0 \)` and `\( \Sigma \text{F}_y = 0 \)`). Pay close attention to unit conversions.

Question 14. A charge Q is fixed at each of two opposite corners of a square. A charge q is placed at each of the other two corners. If the net electrostatic force on each Q is zero, what is Q in terms of q?

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वर्ग के कोनों पर रखे गए चार आवेशों को दर्शाता है। दो विपरीत कोनों पर आवेश Q हैं, जबकि अन्य दो विपरीत कोनों पर आवेश q हैं। वर्ग की भुजा की लंबाई 'a' है। आवेश Q पर कार्य करने वाले बलों को `\( \text{E}_0 \)` (दूसरे Q से) और `\( \text{E}_1, \text{E}_2 \)` (दो q से) के रूप में दर्शाया गया है। केंद्र O पर बलों का विश्लेषण किया जाता है।
Answer: Two charges `\( \text{Q} \)` are placed at opposite corners of a square, and two charges `\( \text{q} \)` are placed at the other two opposite corners. If the total electrostatic force experienced by each `\( \text{Q} \)` charge is zero, we need to find the relationship between `\( \text{Q} \)` and `\( \text{q} \)`.
Let the side length of the square be `\( \text{a} \)`. Consider one charge `\( \text{Q} \)` at a corner. It experiences forces from the other `\( \text{Q} \)` charge (across the diagonal) and from the two `\( \text{q} \)` charges (along the adjacent sides).
1. Force from the other `\( \text{Q} \)` charge (`\( \text{F}_{\text{QQ}} \)`): The distance between these two `\( \text{Q} \)` charges is the diagonal of the square, `\( \text{a}\sqrt{2} \)`.
`\( \text{F}_{\text{QQ}} = \frac{1}{4\pi\varepsilon_0} \frac{\text{Q}^2}{(\text{a}\sqrt{2})^2} = \frac{1}{4\pi\varepsilon_0} \frac{\text{Q}^2}{2\text{a}^2} \)`
This force is directed along the diagonal, away from the other `\( \text{Q} \)` if `\( \text{Q} \)` is positive (repulsive).
2. Forces from the two `\( \text{q} \)` charges (`\( \text{F}_{\text{Qq1}} \)` and `\( \text{F}_{\text{Qq2}} \)`): The distance to each `\( \text{q} \)` charge is `\( \text{a} \)`.
`\( \text{F}_{\text{Qq1}} = \frac{1}{4\pi\varepsilon_0} \frac{\text{Q}\text{q}}{\text{a}^2} \)`
`\( \text{F}_{\text{Qq2}} = \frac{1}{4\pi\varepsilon_0} \frac{\text{Q}\text{q}}{\text{a}^2} \)`
These two forces (`\( \text{F}_{\text{Qq1}} \)` and `\( \text{F}_{\text{Qq2}} \)` ) are perpendicular to each other. Their resultant force, `\( \text{F}_{\text{Qq, resultant}} \)`, is directed along the diagonal.
`\( \text{F}_{\text{Qq, resultant}} = \sqrt{\text{F}_{\text{Qq1}}^2 + \text{F}_{\text{Qq2}}^2} = \sqrt{2} \times \frac{1}{4\pi\varepsilon_0} \frac{|\text{Q}\text{q}|}{\text{a}^2} \)`
For the net force on `\( \text{Q} \)` to be zero, `\( \text{F}_{\text{QQ}} \)` must be equal in magnitude and opposite in direction to `\( \text{F}_{\text{Qq, resultant}} \)`. This implies that `\( \text{Q} \)` and `\( \text{q} \)` must have opposite signs for the forces to cancel out. If `\( \text{Q} \)` and `\( \text{q} \)` were of the same sign, all forces would be repulsive, and the net force would not be zero.
Therefore, we set the magnitudes equal:
`\( \frac{1}{4\pi\varepsilon_0} \frac{\text{Q}^2}{2\text{a}^2} = \sqrt{2} \frac{1}{4\pi\varepsilon_0} \frac{|\text{Q}\text{q}|}{\text{a}^2} \)`
Cancel common terms (`\( \frac{1}{4\pi\varepsilon_0\text{a}^2} \)` and one `\( \text{Q} \)`):
`\( \frac{\text{Q}}{2} = \sqrt{2} |\text{q}| \)`
`\( \text{Q} = 2\sqrt{2} |\text{q}| \)`
Since `\( \text{Q} \)` and `\( \text{q} \)` must have opposite signs, we can express this relationship including the sign:
`\( \text{Q} = -2\sqrt{2} \text{q} \)`In simple words: Imagine a square with big charges (Q) on two opposite corners and small charges (q) on the other two. For a big charge (Q) to feel no total push or pull, the forces from the other charges must cancel out. This happens if the big charges and small charges have opposite types of electricity. The big charge Q will be `\(-2\sqrt{2}\)` times the small charge q.

🎯 Exam Tip: This problem requires careful vector addition of forces. Ensure you correctly determine the direction of forces (attractive/repulsive) and their vector components for proper summation to zero. The sign of the charges is crucial.

Question 15. It is safe to sit inside a vehicle, whenever there is lightning and thunder. Why?

Answer: During a lightning storm, it is generally safe to be inside a vehicle because of a phenomenon called electrostatic screening or the Faraday cage effect. The metallic body of the vehicle acts as a conductor, and any external electric charges from a lightning strike redistribute on its outer surface, ensuring that the electric field inside remains zero. This protects the occupants from the high voltage.In simple words: Sitting inside a car during a lightning storm is safe because the metal body of the car acts like a shield. The lightning hits the outside and the electricity goes around you, not through you.

🎯 Exam Tip: Understand the concept of electrostatic shielding (Faraday cage) and its real-world applications for safety. This principle is fundamental to protecting sensitive environments from electric fields.

Question 16. Equipment very sensitive to electric field is to be transferred from one point to another. What precautions you will do?

Answer: To transport delicate equipment that is sensitive to electric fields, the best precaution is to place it inside a metal enclosure or cage. This metal cage will provide electrostatic shielding (Faraday cage effect), preventing any external electric fields from penetrating and affecting the sensitive instruments inside.In simple words: To move sensitive electrical equipment safely, put it inside a metal box. The metal box will block any outside electric fields from messing with the equipment.

🎯 Exam Tip: Recognize how electrostatic shielding protects sensitive electronics from unwanted electric fields, which is a practical application of electrostatic principles.

Question 17. An electric flux of `\( -6 \times 10^3 \text{ Nm}^2\text{C}^{-1} \)` passes normally through a spherical Gaussian surface of a radius of 10 cm, due to a point of charge placed at the centre.
(i) What is the charge enclosed by the Gaussian surface?
(ii) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?

Answer:
(i) The given electric flux `\( \phi \)` is `\( -6 \times 10^3 \text{ Nm}^2\text{C}^{-1} \)`.
According to Gauss's Theorem, the electric flux through a closed surface is equal to the total charge enclosed (`\( \text{q} \)`) divided by the permittivity of free space (`\( \varepsilon_{0} \)`):
`\( \phi = \frac{\text{q}}{\varepsilon_{0}} \)`
Rearranging the formula to find the charge `\( \text{q} \)`:
`\( \text{q} = \varepsilon_{0} \times \phi \)`
Substituting the known values (`\( \varepsilon_{0} = 8.85 \times 10^{-12} \text{ C}^2\text{N}^{-1}\text{m}^{-2} \)`):
`\( \text{q} = (8.85 \times 10^{-12} \text{ C}^2\text{N}^{-1}\text{m}^{-2}) \times (-6 \times 10^3 \text{ Nm}^2\text{C}^{-1}) \)`
`\( \text{q} = -5.31 \times 10^{-8} \text{ C} \)`
(ii) Gauss's Theorem states that the net electric flux through any closed surface depends only on the net charge enclosed within it, not on the size or shape of the surface (as long as it encloses the same charge). Since the charge enclosed remains the same even if the radius of the Gaussian surface is doubled, the electric flux passing through the surface will also remain the same.
Therefore, the flux will still be `\( -6 \times 10^3 \text{ Nm}^2\text{C}^{-1} \)`.In simple words: A charged object makes an electric field that passes through a surrounding imaginary sphere.
(i) Using a rule called Gauss's theorem, we can find the amount of charge inside the sphere by multiplying the flux by a special constant. Here, the charge is `\( -5.31 \times 10^{-8} \text{ C} \)`.
(ii) If you make the imaginary sphere bigger, but the charge inside stays the same, the total electric field passing through it (flux) also stays the same. So the flux is still `\( -6 \times 10^3 \text{ Nm}^2\text{C}^{-1} \)`.

🎯 Exam Tip: Gauss's Law is crucial for calculating electric fields and charges. Remember that the net flux depends only on the *enclosed* charge, not the size or shape of the Gaussian surface, which is a common conceptual question.