Maharashtra Board Class 11 Physics Chapter 10 Electrostatics Solutions

1. Choose The Correct Option.

Question 1. A positively charged glass rod is brought close to a metallic rod isolated from ground. The charge on the side of the metallic rod away from the glass rod will be
(A) same as that on the glass rod and equal in quantity
(B) opposite to that on the glass of and equal in quantity
(C) same as that on the glass rod but lesser in quantity
(D) same as that on the glass rod but more in quantity
Answer: (A) same as that on the glass rod and equal in quantity
In simple words: When a positively charged rod is brought near an isolated metallic rod, charges redistribute by induction. The side away from the rod develops a charge of the same sign as the inducing rod, but of equal magnitude due to overall charge conservation in the isolated metallic rod.

🎯 Exam Tip: Understanding charge induction and redistribution in isolated conductors is key for such conceptual questions. Remember that the net charge of an isolated system remains constant.

Question 2. An electron is placed between two parallel plates connected to a battery. If the battery is switched on, the electron will
(A) be attracted to the +ve plate
(B) be attracted to the -ve plate
(C) remain stationary
(D) will move parallel to the plates
Answer: (A) be attracted to the +ve plate
In simple words: An electron carries a negative charge. When parallel plates are connected to a battery, an electric field is established, with the positive plate attracting the negatively charged electron.

🎯 Exam Tip: Recall that opposite charges attract. Electrons, being negative, will always move towards a positive potential or plate in an electric field.

Question 3. A charge of \(+ 7 \text{ µC}\) is placed at the centre of two concentric spheres with radius 2.0 cm and 4.0 cm respectively. The ratio of the flux through them will be
(A) 1:4
(B) 1:2
(C) 1:1
(D) 1:16
Answer: (C) 1:1
In simple words: According to Gauss's Law, the electric flux through any closed surface depends only on the net charge enclosed within that surface, not on the size or shape of the surface. Since both concentric spheres enclose the same charge \(+ 7 \text{ µC}\), the flux through them will be equal.

🎯 Exam Tip: Gauss's Law (\( \Phi = Q_{enc}/\epsilon_0 \)) is fundamental here. The flux is independent of the Gaussian surface's radius as long as it encloses the same net charge.

Question 4. Two charges of 1.0 C each are placed one meter apart in free space. The force between them will be
(A) 1.0 N
(B) \(9 \times 10^9\) N
(C) \(9 \times 10^{-9}\) N
(D) 10 N
Answer: (B) \(9 \times 10^9\) N
In simple words: Coulomb's Law calculates the force between two point charges. For unit charges (1 C) placed one meter apart, the force constant \( \frac{1}{4\pi\epsilon_0} \) which is \(9 \times 10^9 \text{ N m}^2/\text{C}^2\) directly gives the force.

🎯 Exam Tip: Remember Coulomb's constant \(k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2\). For \(q_1=1\text{ C}\), \(q_2=1\text{ C}\), and \(r=1\text{ m}\), the force \(F = k \frac{q_1 q_2}{r^2}\) simplifies to \(k\).

Question 5. Two point charges of \(+5 \text{ µC}\) are so placed that they experience a force of \(80 \times 10^{-3}\) N. They are then moved apart, so that the force is now \(2.0 \times 10^{-3}\) N. The distance between them is now
(A) 1/4 the previous distance
(B) double the previous distance
(C) four times the previous distance
(D) half the previous distance
Answer: (B) double the previous distance
In simple words: Coulomb's force is inversely proportional to the square of the distance between charges (\(F \propto 1/r^2\)). If the force decreases from \(80 \times 10^{-3}\) N to \(2 \times 10^{-3}\) N (a factor of 40), then the distance must have increased by a factor of \( \sqrt{40} \). Re-evaluating the options with the exact numbers: if force becomes \(1/4\) (e.g., from 80 to 20), distance doubles. If force becomes \(1/16\), distance quadruples. Here, \(80 / 2 = 40\). This implies distance changes by \( \sqrt{40} \). Let's recheck the question numbers. Force changes from 80 to 2. This is a factor of 40. This is an unexpected value for options given. Let me recheck the calculation of the force. If force reduces from \(80 \times 10^{-3}\) N to \(2.0 \times 10^{-3}\) N, this is a reduction by a factor of \(80/2 = 40\). Since \(F \propto 1/r^2\), then \(r^2 \propto 1/F\). So \(r_{new}^2 / r_{old}^2 = F_{old} / F_{new} = 80/2 = 40\). Thus \(r_{new} = \sqrt{40} r_{old}\). None of the options precisely match \(\sqrt{40}\) times the previous distance. However, if the force was \(80 \times 10^{-3}\) N and reduced to \(20 \times 10^{-3}\) N (a factor of 4), the distance would double. Let's assume there might be a typo in the question or options or the provided answer for the context of common textbook problems, and if the force became \(1/4\) of the original, the distance would be double. Given the answer is (B) double the previous distance, it implies the force reduction was 4 times. So, \(F_1 = k \frac{q_1q_2}{r_1^2}\) and \(F_2 = k \frac{q_1q_2}{r_2^2}\). If \(F_1/F_2 = 4\), then \(r_2^2/r_1^2 = 4\), so \(r_2 = 2r_1\). This aligns with a force decrease from 80 to 20, not 2. This suggests a potential typo in the OCR'd numerical values. Sticking to the provided answer, if the distance is doubled, the force becomes one-fourth. So, \(F_{new} = F_{old}/4\). If \(F_{old} = 80 \times 10^{-3}\text{ N}\), then \(F_{new} = (80 \times 10^{-3}\text{ N})/4 = 20 \times 10^{-3}\text{ N}\). The provided new force is \(2.0 \times 10^{-3}\text{ N}\). There is a clear discrepancy. Given the answer is "double the previous distance", it implies the force reduced by a factor of 4. Let's assume the new force *should have been* \(20 \times 10^{-3}\text{ N}\) for option B to be correct. I will proceed by stating the logical implication of the *given* answer. If the distance is doubled, the force reduces by a factor of four. The question states the force reduced from \(80 \times 10^{-3}\text{ N}\) to \(2.0 \times 10^{-3}\text{ N}\). This is a reduction factor of 40. If this is the case, the distance would be \( \sqrt{40} \) times the previous distance. However, since the provided answer is (B) double the previous distance, it implies that the force decreased by a factor of 4. Therefore, it is assumed the new force should be \(20 \times 10^{-3}\) N for the option to be correct.

🎯 Exam Tip: Force is inversely proportional to the square of the distance (\(F \propto 1/r^2\)). If distance is \(2r\), force is \(F/4\). If distance is \(r/2\), force is \(4F\).

Question 6. A metallic sphere A isolated from ground is charged to \(+50 \text{ µC}\). This sphere is brought in contact with other isolated metallic sphere B of half the radius of sphere A. The charge on the two sphere will be now in the ratio
(A) 1:2
(B) 2:1
(C) 4:1
(D) 1:1
Answer: (D) 1:1
In simple words: When two identical conducting spheres touch, the charge distributes equally between them. However, if they have different radii, the charge distributes such that their electric potentials become equal, meaning the charge ratio will be proportional to their radii. If they are in contact, they form a single conductor, and charge will redistribute until the surface charge densities are not necessarily equal, but the potential is equal. For two spheres in contact, charge distributes such that \(Q_A/R_A = Q_B/R_B\). Since sphere B has half the radius of sphere A (\(R_B = R_A/2\)), then \(Q_B = Q_A (R_B/R_A) = Q_A/2\). The total charge is \(+50 \text{ µC}\). So, \(Q_A + Q_B = 50\text{ µC}\). This means \(Q_A + Q_A/2 = 50\text{ µC}\), or \(3Q_A/2 = 50\text{ µC}\). So \(Q_A = (100/3)\text{ µC}\) and \(Q_B = (50/3)\text{ µC}\). The ratio \(Q_A:Q_B\) would be \( (100/3):(50/3) = 2:1 \). The provided answer (D) 1:1 is only correct if the spheres are identical in size. Given that sphere B has *half the radius* of sphere A, they are not identical. This indicates a potential discrepancy between the question's premise and the provided answer. If the question implies that *after contact, the system behaves as one and then separates*, and somehow the question is misinterpreted as *potential ratio* or *if they were identical*, then 1:1 could be a distracter. However, charge distribution on *unequal* spheres in contact is proportional to their radii. So, the ratio should be 2:1. I will provide an answer assuming the question implies identical spheres, or state the correct physical principle and highlight the discrepancy. For the sake of following the provided answer, let's assume the question implicitly considers them effectively equal for charge *density* or if the question meant the *potential* becomes equal. If the answer is 1:1, it implies that the spheres are considered identical, or that the question is asking for something else. Let's go with the direct interpretation of charge distribution on unequal spheres in contact. When two conducting spheres of radii \(R_A\) and \(R_B\) are brought into contact, the total charge \(Q\) is redistributed such that their electric potentials become equal. \(V_A = V_B\) \(k \frac{Q_A}{R_A} = k \frac{Q_B}{R_B}\) \(\frac{Q_A}{R_A} = \frac{Q_B}{R_B}\) Given \(R_B = R_A/2\). \(\frac{Q_A}{R_A} = \frac{Q_B}{R_A/2}\) \(Q_A = 2Q_B\) Also, \(Q_A + Q_B = 50 \text{ µC}\) (total charge) \(2Q_B + Q_B = 50 \text{ µC}\) \(3Q_B = 50 \text{ µC}\) \(Q_B = 50/3 \text{ µC}\) \(Q_A = 2 \times (50/3) = 100/3 \text{ µC}\) The ratio of charges \(Q_A : Q_B = (100/3) : (50/3) = 2:1\). Therefore, the provided answer (D) 1:1 is incorrect based on the problem description. I will state the correct physical principle. If I *must* output (D), I will add a disclaimer. The instructions state "Extract every word exactly as written." and "answer line formatting". This implies I should transcribe the answer, but the "In simple words" allows me to explain the physics. I will explain the physics correctly.

🎯 Exam Tip: When unequal conducting spheres are in contact, charge redistributes until their electric potentials are equal. The charge on each sphere will then be proportional to its radius (\(Q \propto R\)).

Question 7. Which of the following produces uniform electric field?
(A) point charge
(B) linear charge
(C) two parallel plates
(D) charge distributed an circular any
Answer: (C) two parallel plates
In simple words: A uniform electric field, where the field lines are parallel and equally spaced, is typically produced between two large, closely spaced parallel conducting plates carrying equal and opposite charges.

🎯 Exam Tip: Point charges and linear charges produce non-uniform electric fields. Only idealized parallel plates or very long charged cylinders/planes can approximate a uniform field.

Question 8. Two point charges of A = \(+5.0 \text{ µC}\) and B = \(-5.0 \text{ µC}\) are separated by 5.0 cm. A point charge C = \(1.0 \text{ µC}\) is placed at 3.0 cm away from the centre on the perpendicular bisector of the line joining the two point charges. The charge at C will experience a force directed towards
(A) point A
(B) point B
(C) a direction parallel to line AB
(D) a direction along the perpendicular bisector
Answer: (C) a direction parallel to line AB
In simple words: The two charges A (\(+5.0 \text{ µC}\)) and B (\(-5.0 \text{ µC}\)) form an electric dipole. A positive test charge C (\(1.0 \text{ µC}\)) placed on the perpendicular bisector will be repelled by A and attracted by B. Due to symmetry, the components of these forces perpendicular to the line AB will cancel out, and the components parallel to the line AB will add up, resulting in a net force parallel to the line joining A and B. Specifically, the force will be directed from A towards B, i.e., parallel to the line AB.

🎯 Exam Tip: For an electric dipole, the electric field (and thus force on a test charge) on the perpendicular bisector points parallel to the dipole axis (from positive to negative charge). The components perpendicular to the dipole axis cancel out due to symmetry.

2. Answer The Following Questions.

Question 1. What is the magnitude of charge on an electron?
Answer: The magnitude of charge on an electron is \(1.6 \times 10^{-19}\) C
In simple words: An electron is a fundamental particle carrying the smallest unit of negative electric charge, which has a specific constant value of \(1.6 \times 10^{-19}\) Coulombs.

🎯 Exam Tip: This is a basic constant in physics. Remember the value \(1.6 \times 10^{-19}\) C and its negative sign for an electron's charge.

Question 2. State the law of conservation of charge.
Answer: In any given physical process, charge may get transferred from one part of the system to another, but the total charge in the system remains constant" OR For an isolated system, total charge cannot be created nor destroyed.
In simple words: The law of conservation of charge states that the total amount of electric charge in an isolated system remains constant over time; it can only be transferred from one object to another, not created or destroyed.

🎯 Exam Tip: This law is as fundamental as the conservation of energy and momentum. It means that in any reaction or process, the sum of charges before and after must be the same.

Question 3. Define a unit charge.
Answer: Unit charge (one coulomb) is the amount of charge which, when placed at a distance of one metre from another charge of the same magnitude in vacuum, experiences a force of \(9.0 \times 10^9\) N.
In simple words: A unit charge, or one Coulomb (1 C), is defined as the amount of charge that, when placed one meter away from an identical charge in a vacuum, experiences an electrostatic force of \(9 \times 10^9\) Newtons.

🎯 Exam Tip: This definition is directly derived from Coulomb's Law, where \(F = k \frac{q_1 q_2}{r^2}\). If \(q_1=q_2=1\text{ C}\) and \(r=1\text{ m}\), then \(F = k = 9 \times 10^9\text{ N}\).

Question 4. Two parallel plates have a potential difference of \(10\text{V}\) between them. If the plates are 0.5 mm apart, what will be the strength of electric charge.
Answer: Given: \(V=10\text{V}\) \(d = 0.5 \text{ mm} = 0.5 \times 10^{-3}\) m To find: The strength of electric field (E) Formula: \(E = \frac{V}{d}\) Calculation: From formula,
\(E = \frac{10}{0.5 \times 10^{-3}}\)
\(20 \times 10^3\text{ V/m}\)
In simple words: The electric field strength between two parallel plates is calculated by dividing the potential difference across them by the distance separating them.

🎯 Exam Tip: The relationship between electric field (E), potential difference (V), and distance (d) is \(E = V/d\). Ensure units are consistent (meters for distance, volts for potential, resulting in V/m or N/C for E).

Question 5. What is uniform electric field?
Answer: A uniform electric field is a field whose magnitude and direction are same at all points. For example, field between two parallel plates as shown in the diagram.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एकसमान विद्युत क्षेत्र को दर्शाता है जहाँ दो समांतर प्लेटों के बीच विद्युत क्षेत्र रेखाएँ धनात्मक प्लेट से निकलकर ऋणात्मक प्लेट पर समाप्त होती हैं, जो विद्युत क्षेत्र के परिमाण और दिशा में निरंतरता को इंगित करती हैं।
In simple words: A uniform electric field is a region where the electric field strength has the same magnitude and direction everywhere, typically represented by parallel, equally spaced field lines.

🎯 Exam Tip: Recognize uniform electric fields by parallel, equally spaced electric field lines. Such fields are idealizations often found between large, oppositely charged parallel plates.

Question 6. If two lines of force intersect of one point. What does it mean?
Answer: If two lines of force intersect of one point, it would mean that electric field has two directions at a single point.
In simple words: Electric field lines can never intersect because, at any given point, the electric field must have a unique direction; if lines intersected, it would imply two possible directions for the field at that single point, which is physically impossible.

🎯 Exam Tip: This is a fundamental property of electric field lines. Their non-intersection ensures that the electric field at any point in space is uniquely defined.

Question 7. State the units of linear charge density.
Answer: SI unit of \( \lambda \) is (C / m).
In simple words: Linear charge density (\( \lambda \)) represents the amount of electric charge distributed per unit length of a line or rod, and its standard unit is Coulombs per meter (C/m).

🎯 Exam Tip: Understand that charge density units reflect its definition: linear charge density is charge per unit length, surface charge density is charge per unit area, and volume charge density is charge per unit volume.

Question 8. What is the unit of dipole moment?
Answer: i. Strength of a dipole is measured in terms of a quantity called the dipole moment.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक विद्युत द्विध्रुव को दर्शाता है जिसमें दो समान और विपरीत आवेश (-q और +q) 2l दूरी पर रखे गए हैं। यह आरेख द्विध्रुव आघूर्ण की परिभाषा और उसकी दिशा को समझाता है। ii. Let q be the magnitude of each charge and \(2\vec{l}\) be the distance from negative charge to positive charge. Then, the product \(q(2\vec{l})\) is called the dipole moment \( \vec{p} \). iii. Dipole moment is defined as \( \vec{p} = q(2\vec{l}) \) iv. A dipole moment is a vector whose magnitude is \(q (2l)\) and the direction is from the negative to the positive charge. v. The unit of dipole moment is coulomb-metre (C m) or debye (D).
In simple words: Dipole moment quantifies the strength of an electric dipole, which consists of two equal and opposite charges separated by a distance; its unit is coulomb-metre (C m), calculated as the product of the charge magnitude and the separation distance.

🎯 Exam Tip: Remember that dipole moment (\( \vec{p} \)) is a vector quantity, its direction is from negative to positive charge, and its SI unit is C m. Debye (D) is a common non-SI unit.

Question 9. What is relative permittivity?
Answer: i. Relative permittivity or dielectric constant is the ratio of absolute permittivity of a medium to the permittivity of free space. It is denoted as K or \( \epsilon_r \). i.e., \( K \text{ or } \epsilon_r = \frac{\epsilon}{\epsilon_0} \) ii. It is the ratio of the force between two point charges placed a certain distance apart in free space or vacuum to the force between the same two point charges when placed at the same distance in the given medium. i.e., \( K \text{ or } \epsilon_r = \frac{F_{\text{vacuum}}}{F_{\text{medium}}} \) iii. It is also called as specific inductive capacity or dielectric constant.
In simple words: Relative permittivity, also known as the dielectric constant, is a dimensionless quantity that describes how an electric field affects, and is affected by, a dielectric medium; it is the ratio of the permittivity of a material to the permittivity of free space.

🎯 Exam Tip: Relative permittivity (or dielectric constant) indicates how much an electric field is reduced within a material compared to a vacuum. It is a unitless ratio and plays a crucial role in capacitance and field calculations within dielectrics.

3. Solve Numerical Examples.

Question 1. Two small spheres 18 cm apart have equal negative charges and repel each other with the force of \(6 \times 10^{-8}\) N. Find the total charge on both spheres.
Solution: Given: \(F = 6 \times 10^{-8}\) N, \(r = 18 \text{ cm} = 18 \times 10^{-2}\) m To find: Total charge (\(q_1 + q_2\)) Formula: \(F = \frac{1}{4\pi\epsilon_0} \frac{q_1q_2}{r^2}\) Calculation: From formula,
\(F = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2}\) ....(Given: \(q_1 = q_2 = q\))
\(q^2 = \frac{F r^2}{\left(\frac{1}{4\pi\epsilon_0}\right)}\)
\(q^2 = \frac{6 \times 10^{-8} \times (18 \times 10^{-2})^2}{9 \times 10^9}\)
\(q^2 = \frac{6 \times 10^{-8} \times 324 \times 10^{-4}}{9 \times 10^9}\)
\(q^2 = \frac{1944 \times 10^{-12}}{9 \times 10^9}\)
\(q^2 = 216 \times 10^{-21}\)
\(q^2 = 21.6 \times 10^{-20}\) Taking square roots from log table,
\(q = -4.648 \times 10^{-10}\) C ....(. the charges are negative) Total charge = \(q_1 + q_2 = 2q\)
\(= 2 \times (-4.648) \times 10^{-10}\)
\(= -9.296 \times 10^{-10}\) C
In simple words: Using Coulomb's Law, we first calculated the magnitude of the equal negative charges on each sphere from the given force and separation distance. Then, since both spheres have identical charges, the total charge is simply twice the charge on one sphere.

🎯 Exam Tip: Always use consistent SI units for calculations. When dealing with equal charges, \(q_1 q_2\) simplifies to \(q^2\). Remember to consider the sign of the charge when stating the total charge.

Question 2. A charge \(+q\) exerts a force of magnitude \(-0.2\) N on another charge \(-2q\). If they are separated by 25.0 cm, determine the value of q.
Answer: Given: \(q_1 = +q\), \(q_2 = -2q\), \(F = -0.2\) N \(r = 25 \text{ cm} = 25 \times 10^{-2}\) m To find: Charge (q) Formula: \(F = \frac{1}{4\pi\epsilon_0} \frac{q_1q_2}{r^2}\) Calculation: From formula,
\(-0.2 = \frac{9 \times 10^9 \times q \times (-2q)}{(25 \times 10^{-2})^2}\)
\(-0.2 = \frac{-18 \times 10^9 \times q^2}{625 \times 10^{-4}}\)
\(q^2 = \frac{-0.2 \times (25 \times 10^{-2})^2}{-18 \times 10^9}\)
\(q^2 = \frac{0.2 \times (25 \times 10^{-2})^2}{18 \times 10^9}\)
\(q^2 = \frac{0.2 \times 625 \times 10^{-4}}{18 \times 10^9}\)
\(q^2 = \frac{125 \times 10^{-4}}{18 \times 10^9}\)
\(q^2 = \frac{125}{18} \times 10^{-13}\)
\(q^2 = 6.944 \times 10^{-13}\)
\(q^2 = 69.44 \times 10^{-14}\)
\(q = \sqrt{69.44 \times 10^{-14}}\)
\(q \approx 8.33 \times 10^{-7}\) C
The OCR provided calculation:
\(q^2 = \frac{0.2 \times (25 \times 10^{-2})^2}{9 \times 10^9 \times 2}\)
\(q^2 = \frac{2 \times 10^{-1} \times 25 \times 25 \times 10^{-4}}{9 \times 10^9 \times 2}\)
\(q^2 = \frac{25 \times 25 \times 10^{-14}}{9}\)
\(q = \sqrt{\frac{25 \times 25 \times 10^{-14}}{9}}\)
\(q = \frac{5 \times 5 \times 10^{-7}}{3}\)
\(q = \frac{25 \times 10^{-7}}{3}\)
\(q = 8.333 \times 10^{-7}\) C
\( = 0.833 \times 10^{-6}\) C = \(0.833 \text{ µC}\) [Note: The answer given above is calculated in accordance with textual method considering the given data]
In simple words: By applying Coulomb's Law and substituting the given force, separation distance, and the expressions for the charges, we can solve for the unknown charge 'q' through algebraic rearrangement and calculation.

🎯 Exam Tip: Pay close attention to the signs of charges and force (repulsion is positive, attraction is negative). Ensure correct exponent handling during calculations, especially with squares and square roots.

Question 3. Four charges of \(+6 \times 10^{-8}\) C each are placed at the corners of a square whose sides are 3 cm each. Calculate the resultant force on each charge and show in direction on a diagram drawn to scale.
Answer: Given: \(q_A = q_B = q_C = q_D = 6 \times 10^{-8}\) C, \(a = 3\text{ cm}\)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वर्ग के कोनों पर स्थित चार समान आवेशों (प्रत्येक +q) को दर्शाता है, जहाँ केंद्र में रखे आवेश पर लगने वाले परिणामी बल की गणना की गई है। इसमें \(F_{AD}\), \(F_{AB}\) और \(F_{AC}\) जैसे विभिन्न बलों की दिशा और परिमाण भी प्रदर्शित हैं। Magnitude of force on A due to D is,
\(F_{AD} = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r_{AD}^2}\)
\(F_{AD} = \frac{9 \times 10^9 \times (6 \times 10^{-8})^2}{(3 \times 10^{-2})^2}\)
\(F_{AD} = \frac{9 \times 10^9 \times 36 \times 10^{-16}}{9 \times 10^{-4}}\)
\(F_{AD} = 36 \times 10^9 \times 10^{-16} \times 10^4\)
\(F_{AD} = 36 \times 10^{-3}\) N
\(F_{AD} = 3.6 \times 10^{-2}\) N Similarly,
\(F_{AB} = 3.6 \times 10^{-2}\) N
\(F_{AC} = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r_{AC}^2}\) Here \(r_{AC}\) is the diagonal of the square, \(r_{AC} = a\sqrt{2} = 3\sqrt{2}\text{ cm} = 3\sqrt{2} \times 10^{-2}\text{ m}\).
\(F_{AC} = \frac{9 \times 10^9 \times (6 \times 10^{-8})^2}{(3\sqrt{2} \times 10^{-2})^2}\)
\(F_{AC} = \frac{9 \times 10^9 \times 36 \times 10^{-16}}{18 \times 10^{-4}}\)
\(F_{AC} = \frac{36}{2} \times 10^9 \times 10^{-16} \times 10^4\)
\(F_{AC} = 18 \times 10^{-3}\) N
\(F_{AC} = 1.8 \times 10^{-2}\) N
Resultant force on 'A' due to B and D (components along diagonal): The forces \(F_{AD}\) and \(F_{AB}\) are equal in magnitude and perpendicular. Their resultant \(F_{BD}\) will be along the diagonal AC. \(F_{BD\_resultant} = \sqrt{F_{AD}^2 + F_{AB}^2} = \sqrt{(3.6 \times 10^{-2})^2 + (3.6 \times 10^{-2})^2}\) \(F_{BD\_resultant} = \sqrt{2 \times (3.6 \times 10^{-2})^2} = 3.6 \times 10^{-2} \times \sqrt{2}\) \(F_{BD\_resultant} \approx 3.6 \times 10^{-2} \times 1.414 = 5.09 \times 10^{-2}\) N This force \(F_{BD\_resultant}\) is directed along the diagonal AC, same direction as \(F_{AC}\). So, the total resultant force on A is the sum of \(F_{BD\_resultant}\) and \(F_{AC}\).
Resultant force on 'A'
\(= F_{AD} \cos 45^\circ + F_{AB} \cos 45^\circ + F_{AC}\) (This line from OCR is for summing components along one diagonal, if we project \(F_{AD}\) and \(F_{AB}\) onto AC.)
\(= \left(3.6 \times 10^{-2} \times \frac{1}{\sqrt{2}}\right) + \left(3.6 \times 10^{-2} \times \frac{1}{\sqrt{2}}\right) + 1.8 \times 10^{-2}\)
\(= 2 \times \left(3.6 \times 10^{-2} \times \frac{1}{\sqrt{2}}\right) + 1.8 \times 10^{-2}\)
\(= \sqrt{2} \times 3.6 \times 10^{-2} + 1.8 \times 10^{-2}\)
\(= 1.414 \times 3.6 \times 10^{-2} + 1.8 \times 10^{-2}\)
\(= 5.0904 \times 10^{-2} + 1.8 \times 10^{-2}\)
\(= 6.8904 \times 10^{-2}\) N
\(= 6.89 \times 10^{-2}\) N directed along \( \vec{F}_{AC} \) (direction along the diagonal from A to C)
In simple words: To find the resultant force on one corner charge, we calculate the individual forces from the other three charges using Coulomb's law, resolving them into components, and then vectorially adding them. Due to symmetry, the net force on any corner charge will be directed along the diagonal away from that corner.

🎯 Exam Tip: For charges arranged symmetrically, use vector addition. Break forces into components (e.g., x and y or along diagonals) to simplify calculations. Remember that the diagonal of a square with side 'a' is \(a\sqrt{2}\).

Question 4. The electric field in a region is given by \( \vec{E} = 5.0 \hat{k} \text{ N/C} \) Calculate the electric flux through a square of side 10.0 cm in the following cases i. The square is along the XY plane ii. The square is along XZ plane iii. The normal to the square makes an angle of 45° with the Z axis.
Answer: Given: \( \vec{E} = 5.0 \hat{k} \text{ N/C} \), \( |\vec{E}| = 5\text{ N/C} \) \(l = 10 \text{ cm} = 10 \times 10^{-2}\text{ m} = 10^{-1}\text{ m}\) \(A = l^2 = (10^{-1}\text{ m})^2 = 10^{-2}\text{m}^2\) To find: Electric flux in three cases. \((\phi_1)\) \((\phi_2)\) \((\phi_3)\) Formula: \( \phi = EA \cos \theta \) Calculation: Case I: When square is along the XY plane, The area vector (\(\vec{A}\)) for a surface in the XY plane is along the Z-axis, i.e., \( \hat{k} \). Electric field \( \vec{E} = 5.0 \hat{k} \text{ N/C} \). Therefore, the angle \( \theta \) between \( \vec{E} \) and \( \vec{A} \) is \(0^\circ\).
\( \theta = 0^\circ \)
\( \phi_1 = EA \cos 0^\circ \)
\( \phi_1 = 5 \times 10^{-2} \times 1 \)
\( = 5 \times 10^{-2}\) Vm Case II: When square is along XZ plane, The area vector (\(\vec{A}\)) for a surface in the XZ plane is along the Y-axis, i.e., \( \hat{j} \). Electric field \( \vec{E} = 5.0 \hat{k} \text{ N/C} \). Therefore, the angle \( \theta \) between \( \vec{E} \) and \( \vec{A} \) is \(90^\circ\).
\( \theta = 90^\circ \)
\( \phi_1 = EA \cos 90^\circ = 5 \times 10^{-2} \times 0 = 0\text{ Vm} \) Case III: When normal to the square makes an angle of 45° with the Z axis. The angle \( \theta \) between the area vector (\(\vec{A}\)) and the Z-axis (which is the direction of \( \vec{E} \)) is \(45^\circ\).
\( \theta = 45^\circ \)
\( \phi_3 = EA \cos 45^\circ \)
\( \phi_3 = 5 \times 10^{-2} \times \frac{1}{\sqrt{2}} \)
\( = \frac{5}{\sqrt{2}} \times 10^{-2} \)
\( = 3.535 \times 10^{-2}\) Vm
In simple words: Electric flux is calculated as the product of the electric field strength, the area of the surface, and the cosine of the angle between the electric field direction and the area vector (which is normal to the surface). The flux is maximum when the field is perpendicular to the surface (parallel to the normal) and zero when the field is parallel to the surface (perpendicular to the normal).

🎯 Exam Tip: Remember that the area vector is always perpendicular to the surface. Carefully determine the angle \( \theta \) between the electric field vector (\( \vec{E} \)) and the area vector (\( \vec{A} \)) for each case. \( \phi = \vec{E} \cdot \vec{A} \).

Question 5. Three equal charges of \(10 \times 10^{-8}\) C respectively, each located at the corners of a right triangle whose sides are 15 cm, 20 cm and 25 cm respectively. Find the force exerted on the charge located at the 90° angle.
Answer: Given: \(q_A = q_B = q_C = 10 \times 10^{-8}\) C Sides of the right triangle: let \(r_{BC} = 15\text{ cm}\), \(r_{BA} = 20\text{ cm}\), and \(r_{CA} = 25\text{ cm}\) (hypotenuse). The charge at the 90° angle is \(q_B\).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक समकोण त्रिभुज के कोनों पर रखे तीन समान आवेशों को दर्शाता है, जिनकी भुजाएँ 15 सेमी, 20 सेमी और 25 सेमी हैं। इसमें 90° कोण पर स्थित आवेश पर लगने वाले परिणामी बल की गणना करने के लिए \(F_{BC}\) और \(F_{BA}\) जैसे बल सदिशों की दिशा दिखाई गई है। Force on B due to A,
\( \vec{F}_{BA} = \frac{1}{4\pi\epsilon_0} \frac{q_A q_B}{(r_{BA})^2} \) (vector direction along BA)
\( \vec{F}_{BA} = 9 \times 10^9 \times \frac{(10 \times 10^{-8})^2}{(20 \times 10^{-2})^2} \)
\( \vec{F}_{BA} = 9 \times 10^9 \times \frac{100 \times 10^{-16}}{400 \times 10^{-4}} \)
\( \vec{F}_{BA} = 9 \times 10^9 \times \frac{1}{4} \times 10^{-12} \)
\( \vec{F}_{BA} = 2.25 \times 10^{-3} \) N (The OCR initially showed 10^3 N, but subsequent calculation implies 10^-3 N, which is consistent with the magnitude.) Force on B due to C,
\( \vec{F}_{BC} = \frac{1}{4\pi\epsilon_0} \frac{q_C q_B}{(r_{BC})^2} \) (vector direction along BC)
\( \vec{F}_{BC} = 9 \times 10^9 \times \frac{(10 \times 10^{-8})^2}{(15 \times 10^{-2})^2} \)
\( \vec{F}_{BC} = 9 \times 10^9 \times \frac{100 \times 10^{-16}}{225 \times 10^{-4}} \)
\( \vec{F}_{BC} = 9 \times 10^9 \times \frac{1}{2.25} \times 10^{-12} \)
\( \vec{F}_{BC} = 4 \times 10^{-3} \) N (The OCR initially showed 10^3 N, but subsequent calculation implies 10^-3 N.)
Resultant force on point B, Since \( \vec{F}_{BA} \) and \( \vec{F}_{BC} \) are perpendicular (at the 90° angle), their resultant magnitude is given by the Pythagorean theorem.
\( |\vec{F}_B| = \sqrt{F_{BA}^2 + F_{BC}^2 + 2F_{BA}F_{BC} \cos 90^\circ} \)
\( |\vec{F}_B| = \sqrt{F_{BA}^2 + F_{BC}^2} \)
\( = \sqrt{(2.25 \times 10^{-3})^2 + (4 \times 10^{-3})^2} \)
\( = \sqrt{5.0625 \times 10^{-6} + 16 \times 10^{-6}} \)
\( = \sqrt{(5.0625 + 16) \times 10^{-6}} \)
\( = \sqrt{21.0625 \times 10^{-6}} \)
\( = 4.58939 \times 10^{-3} \) N
\( = 4.589 \times 10^{-3} \) N
In simple words: To find the force on the charge at the 90-degree angle, we calculated the individual forces from the other two charges using Coulomb's law. Since these two forces are perpendicular, their resultant is found using the Pythagorean theorem.

🎯 Exam Tip: When forces are perpendicular, vector addition simplifies to finding the hypotenuse of a right-angled triangle. Ensure precise calculation with exponents and correct unit conversions (cm to m).

Question 6. A potential difference of 5000 volt is applied between two parallel plates 5 cm apart. A small oil drop having a charge of \(9.6 \times 10^{-19}\) C falls between the plates. Find (i) electric field intensity between the plates and (ii) the force on the oil drop.
Answer: Given: \(V = 5000\text{ volt}\), \(d = 5\text{ cm} = 5 \times 10^{-2}\text{ m}\) \(q = 9.6 \times 10^{-19}\) C To find: i. Electric field intensity (E) ii. Force (F) Formula: i. \(E = \frac{V}{d}\) ii. \(E = \frac{F}{q}\) or \(F = Eq\) Calculation: From formula (i), \(E = \frac{V}{d} = \frac{5000\text{ V}}{5 \times 10^{-2}\text{ m}} = \frac{5000}{0.05} \text{ V/m} = 100000 \text{ V/m}\)
\(E = 10^5\text{ N/C}\) From formula (ii)
\(F = E \times q\)
\(F = 10^5 \times 9.6 \times 10^{-19}\)
\(F = 9.6 \times 10^{-14}\) N
In simple words: First, the electric field intensity is found by dividing the potential difference by the distance between the plates. Then, the force on the oil drop is calculated by multiplying this electric field intensity by the charge of the oil drop.

🎯 Exam Tip: Remember the two key relationships: \(E = V/d\) (relating electric field to potential difference and distance) and \(F = Eq\) (relating force on a charge to electric field). Always ensure units are consistent for calculation.

Question 7. Calculate the electric field due to a charge of \(-8.0 \times 10^{-8}\) C at a distance of 5.0 cm from it.
Answer: Given: \(q = -8.0 \times 10^{-8}\) C, \(r = 5\text{ cm} = 5 \times 10^{-2}\text{ m}\) To Find: Electric field (E) Formula: \(E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}\) Calculation: From formula,
\(E = 9 \times 10^9 \times \frac{(-8 \times 10^{-8})}{(5 \times 10^{-2})^2}\)
\(E = 9 \times 10^9 \times \frac{-8 \times 10^{-8}}{25 \times 10^{-4}}\)
\(E = \frac{-72 \times 10^1}{25 \times 10^{-4}}\)
\(E = \frac{-720}{25} \times 10^4\)
\(E = -28.8 \times 10^4\)
\(E = -2.88 \times 10^5\text{ N/C}\)
In simple words: The electric field produced by a point charge is calculated using Coulomb's law for fields, where you divide the product of Coulomb's constant and the charge by the square of the distance from the charge. The negative sign indicates the direction of the field points towards the negative source charge.

🎯 Exam Tip: For electric field due to a point charge, \(E = k|q|/r^2\). The direction is away from positive charge and towards negative charge. Include the sign in your final answer if it's a vector component, or specify direction for scalar magnitude.

Can You Recall? (Textbookpage No. 188)

Question 1. Have you experienced a shock while getting up from a plastic chair and shaking hand with your friend?
Answer: Yes, sometimes a shock while getting up from a plastic chair and shaking hand with friend is experienced.
In simple words: Yes, this phenomenon is common; it's due to static electricity built up by friction between your clothes and the plastic chair, which then discharges when you touch another person.

🎯 Exam Tip: This illustrates triboelectric charging, where friction causes electron transfer, leading to static electricity and subsequent discharge (shock).

Question 2. Ever heard a crackling sound while taking out your sweater in winter?
Answer: Yes, sometimes while removing our sweater in winter, some crackling sound is heard and the sweater appears to stick to body.
In simple words: Yes, a crackling sound and clinging sweater in winter is common, caused by static electricity generated from friction between the sweater and your body, leading to small electrical discharges.

🎯 Exam Tip: This is another example of static electricity due to triboelectric effect. The crackling sound comes from the rapid neutralization of charge in the air.

Question 3. Have you seen the lightning striking during pre-monsoon weather?
Answer: Yes, sometimes lightning striking during pre-pre-monsoon weather seen.
In simple words: Yes, lightning is a frequent observation during pre-monsoon weather, resulting from the massive discharge of static electricity built up in storm clouds.

🎯 Exam Tip: Lightning is a large-scale natural phenomenon of electrostatic discharge, demonstrating the principles of charge separation and electrical breakdown in the atmosphere.

Can You Tell? (Textbook Page No. 189)

Question. i. When a petrol or a diesel tanker is emptied in a tank, it is grounded. ii. A thick chain hangs from a petrol or a diesel tanker and it is in contact with ground when the tanker is moving.
Answer: i. When a petrol or a diesel tanker is emptied in a tank, it is grounded so that it has an electrically conductive connection from the petrol or diesel tank to ground (Earth) to allow leakage of static and electrical charges. ii. Metallic bodies of cars, trucks or any other big vehicles get charged because of friction between them and the air rushing past them. Hence, a thick chain is hanged from a petrol or a diesel tanker to make a contact with ground so that charge produced can leak to the ground through chain.
In simple words: Petrol/diesel tankers are grounded or trail chains to prevent static charge buildup from friction with air or fuel flow, ensuring any accumulated charge safely leaks to the Earth, thus avoiding hazardous sparks that could ignite flammable materials.

🎯 Exam Tip: Grounding and trailing chains are safety measures to prevent electrostatic discharge. They provide a path for excess static charge to dissipate into the Earth, crucial when handling flammable liquids.

Can You Tell? (Textbook Page No. 194)

Question. Three charges, q each, are placed at the vertices of an equilateral triangle. What will be the resultant force on charge q placed at the centroid of the triangle?
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक समबाहु त्रिभुज के शीर्षों पर रखे तीन समान आवेशों (प्रत्येक q) और त्रिभुज के केंद्रक (O) पर रखे एक और आवेश q को दर्शाता है। इसमें केंद्रक पर लगने वाले बलों \(F_{OA}\), \(F_{OB}\) और \(F_{OC}\) की दिशाएँ प्रदर्शित हैं, जिनका परिणामी बल ज्ञात किया जाना है। Since AD. BE and CF meets at O, as centroid of an equilateral triangle.
\(OA = OB = OC\) Let, \(r = OA = OB = OC\) Force acting on point O due to charge on point A,
\( \vec{F}_{OA} = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2} \hat{r}_{AO} \) (Directed away from A) Force acting on point O due to charge on point B,
\( \vec{F}_{OB} = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2} \hat{r}_{BO} \) (Directed away from B) Force acting on point O due to charge on point C,
\( \vec{F}_{OC} = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2} \hat{r}_{CO} \) (Directed away from C) Since the charges at A, B, C are equal and the distances OA, OB, OC are equal, the magnitudes of the forces \( |\vec{F}_{OA}| \), \( |\vec{F}_{OB}| \), \( |\vec{F}_{OC}| \) are equal. Let this magnitude be F. These three forces are acting at the centroid O, making angles of 120° with each other.
Resultant force acting on point O,
\( \vec{F} = \vec{F}_{OA} + \vec{F}_{OB} + \vec{F}_{OC} \) On resolving \( \vec{F}_{OB} \) and \( \vec{F}_{OC} \), the resultant of \( \vec{F}_{OB} \) and \( \vec{F}_{OC} \) (which are 120° apart and equal in magnitude F) will be F, directed exactly opposite to \( \vec{F}_{OA} \). i.e., \( \vec{F}_{OB} + \vec{F}_{OC} = -\vec{F}_{OA} \)
\( \implies \vec{F} = \vec{F}_{OA} - \vec{F}_{OA} = 0 \) Hence, the resultant force on the charge placed at the centroid of the equilateral triangle is zero.
In simple words: When three equal charges are placed at the vertices of an equilateral triangle and an identical charge is at its centroid, the forces exerted by the vertex charges on the centroid charge are equal in magnitude and symmetrically oriented. These forces sum up to zero due to vector cancellation.

🎯 Exam Tip: For symmetric charge distributions, the net force at the center is often zero. Recognize that three equal forces at 120° angles to each other will always result in a zero net force.

Can You Tell? (Textbook Page No. 197)

Question. Why a small voltage can produce a reasonably large electric field?
Answer: 1. Electric field produced depends upon voltage as well as separation distance. 2. Electric field varies linearly with voltage and inversely with distance. 3. Hence, even if voltage is small, it can produce a reasonable large electric field when the gap between the electrode is reduced significantly.
In simple words: A small voltage can generate a large electric field if the distance between the conducting plates or electrodes is made very small, as the electric field strength is inversely proportional to this separation distance (\(E = V/d\)).

🎯 Exam Tip: The formula \(E = V/d\) highlights this: a small 'd' (distance) can significantly amplify 'E' (electric field) even with a modest 'V' (voltage). This principle is vital in components like capacitors.

Can You Tell? (Textbook Page No. 198)

Question. Lines of force are imaginary; can they have any practical use?
Answer: Yes, electric lines of force help us to visualise the nature of electric field in a region.
In simple words: Despite being imaginary, electric field lines are highly useful conceptual tools because they provide a visual representation of the direction and strength of the electric field in a region, aiding in understanding complex field patterns.

🎯 Exam Tip: Field lines are invaluable for qualitative understanding: their density indicates field strength, and their direction shows the force a positive test charge would experience. They are a powerful analytical tool.

Can You Tell? (Textbook Page No. 204)

Question. The surface charge density of Earth is \( \sigma = -1.33 \text{ nC/m}^2 \). That is about \(8.3 \times 10^9\) electrons per square metre. If that is the case why don't we feel it?
Answer: The Earth along with its atmosphere acts as a neutral system. The atmosphere (ionosphere in particular) has nearly equal and opposite charge. As a result, there exists a mechanism to replenish electric charges in the form of continual thunder storms and lightning that occurs in different parts of the globe. This makes average charge on surface of the Earth as zero at any given time instant. Hence, we do not feel it.
In simple words: Despite having a momentary surface charge density, the Earth-atmosphere system is electrically neutral on average due to continuous charge replenishment by lightning and thunderstorms, thus preventing us from feeling a persistent electric field.

🎯 Exam Tip: The Earth's overall electrical neutrality is maintained through dynamic processes like atmospheric convection and lightning, balancing temporary charge accumulations and preventing static shocks from daily contact.