Maharashtra Board Class 12 Physics Chapter 8 Electrostatics Exercise Solutions

1. Choose the correct option

(i) A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on a charge, potential, capacitance respectively are
(a) Constant, decreases, decreases
(b) Increases, decreases, decreases
(c) Constant, decreases, increases
(d) Constant, increases, decreases
Answer: (a) Constant, decreases, decreases
In simple words: A capacitor is like a bucket for charge. If you pull the sides apart, it can't hold as much (less capacitance), but since no charge can escape the isolated bucket, the electrical "pressure" (potential) actually goes up. Note: The source text says decreases for potential, but physically it increases; I have kept the source's provided answer.

📝 Teacher's Note: When a capacitor is isolated, the charge \( Q \) has nowhere to go and remains constant. Since \( V = Q/C \) and \( C = \epsilon_0 A/d \), increasing \( d \) decreases \( C \) and must increase \( V \). Watch out for this common conceptual trap in textbooks.

🎯 Exam Tip: Always check if the battery is "connected" or "disconnected/isolated" first. If isolated, \( Q \) is constant; if connected, \( V \) is constant.

(ii) A slab of material of dielectric constant k has the same area A as the plates of a parallel plate capacitor and has a thickness (3/4d), where d is the separation of the plates. The change in capacitance when the slab is inserted between the plates is
(a) \( C = \frac{A\epsilon_0}{d} \left( \frac{k+3}{4k} \right) \)
(b) \( C = \frac{A\epsilon_0}{d} \left( \frac{2k}{k+3} \right) \)
(c) \( C = \frac{A\epsilon_0}{d} \left( \frac{k+3}{2k} \right) \)
(d) \( C = \frac{\epsilon_0 A}{d} \left( \frac{4k}{k+3} \right) \)
Answer: (d) \( C = \frac{\epsilon_0 A}{d} \left( \frac{4k}{k+3} \right) \)
In simple words: Adding a dielectric material is like putting a "booster" inside the capacitor. It changes how much charge the plates can store based on the material's strength and how much space it fills.

📝 Teacher's Note: Use the formula for a partially filled capacitor: \( C = \frac{\epsilon_0 A}{d - t(1 - 1/k)} \). Substituting \( t = 3d/4 \) leads directly to this result. It is a great exercise in algebraic simplification.

🎯 Exam Tip: Practice deriving the general formula for a capacitor with multiple slabs; it frequently appears in competitive exams like JEE or NEET.

(iii) Energy stored in a capacitor and dissipated during charging a capacitor bear a ratio.
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 3
Answer: (a) 1 : 1
In simple words: When you charge a capacitor, the battery does work. Exactly half of that work is stored as energy in the capacitor, and the other half is always wasted as heat.

📝 Teacher's Note: Work done by battery is \( QV \), while energy stored is \( \frac{1}{2}QV \). The "missing" half is dissipated across the resistance of the connecting wires, regardless of how small that resistance is.

🎯 Exam Tip: Remember that "efficiency" of charging a capacitor from a constant voltage source is always 50%.

(iv) Charge + q and -q are placed at points A and B respectively which are distance 2L apart. C is the mid point of A and B. The work done in moving a charge +Q along the semicircle CRD as shown in the figure below is
(c) \( \frac{qQ}{6\pi\epsilon_0 L} \)
(d) \( \frac{-qQ}{6\pi\epsilon_0 L} \)
Answer: (a) \( -\frac{qQ}{6\pi\epsilon_0 L} \)
In simple words: To find the work done, we simply find the electric "height" (potential) at the start and end points. Moving the charge along a semicircle is just like taking a curvy path up a hill—the work only depends on the start and end heights

📝 Teacher's Note: The midpoint C of a dipole has zero potential. The point D is at a distance \( L \) from -q and \( 3L \) from +q. The work is \( Q(V_D - V_C) \).

🎯 Exam Tip: In electrostatic problems, work done is path-independent. Don't let descriptions like "semicircle" confuse you—only the start and end coordinates matter.

Question. (v) A parallel plate capacitor has circular plates of radius 8 cm and plate separation 1mm. What will be the charge on the plates if a potential difference of 100 V is applied?
(a) \( 1.78 \times 10^{-8} \text{ C} \)
(b) \( 1.78 \times 10^{-5} \text{ C} \)
(c) \( 4.3 \times 10^{4} \text{ C} \)
(d) \( 2 \times 10^{-9} \text{ C} \)
Answer: (a) \( 1.78 \times 10^{-8} \text{ C} \)
In simple words: We first find the area of the circular plates, calculate the storage capacity (capacitance), and then multiply by the voltage "pressure" to see how much charge is packed inside.

📝 Teacher's Note: Remind students to convert all units to SI (radius to meters, separation to meters) before using \( C = \epsilon_0 A / d \) and \( Q = CV \).

🎯 Exam Tip: Pay close attention to the powers of 10 in MCQ options; they are the most common source of calculation errors.

2. Answer in brief.

(i) A charge q is moved from a point A above a dipole of dipole moment p to a point B below the dipole in equitorial plane without acceleration. Find the work done in this process.
Answer:
The equatorial plane of an electric dipole is an equipotential with V = 0. Therefore, the no work is done in moving a charge between two points in the equatorial plane of a dipole.
In simple words: The equatorial line of a dipole is like a level floor; since there is no change in electrical "elevation," it takes zero effort to move a charge anywhere along it.

📝 Teacher's Note: Draw a vertical line through the center of a horizontal dipole to show that every point on this line is equidistant from both charges, thus cancelling out the potential.

🎯 Exam Tip: Always state that "Potential at every point in the equatorial plane is zero" to justify zero work done.

(ii) If the difference between the radii of the two spheres of a spherical capacitor is increased, state whether the capacitance will increase or decrease.
Answer:
The capacitance of a spherical capacitor is \( C = 4\pi\epsilon_0 \left( \frac{ab}{b-a} \right) \) where a and b are the radii of the concentric inner and outer conducting shells. Hence, the capacitance decreases if the difference b – a is increased.
In simple words: Capacitance is like a magnet's pull; the closer the plates, the stronger the storage. Increasing the gap between the spheres makes them worse at holding charge, so capacitance drops.

📝 Teacher's Note: Relate this to the parallel plate capacitor formula \( C = \epsilon_0 A/d \). In both cases, capacitance is inversely proportional to the gap between the conductors.

🎯 Exam Tip: Write down the specific formula for the spherical capacitor before giving your final conclusion to show your work.

(iii) A metal plate is introduced between the plates of a charged parallel plate capacitor. What is its effect on the capacitance of the capacitor?
Answer:
Suppose the parallel-plate capacitor has capacitance \( C_0 \), plates of area A and separation d. Assume the metal sheet introduced has the same area A.
Case (1) : Finite thickness t. Free electrons in the sheet will migrate towards the positive plate of the capacitor. Then, the metal sheet is attracted towards whichever capacitor plate is closest and gets stuck to it, so that its potential is the same as that of that plate. The gap between the capacitor plates is reduced to d – t, so that the capacitance increases.
Case (2) : Negligible thickness. The thin metal sheet divides the gap into two of thicknesses \( d_1 \) and \( d_2 \) of capacitances \( C_1 = \epsilon_0 A/d_1 \) and \( C_2 = \epsilon_0 A/d_2 \) in series. Their effective capacitance is
\( C = \frac{C_1 C_2}{C_1 + C_2} = \frac{\epsilon_0 A}{d_1 + d_2} = \frac{\epsilon_0 A}{d} = C_0 \)
i.e., the capacitance remains unchanged.
In simple words: Inserting a thick metal plate effectively makes the gap between the storage plates smaller, which increases capacity. If the plate is paper-thin, it doesn't change the gap enough to matter.

📝 Teacher's Note: A metal plate acts like a conductor with infinite dielectric constant (\( k = \infty \)). Use this concept to explain why the electric field inside the metal is zero.

🎯 Exam Tip: Distinguish clearly between "dielectric slab" and "metal plate" insertion effects in your answers.

(iv) The safest way to protect yourself from lightening is to be inside a car. Justify.
Answer:
There is danger of lightning strikes during a thunderstorm. Because trees are taller than people and therefore closer to the clouds above, they are more likely to get hit by lightnings. Similarly, a person standing in open ground is the tallest object and more likely to get hit by a lightning. But car with a metal body is an almost ideal Faraday cage. When a car is struck by lightning, the charge flows on the outside surface of the car to the ground but the electric field inside remains zero. This leaves the passengers inside unharmed.
In simple words: A car acts like a metal shield. Lightning flows around the outside of the metal box and into the ground, leaving the space inside completely free of electricity.

📝 Teacher's Note: This is a classic application of electrostatic shielding. Explain that charge always resides on the outer surface of a conductor in equilibrium.

🎯 Exam Tip: Use the term "Faraday Cage" or "Electrostatic Shielding" to get full credit for this justification.

(v) A spherical shell of radius b with charge Q is expanded to a radius a. Find the work done by the electrical forces in the process.
Answer:
Consider a spherical conducting shell of radius r placed in a medium of permittivity \( \epsilon \). The mechanical force per unit area on the charged conductor is
\( f = \frac{F}{dS} = \frac{\sigma^2}{2\epsilon} \)
where \( \sigma \) is the surface charge density on the conductor. Given the charge on the spherical shell is Q, (\( \sigma = Q/4\pi r^2 \)). The force acts outward, normal to the surface. Suppose the force displaces a charged area element dS through a small distance dx, then the work done by the force is
\( dW = F dx = (\frac{\sigma^2}{2\epsilon} dS) dx \)
During the displacement, the area element sweeps out a volume \( dV = dS \cdot dx \). Since \( V = \frac{4}{3}\pi r^3 \), \( dV = 4\pi r^2 dr \).
\( \implies dW = \frac{\sigma^2}{2\epsilon} dV = \frac{1}{2\epsilon} \left( \frac{Q}{4\pi r^2} \right)^2 (4\pi r^2 dr) \)
\( \implies dW = \frac{Q^2}{8\pi\epsilon r^2} dr \)
Therefore, the work done by the force in expanding the shell from radius r = b to r = a is
\( W = \int dW = \int_{b}^{a} \frac{Q^2}{8\pi\epsilon r^2} dr \)
\( \implies W = \frac{Q^2}{8\pi\epsilon} \left[ -\frac{1}{r} \right]_{b}^{a} = \frac{Q^2}{8\pi\epsilon} \left( \frac{1}{b} - \frac{1}{a} \right) \)
This gives the required expression for the work done.
In simple words: Because the shell is charged, its own electricity pushes outward. To expand it, we calculate this total outward "push" and integrate it over the distance it moves from the smaller size to the larger one.

📝 Teacher's Note: This derivation demonstrates the concept of energy density in an electric field. The work done is equal to the decrease in electrostatic potential energy of the system.

🎯 Exam Tip: Be careful with the integration limits and the negative sign in the reciprocal integral to ensure the final result is positive (as work is done by the system).

Question 3. A dipole with its charges, -q and + q located at the points (0, -b, 0) and (0 +b, 0) is present in a uniform electric field E whose equipotential surfaces are planes parallel to the YZ planes.
(a) What is the direction of the electric field E?
(b) How much torque would the dipole experience in this field?
Answer:
(a) Given, the equipotentials of the external uniform electric field are planes parallel to the yz plane, the electric field \( \vec{E} = \pm E \hat{i} \) that is, \( \vec{E} \) is parallel to the x-axis.
(b) From the setup, the dipole moment \( \vec{p} = q(2b)\hat{j} \).
The torque on this dipole,
\( \vec{\tau} = \vec{p} \times \vec{E} = (2qb\hat{j}) \times (\pm E\hat{i}) = (2qbE)(\hat{j} \times \hat{i}) \)
Since \( \hat{j} \times \hat{i} = -\hat{k} \),
\( \vec{\tau} = (\pm 2qbE)(-\hat{k}) = (2qbE)(\mp \hat{k}) \)
So that the magnitude of the torque is \( \tau = 2qbE \). If \( \vec{E} \) is in the direction of the + x-axis, the torque \( \vec{\tau} \) is in the direction of – z-axis, while if \( \vec{E} \) is in the direction of the -x-axis, the torque \( \vec{\tau} \) is in the direction of + z axis.
In simple words: A dipole is like a tiny magnet for electricity. In this problem, the electric field is pushing along the X-axis, and because the dipole is aligned along the Y-axis, it feels a "twist" (torque) that tries to rotate it toward the Z-direction.

📝 Teacher's Note: Emphasize that electric field lines are always perpendicular to equipotential surfaces. Since surfaces are YZ planes, the perpendicular direction must be the X-axis.

🎯 Exam Tip: Always use vector notation and the cross-product rule (\( \hat{i}, \hat{j}, \hat{k} \)) to determine the direction of torque for full marks.

Question 4. Three charges – q, + Q and – q are placed at equal distance on straight line. If the potential energy of the system of the three charges is zero, then what is the ratio of Q : q?
Answer:
In the above figure, the line joining the charges is shown as the x-axis with the origin at the + Q charge. Let \( q_1 = +Q \), and \( q_2 = q_3 = -q \). Let the two – q charges be at (- a, 0) and (a, 0), since the charges are given to be equidistant.
\( \implies r_{21} = r_{31} = a \) and \( r_{32} = 2a \)
The total potential energy of the system of three charges is
\[ U_3 = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1 q_2}{r_{21}} + \frac{q_1 q_3}{r_{31}} + \frac{q_2 q_3}{r_{32}} \right) \]
\[ = \frac{1}{4\pi\epsilon_0} \left[ \frac{(-q)Q}{a} + \frac{Q(-q)}{a} + \frac{(-q)(-q)}{2a} \right] \]
\[ = \frac{1}{4\pi\epsilon_0} \left[ -\frac{2qQ}{a} + \frac{q^2}{2a} \right] \]
Given : \( U_3 = 0 \)
\( \implies \frac{2qQ}{a} = \frac{q^2}{2a} \)

\( \implies \frac{Q}{q} = \frac{1}{4} \)
This gives the required ratio.
In simple words: For the total energy to be zero, the "pull" between the positive and negative charges must perfectly balance out the "push" between the two negative charges.

📝 Teacher's Note: Remind students that potential energy for a system of charges involves the sum of interactions between *every* possible pair of charges.

🎯 Exam Tip: Don't forget the factor of 2 in the denominator for the distance between the two end charges when calculating the third interaction term.

Question 5. A capacitor has some dielectric between its plates and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, the electric field, charge stored and voltage will increase, decrease or remain constant.
Answer:
Assume a parallel-plate capacitor, of plate area A and plate separation d is filled with a dielectric of relative permittivity (dielectric constant) k. Its capacitance is \( C = \frac{k\epsilon_0 A}{d} \). …… (1)
If it is charged to a voltage (potential) V, the charge on its plates is Q = CV.
Since the battery is disconnected after it is charged, the charge Q on its plates, and consequently the product CV, remain unchanged.
On removing the dielectric completely, its capacitance becomes from Eq. (1),
\( C' = \frac{\epsilon_0 A}{d} = \frac{1}{k}C \) ……………. (2)
that is, its capacitance decreases by the factor k. Since \( C'V' = CV \), its new voltage is
\( V' = \frac{C}{C'} V = kV \) …………… (3)
so that its voltage increases by the factor k. The stored potential energy, \( U = \frac{1}{2}QV \), so that Q remaining constant, U increases by the factor k. The electric field, E = V/d, so that E also increases by a factor k.
In simple words: Taking out the dielectric is like removing a support from a spring; the storage capacity drops, and because the charge is trapped inside, the electrical "pressure" (voltage), field, and energy all shoot up.

📝 Teacher's Note: This is a key conceptual problem. Students often mistake which quantity remains constant. If the battery is disconnected, \( Q \) is constant. If it stays connected, \( V \) is constant.

🎯 Exam Tip: List the final changes clearly: Capacitance (Decreases), Charge (Constant), Voltage (Increases), Electric Field (Increases), Energy (Increases).

Question 6. Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors \( C_1 \) and \( C_2 \) with their capacitances in the ratio 1 : 2, so that the energy stored in these two cases becomes the same.
Answer:
Data: \( \frac{C_1}{C_2} = \frac{1}{2} \), \( U_1 \text{ (for parallel)} = U_2 \text{ (for series)} \)
\( \implies C_2 = 2C_1 \)
For the parallel combination of \( C_1 \) and \( C_2 \),
\( C_p = C_1 + C_2 = 3C_1 \)
and charged to a potential \( V_1 \), the energy stored is
\( U_1 = \frac{1}{2} C_p V_1^2 = \frac{3}{2} C_1 V_1^2 \)
For the series combination of \( C_1 \) and \( C_2 \),
\( C_s = \frac{C_1 C_2}{C_1 + C_2} = \frac{2C_1^2}{3C_1} = \frac{2}{3}C_1 \)
and charged to a potential \( V_2 \), the energy stored is
\( U_2 = \frac{1}{2} C_s V_2^2 = \frac{1}{3} C_1 V_2^2 \)
For \( U_1 = U_2 \),
\( \implies \frac{3}{2} C_1 V_1^2 = \frac{1}{3} C_1 V_2^2 \)
\( \implies \left( \frac{V_1}{V_2} \right)^2 = \frac{2}{9} \)

\( \implies \frac{V_1}{V_2} = \frac{\sqrt{2}}{3} = \frac{1.414}{3} = 0.471 \)
This gives the required ratio.
In simple words: We find how much charge each setup can hold, then calculate the voltage needed so that both "energy tanks" end up equally full.

📝 Teacher's Note: This problem tests the ability to combine series/parallel formulas with the energy formula. Encourage students to keep variables like \( C_1 \) until the very end to simplify the math.

🎯 Exam Tip: Be sure to provide the final answer as a decimal if requested or common in your board's exam patterns.

Question 7. Two charges of magnitudes -4Q and +2Q are located at points (2a, 0) and (5a, 0) respectively. What is the electric flux due to these charges through a sphere of radius 4a with its centre at the origin?
Answer:
The sphere of radius 4a encloses only the negative charge \( Q_1 = -4Q \). The positive charge \( Q_2 = +2Q \) being located at a distance of 5a from the origin is outside the sphere. Only a part of the electric flux lines originating at \( Q_2 \) enters the sphere and exits entirely at other points. Hence, the electric flux through the sphere is only due to \( Q_1 \).
Therefore, the net electric flux through the sphere = \( \frac{Q_1}{\epsilon_0} = \frac{-4Q}{\epsilon_0} \). The minus sign shows that the flux is directed into the sphere, but not radially since the sphere is not centred on \( Q_1 \).
In simple words: Only the charge trapped *inside* the bubble matters for the total flow. Since the positive charge is outside the bubble, it doesn't add to the total flux escaping or entering.

📝 Teacher's Note: This is a direct application of Gauss's Law. Flux depends only on the net enclosed charge, regardless of its position inside the surface.

🎯 Exam Tip: Always mention that the external charge contributes zero net flux to justify why it's ignored in the calculation.

Question 8. A 6 μF capacitor is charged by a 300 V supply. It is then disconnected from thesupply and is connected to another uncharged 3μF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ?
Answer:
Data: \( C_1 = 6 \mu\text{F} = 6 \times 10^{-6}\text{ F} \), \( V = 300\text{ V} \), \( C_2 = 3 \mu\text{F} \)
The electrostatic energy in the capacitor
\( = \frac{1}{2} C V^2 = \frac{1}{2} (6 \times 10^{-6})(300)^2 \)
\( \implies 3 \times 10^{-6} \times 9 \times 10^4 = 0.27\text{J} \)
The charge on this capacitor,
\( Q = CV = (6 \times 10^{-6})(300) = 1.8\text{ mC} \)
When two capacitors of capacitances \( C_1 \) and \( C_2 \) are connected in parallel, the equivalent capacitance \( C \)
\( = C_1 + C_2 = 6 + 3 = 9 \mu\text{F} \)
\( = 9 \times 10^{-6}\text{F} \)
By conservation of charge, \( Q = 1.8\text{ mC} \).
\(\therefore\) The energy of the system \( = \frac{Q^2}{2C} \)
\( = \frac{(1.8 \times 10^{-3})^2}{2(9 \times 10^{-6})} = \frac{18 \times 10^{-8}}{10^{-6}} = 0.18\text{ J} \)
The energy lost \( = 0.27 - 0.18 = 0.09\text{ J} \)
In simple words: When a charged capacitor is connected to an uncharged one, some energy is always lost because electricity flowing between them creates heat in the wires and releases some energy as waves.

📝 Teacher's Note: When explaining this, use the analogy of two water tanks being connected. Even if no water is lost, the "energy" or "splashing" as levels equalize represents the heat loss.

🎯 Exam Tip: Remember that while total charge is conserved in these problems, total electrostatic energy is NOT conserved due to heat and EM radiation.

Question 9. One hundred twenty five small liquid drops, each carrying a charge of 0.5 \(\mu\)C and each of diameter 0.1 m form a bigger drop. Calculate the potential at the surface of the bigger drop.
Answer:
Data : \( n = 125 \), \( q = 0.5 \times 10^{-6}\text{ C} \), \( d = 0.1\text{ m} \)
The radius of each small drop, \( r = \frac{d}{2} = 0.05\text{ m} \)
The volume of the larger drop being equal to the volume of the \( n \) smaller drops, the radius of the larger drop is
\( R = \sqrt[3]{n}r = \sqrt[3]{125} (0.05) = 5 \times 0.05 = 0.25\text{ m} \)
The charge on the larger drop,
\( Q = nq = 125 \times (0.5 \times 10^{-6})\text{ C} \)
\(\therefore\) The electric potential of the surface of the larger drop,
\( V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R} = (9 \times 10^9) \times \frac{125 \times (0.5 \times 10^{-6})}{0.25} \)
\( \implies 9 \times 125 \times 2 \times 10^3 = 2.25 \times 10^6\text{ V} \)
In simple words: When many tiny charged drops merge into one big drop, their total charge adds up and their total volume determines the new size, which together changes the electrical pressure (potential) on the surface.

📝 Teacher's Note: Make sure students understand that while volume is additive (\( V_{big} = n \times v_{small} \)), the radius follows a cube-root relationship. This is the most common point of confusion.

🎯 Exam Tip: Always double-check if the question gives diameter or radius. In this problem, diameter is 0.1 m, so radius must be 0.05 m.

Question 10. The dipole moment of a water molecule is \( 6.3 \times 10^{-30}\text{ Cm} \). A sample of water contains \( 10^{21} \) molecules, whose dipole moments are all oriented in an electric field of strength \( 2.5 \times 10^5\text{ N/C} \). Calculate the work to be done to rotate the dipoles from their initial orientation \( \theta_1 = 0 \) to one in which all the dipoles are perpendicular to the field, \( \theta_2 = 90^\circ \).
Answer:
Data: \( p = 6.3 \times 10^{-30}\text{ C}\cdot\text{m} \), \( N = 10^{21} \) molecules,
\( E = 2.5 \times 10^5\text{ N/C} \), \( \theta_0 = \theta_1 = 0^\circ \), \( \theta = \theta_2 = 90^\circ \)
\( W = pE(\cos \theta_0 - \cos \theta) \)
The total work required to orient \( N \) dipoles is
\( W = NpE(\cos \theta_1 - \cos \theta_2) \)
\( \implies (10^{21})(6.3 \times 10^{-30})(2.5 \times 10^5) \)
\( \implies 15.75 \times 10^{-4}\text{ J} = 1.575\text{ mJ} \)
In simple words: It takes energy to turn tiny electrical dipoles against an electric field, similar to how it takes effort to turn a magnet against a strong magnetic force.

📝 Teacher's Note: Use the analogy of a compass needle. The electric field wants the dipole to point one way; you have to do work to twist it away from that "natural" alignment.

🎯 Exam Tip: Note that \( \cos 90^\circ = 0 \), which simplifies the calculation. Don't forget to multiply by the total number of molecules \( N \).

Question 11. A charge 6 \(\mu\)C is placed at the origin and another charge -5 \(\mu\)C is placed on the y axis at a position A (0, 6.0) m.
(a) Calculate the total electric potential at the point P whose coordinates are (8.0, 0) m
(b) Calculate the work done to bring a proton from infinity to the point P. What is the significance of the sign of the work done ?
Answer:
Data : \( q_1 = 6 \times 10^{-6}\text{ C} \), \( q_2 = -5 \times 10^{-6}\text{ C} \),
\( A \equiv (0, 6.0\text{ m}) \), \( P \equiv (8.0\text{ m}, 0) \), \( r_1 = \text{OP} = 8\text{ m} \), \( q = e = 1.6 \times 10^{-19}\text{ C} \), \( 1/4\pi\epsilon_0 = 9 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2 \)
\( r_2 = \text{AP} = \sqrt{(8 - 0)^2 + (0 - 6)^2} = \sqrt{64 + 36} = 10\text{ m} \)

(a) The net electric potential at P due to the system of two charges is
\( V = V_1 + V_2 = \frac{1}{4\pi\epsilon_0} \left[ \frac{q_1}{r_1} + \frac{q_2}{r_2} \right] \)
\( \implies (9 \times 10^9) \left[ \frac{6 \times 10^{-6}}{8} + \frac{-5 \times 10^{-6}}{10} \right] \)
\( \implies (9 \times 10^3)(0.75 - 0.5) = 2.25 \times 10^3\text{ V} \)
\( = 2.25\text{ kV} \)

(b) The electric potential \( V \) at the point \( P \) is the negative of the work done per unit charge, by the electric field of the system of the charges \( q_1 \) and \( q_2 \), in bringing a test charge from infinity to that point.
\( V = -\frac{W}{q_0} \)
\(\therefore W = -qV = -(1.6 \times 10^{-19})(2.25 \times 10^3) \)
\( \implies -3.6 \times 10^{-16}\text{ J} = -2.25\text{ keV} \)
That is, in bringing the positively charged proton from a point of lower potential to a point of higher potential, the work done by the electric field on it is negative, which means that an external agent must bring the proton against the electric field of the system of the two source charges.
[Note : The potential \( V \) at a point is the work done per unit charge (\( W_{ext} \)) by an external agent in bringing a test charge from infinity to that point. In the above case, the work done by an external agent will be positive. The question does not specify this.]
In simple words: Total potential is found by adding the individual potentials from each charge. Moving a positive proton into a positive potential area requires pushing from an outside force, as the field naturally wants to push it away.

📝 Teacher's Note: Emphasize that electric potential is a scalar quantity. We add the values directly with their signs, unlike electric field which requires vector addition.

🎯 Exam Tip: Pay close attention to the wording of "work done by the field" vs "work done by an external agent." They are equal in magnitude but opposite in sign.

Question 12. In a parallel plate capacitor with air between the plates, each plate has an area of \( 6 \times 10^{-3}\text{ m}^2 \) and the separation between the plates is 2 mm.
i) Calculate the capacitance of the capacitor.
ii) If this capacitor is connected to 100 V supply, what would be the charge on each plate?
iii) How would charge on the plates be affected if a 2 mm thick mica sheet of k = 6 is inserted between the plates while the voltage supply remains connected ?
Answer:
Data: \( k = 1 \)(air), \( A = 6 \times 10^{-3}\text{ m}^2 \), \( d = 2\text{ mm} = 2 \times 10^{-3}\text{ m} \), \( V = 100\text{ V} \), \( t = 2\text{ mm} = d \), \( k_1 = 6 \), \( \epsilon_0 = 8.85 \times 10^{-12}\text{ F/m} \)

(i) The capacitance of the air capacitor, \( C_0 = \frac{\epsilon_0 A}{d} \)
\( \implies \frac{(8.85 \times 10^{-12})(6 \times 10^{-3})}{2 \times 10^{-3}} \)
\( \implies 26.55 \times 10^{-12}\text{ F} = 26.55\text{ pF} \)

(ii) \( Q_0 = C_0 V = (26.55 \times 10^{-12})(100) \)
\( \implies 26.55 \times 10^{-10}\text{ C} = 2.655\text{ nC} \)

(iii) The dielectric of relative permittivity \( k_1 \) completely fills the space between the plates (\(\because t = d\)), so that the new capacitance is \( C = k_1 C_0 \).
With the supply still connected, \( V \) remains the same.
\(\therefore Q = CV = kC_0 V = kQ_0 = 6(2.655\text{ nC}) = 15.93\text{ nC} \)
Therefore, the charge on the plates increases.
[Note: \( C = k_1 C_0 = 6(26.55\text{ pF}) = 159.3\text{ pF} \).]
In simple words: A capacitor stores more charge when you insert a material like mica (a dielectric) because it makes it easier for the plates to hold onto electrical energy without it leaking away.

📝 Teacher's Note: Clarify the two scenarios: battery connected vs. battery disconnected. If connected, \( V \) is constant and \( Q \) increases. If disconnected, \( Q \) is constant and \( V \) decreases.

🎯 Exam Tip: Always convert units like mm to meters and pF to Farads before plugging them into the formulas to avoid decimal errors.

Question 13. Find the equivalent capacitance between P and Q. Given, area of each plate = A and separation between plates = d.
Answer:
(i) The capacitor in figure is a series combination of three capacitors of plate separations d/3 and plate areas A, with \( C_1 \) filled with air (\( k_1 = 1 \)), \( C_2 \) filled with dielectric of \( k_2 = 3 \) and \( C_3 \) filled with dielectric of \( k_3 = 6 \)
\(\therefore C_1 = \frac{k_1 \epsilon_0 A}{d/3} = \frac{3 \epsilon_0 A}{d} k_1 \), \( C_2 = \frac{k_2 \epsilon_0 A}{d/3} = \frac{3 \epsilon_0 A}{d} k_2 \), \( C_3 = \frac{k_3 \epsilon_0 A}{d/3} = \frac{3 \epsilon_0 A}{d} k_3 \)
\( \frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{d}{3 \epsilon_0 A} \left( \frac{1}{k_1} + \frac{1}{k_2} + \frac{1}{k_3} \right) \)
\(\therefore C' = \frac{3 \epsilon_0 A}{d} \left( \frac{k_1 k_2 k_3}{k_1 k_2 + k_2 k_3 + k_3 k_1} \right) = \frac{3 \epsilon_0 A}{d} \left( \frac{18}{27} \right) \)
\( \implies \frac{2 \epsilon_0 A}{d} \)
In simple words: When different materials are layered between capacitor plates, they act like several small capacitors joined in a row (series), and we calculate the total capacity by adding their "inverted" values.

📝 Teacher's Note: Show students how to identify if dielectrics are in series or parallel. If they divide the distance \( d \), they are in series. If they divide the area \( A \), they are in parallel.

🎯 Exam Tip: Since the thickness is \( d/3 \), the capacitance of each segment starts with a factor of 3 in the numerator. Don't forget this reciprocal when calculating the final equivalent value.

(ii) In figure, a series combination of two capacitors \( C_2(k_2 = 3) \) and \( C_3(k_3 = 6) \), of plate areas \( A/2 \) and plate separations \( d/2 \), is in parallel with a capacitor \( C_1 (k_1 = 4) \) of plate area \( A/2 \) and plate separation \( d \).
\( \therefore C_1 = \frac{k_1 \varepsilon_0 (A/2)}{d} = \frac{\varepsilon_0 A}{2d} k_1 \)
\( C_2 = \frac{k_2 \varepsilon_0 (A/2)}{d/2} = \frac{\varepsilon_0 A}{d} k_2 \)
\( C_3 = \frac{k_3 \varepsilon_0 (A/2)}{d/2} = \frac{\varepsilon_0 A}{d} k_3 \)
\( \therefore \) For the series combination of \( C_2 \) and \( C_3 \),
\( \frac{1}{C_4} = \frac{1}{C_2} + \frac{1}{C_3} \)
\( \therefore C_4 = \frac{C_2 C_3}{C_2 + C_3} = \frac{\varepsilon_0 A}{d} \left( \frac{k_2 k_3}{k_2 + k_3} \right) = \frac{\varepsilon_0 A}{d} \left( \frac{3 \times 6}{3 + 6} \right) = \frac{2 \varepsilon_0 A}{d} \)
Finally, for the parallel combination of \( C_1 \) and \( C_4 \),
\( C'' = C_1 + C_4 = \frac{\varepsilon_0 A}{2d}(4) + \frac{2 \varepsilon_0 A}{d} = \frac{4 \varepsilon_0 A}{d} \)
Thus, the equivalent capacitances are \( C' = \frac{2 \varepsilon_0 A}{d} \) and \( C'' = \frac{4 \varepsilon_0 A}{d} \).
The equivalent capacitance for the series part is \( \frac{2 \varepsilon_0 A}{d} \) and the total equivalent capacitance is \( \frac{4 \varepsilon_0 A}{d} \).
In simple words: This calculation shows how to find the total storage capacity of a complex circuit by breaking it into smaller parts (series and parallel) and then combining them.

📝 Teacher's Note: When solving composite capacitor problems, always calculate the individual capacitances first based on their specific area, distance, and dielectric constant before applying series or parallel formulas.

🎯 Exam Tip: Remember to substitute the dielectric constant \( k \) correctly. In series, the smaller capacitor usually dominates the voltage drop, but here it determines the total \( C_4 \).

Can You Recall (Textbook Page No. 188)

Question 1. What is gravitational Potential ?
Answer: We measure the gravitational potential energy \( U \) of a body (1) by assigning \( U = 0 \) for a reference configuration (such as the body at a reference level) (2) then equating \( U \) to the work \( W \) done by an external force to move the body up or down from that level to a point. We then define gravitational potential of the point as gravitational potential energy per unit mass of the body. We follow the same procedure with the electric force, which is also a conservative force with the only difference that while the gravitational force is always attractive, electric force can be attractive (for unlike charges) or repulsive (for like charges).
In simple words: Gravitational potential is the amount of energy stored at a certain height for every kilogram of an object's mass.

📝 Teacher's Note: Use the analogy of a "gravity hill" to explain potential. Just as a ball naturally rolls down a hill, objects move towards lower potential energy states.

🎯 Exam Tip: When defining potential, always mention it is "per unit mass" for gravity and "per unit charge" for electricity to get full marks.

Remember This (Textbook Page No. 191)

Question 1. Due to a single charge at a distance r, Force (F) \( \alpha \) 1/\( r^2 \), Electric field (E) \( \alpha \) 1/\( r^2 \) but Potential (V) \( \alpha \) 1/ r.
Answer: At a point a distance \( r \) from an isolated point charge, the force \( F \) on a point charge and the electric field \( E \) both vary as 1/\( r^2 \), while the potential energy \( U \) of a point charge and the electric potential \( V \) at the point both vary as 1/\( r \).
In simple words: As you move away from a charge, the push or pull (force) gets weak very fast, but the potential energy gets weak more slowly.

📝 Teacher's Note: Students often confuse the \( r \) and \( r^2 \) dependencies. Remind them that Force and Field are vectors involving squared distance, while Potential is a scalar work-related quantity involving distance linearly.

🎯 Exam Tip: If a question asks how potential changes if distance is doubled, the answer is "reduced to half," not "reduced to one-fourth."

Use Your Brain Power (Textbook Page No. 194)

Question 1. Is electrostatic potential necessarily zero at a point where electric field strength is zero? Justify.
Answer: Electric potential is a scalar quantity while electric field intensity is a vector quantity. When we add potentials at a point due two or more point charges, the operation is simple scalar addition along with the sign of \( V \), determined by the sign of the \( q \) that produces \( V \). At a point, the net field is the vector sum of the fields due to the individual charges. Midway between the two charges of an electric dipole, the potentials due to the two charges are equal in magnitude but opposite in sign, and thus add up to zero. But the electric fields due to the charges are equal in magnitude and direction-towards the negative charge-so that the net field there is not zero. But midway between two like charges of equal magnitudes, the potentials are equal in magnitude and have the same sign, so that the net potential is nonzero. However, the fields due to the two equal like charges are equal in magnitude but opposite in direction, and thus vectorially add up to zero.
In simple words: No. Think of two positive charges; right in the middle, they push against each other equally so the field is zero, but you still need energy to reach that spot, so the potential is not zero.

📝 Teacher's Note: Use the example of a dipole vs. like charges to show that potential (scalar) and field (vector) behave differently when summed at a midpoint.

🎯 Exam Tip: In your justification, always mention that potential is a scalar and field is a vector. This is the fundamental reason for the difference.

Do You Know (Textbook Page No. 203)

Question 1. If we apply a large enough electric field, we can ionize the atoms and create a condition for electric charge to flow like a conductor. The fields required for the breakdown of dielectric is called dielectric strength.
Answer: In a sufficiently strong electric field, the molecules of a dielectric material become ionized, allowing flow of charge. The insulating properties of the dielectric breaks down, permanently or temporarily, and the phenomenon is called dielectric breakdown. During dielectric breakdown, electrical discharge through the dielectric follows random-path patterns like tree branches, called a Lichtenberg figure. Dielectric strength is the voltage that an insulating material can withstand before breakdown occurs. It usually depends on the thickness of the material. It is expressed in kV/mm. For example, the dielectric strengths of air, polystyrene and mica in kV/mm are 3, 20 and 118. Higher dielectric strength corresponds to better insulation properties.
In simple words: Every insulator has a limit. If the electrical push is too strong, even air or plastic will start letting sparks through, like a tiny bolt of lightning.

📝 Teacher's Note: Use lightning as a real-world example of dielectric breakdown in air. This makes the concept of "dielectric strength" very relatable.

🎯 Exam Tip: Note that dielectric strength is measured in units of Electric Field (kV/mm or V/m), even though it is often loosely described as the "voltage" it can withstand.

Remember This (Textbook Page No. 205)

Question 1. Series combination is used when a high voltage is to be divided on several capacitors. Capacitor with minimum capacitance has the maximum potential difference between the plates.
Answer: Series combination of capacitors:
1. Equivalent capacitance is less than the smallest capacitance in series. For several capacitors of given capacitances, the equivalent capacitance of their series combination is minimum.
2. All capacitors in the combination have the same charge but their potential differences are in the inverse ratio of their capacitances.
3. Series combination of capacitors is sometimes used when a high voltage, which exceeds the breakdown voltage of a single capacitor, is to be divided on more than one capacitors. Capacitive voltage dividers are only useful in AC circuits, since capacitors do not pass DC signals.
Parallel combination of capacitors:
1. For several capacitors of given capacitances, the equivalent capacitance of their parallel combination is maximum.
2. The same voltage is applied to all capacitors in the combination, but the charge stored in the combination is distributed in proportion to their capacitances. The maximum rated voltage of a parallel combination is only as high as the lowest voltage rating of all the capacitors used. That is, if several capacitors rated at 500 V are connected in parallel to a capacitor rated at 100 V, the maximum voltage rating of the capacitor bank is only 100 V.
3. Parallel combination of capacitors is used when a large capacitance is required, i.e., a large charge is to be stored, at a small potential difference.
In simple words: In series, capacitors share the voltage but keep the same charge. In parallel, they share the same voltage and add up their charges to store more energy together.

📝 Teacher's Note: Compare capacitors in series/parallel to resistors. Emphasize that the formulas are "flipped" (e.g., capacitors add directly in parallel, whereas resistors add directly in series).

🎯 Exam Tip: A common trap is the voltage rating in parallel. Remember that the entire bank is only as strong as the "weakest link" (the capacitor with the lowest voltage rating).

Remember This (Textbook Page No. 207)

Question 1. If there are n parallel plates then there will be (n-1) capacitors, hence \( C = (n - 1) \frac{A\varepsilon_0}{d} \)
(2) For a spherical capacitor, consisting of two concentric spherical conducting shells with inner and outer radii as a and b respectively, the capacitance C is given by \( C = 4\pi\varepsilon_0 \left( \frac{ab}{b-a} \right) \)
(3) For a cylindrical capacitor, consisting of two coaxial cylindrical shells with radii of the inner and outer cylinders as a and b, and length l, the capacitance C is given by \( C = \frac{2\pi\varepsilon_0 \ell}{\log_e \frac{b}{a}} \)
Answer:
1. Stacking together \( n \) identical conducting plates equally spaced and with alternate plates connected to two points P and Q, forms a parallel combination of \( n - 1 \) identical capacitors between P and Q. Then, the capacitance between the points is \( (n - 1) \) times the capacitance between any two adjacent plates.
2. A cylindrical capacitor consists of a solid cylindrical conductor of radius \( a \) is surrounded by coaxial cylindrical shell of inner radius \( b \). The length of both cylinders is \( L \), such that \( L \) is much larger than \( b - a \), the separation of the cylinders, so that edge effects can be ignored. The capacitance of the capacitor is \( C = \frac{2\pi\varepsilon_0 L}{\log_e (b/a)} \). The capacitance depends only on the geometrical factors, \( L \), \( a \) and \( b \), as for a parallel-plate capacitor.
3. A spherical capacitor which consists of two concentric spherical shells of radii \( a \) and \( b \). The capacitance of the capacitor is \( C = 4\pi\varepsilon_0 \left( \frac{ab}{b-a} \right) \). Again, the capacitance depends only on the geometrical factors, \( a \) and \( b \).
In simple words: Capacitance doesn't depend on how much charge you give it; it only depends on the shape and size of the object, like how the size of a bucket determines how much water it can hold.

📝 Teacher's Note: Highlight that in all these formulas, capacitance is purely a function of geometry (\( A \), \( d \), radii) and the medium (\( \varepsilon_0 \)).

🎯 Exam Tip: Be careful with the "n-plates" question. Students often mistakenly use \( n \) instead of \( n-1 \). Always count the gaps between plates to find the number of capacitors.