NCERT Solutions Class 12 Physics Electrostatic Potential and Capacitance

NCERT Solutions Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance is an indispensable, comprehensive learning resource designed for CBSE students aiming for top scores in their board examinations and competitive exams like JEE and NEET. In this chapter, students transition from static electric forces and fields to scalar electrostatic potential, potential energy of charge configurations, properties of dielectric materials, and the functioning of capacitors. Our NCERT Solutions step by step guide provides authentic, classroom-tested answers to all textbook exercises (Exercises 2.1 to 2.11) with rigorous derivations, mathematical substitutions, and conceptual insights. As part of our complete NCERT Solutions Physics chapter wise series, this guide serves as a self-study companion for mastering theoretical concepts and numerical problems.

Chapter Overview

Chapter 2 of CBSE Class 12 Physics bridges fundamental electrostatics with circuit theory and practical electrical components. Electrostatic force is a conservative force, which allows us to define the concept of electric potential energy and electrostatic potential at any point in an electric field.

The primary concepts covered in this chapter include:

• Electrostatic Potential and Potential Difference: Work done per unit positive test charge in bringing it from infinity to a point against electrostatic forces ($V = W/q$).
• Potential due to a Point Charge and Electric Dipole: Derivations showing $V \propto 1/r$ for a single point charge and $V \propto \cos\theta/r^2$ for a dipole.
• Equipotential Surfaces: Geometric surfaces with constant potential throughout, their perpendicular relationship with electric field lines ($\vec{E} = -\nabla V$), and the fact that no work is done moving a charge along an equipotential surface.
• Potential Energy of a System of Charges: Calculation of stored electrostatic energy for discrete charge systems and an electric dipole in an external electric field ($U = -\vec{p}\cdot\vec{E}$).
• Electrostatics of Conductors and Dielectrics: Behavior of free charges in conductors, electrostatic shielding, electric polarization, and dielectric breakdown.
• Capacitance and Capacitors: Definition of capacitance ($C = Q/V$), capacitance of a parallel plate capacitor ($C = \varepsilon_0 A/d$), and the effect of inserting dielectric slabs.
• Combination of Capacitors: Equivalent capacitance in series ($1/C_s = \sum 1/C_i$) and parallel ($C_p = \sum C_i$) configurations.
• Energy Stored in a Capacitor: Energy density in an electric field ($u = \frac{1}{2}\varepsilon_0 E^2$) and total energy stored ($U = \frac{1}{2}CV^2$).

In the CBSE Class 12 Physics board examination, Unit 1 (Electrostatics, comprising Chapters 1 and 2) carries approximately 16 marks together with Current Electricity. Chapter 2 regularly features in 2-mark derivations, 3-mark numericals, and 5-mark long-answer questions involving capacitor combinations and dielectric insertions.

Exercise Solutions

Below are the detailed, classroom-style CBSE NCERT solutions for all exercises in Chapter 2, formatted with complete step-by-step methods: Given, To Find, Formula, Substitution, and Answer.

NCERT Exercise 2.1

Question: Two charges $5 \times 10^{-8}\text{ C}$ and $-3 \times 10^{-8}\text{ C}$ are located $16\text{ cm}$ apart. At what points on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

1. Given:
   • $q_1 = 5 \times 10^{-8}\text{ C}$
   • $q_2 = -3 \times 10^{-8}\text{ C}$
   • Distance between charges $d = 16\text{ cm} = 0.16\text{ m}$
2. To Find: Position of points along the line joining the charges where net electric potential $V = 0$.
3. Formula: $V = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r_1} + \frac{q_2}{r_2}\right) = 0$, where $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2$.
4. Step-by-Step Calculation:
   • Case 1: Point P lies between the two charges at distance $x$ from $q_1$ ($0 < x < 0.16\text{ m}$):
     Distance from $q_2$ is $(0.16 - x)\text{ m}$.
     $\frac{1}{4\pi\varepsilon_0}\left[\frac{5 \times 10^{-8}}{x} + \frac{-3 \times 10^{-8}}{0.16 - x}\right] = 0$
     $\frac{5}{x} = \frac{3}{0.16 - x} \implies 5(0.16 - x) = 3x$
     $0.80 - 5x = 3x \implies 8x = 0.80 \implies x = 0.10\text{ m} = 10\text{ cm}$.
   • Case 2: Point P lies outside the line segment on the side of the smaller magnitude charge $q_2$ at distance $x$ from $q_1$ ($x > 0.16\text{ m}$):
     Distance from $q_2$ is $(x - 0.16)\text{ m}$.
     $\frac{5 \times 10^{-8}}{x} + \frac{-3 \times 10^{-8}}{x - 0.16} = 0$
     $\frac{5}{x} = \frac{3}{x - 0.16} \implies 5(x - 0.16) = 3x$
     $5x - 0.80 = 3x \implies 2x = 0.80 \implies x = 0.40\text{ m} = 40\text{ cm}$.
5. Answer: The electric potential is zero at a distance of 10 cm and 40 cm from the positive charge $5 \times 10^{-8}\text{ C}$ along the line joining the two charges.

NCERT Exercise 2.2

Question: A regular hexagon of side $10\text{ cm}$ has a charge $5\ \mu\text{C}$ at each of its vertices. Calculate the potential at the centre of the hexagon.

1. Given:
   • Side of regular hexagon $a = 10\text{ cm} = 0.1\text{ m}$
   • Distance of each vertex from the center $r = a = 0.1\text{ m}$
   • Charge at each vertex $q = 5\ \mu\text{C} = 5 \times 10^{-6}\text{ C}$
   • Number of vertices $n = 6$
2. To Find: Net electric potential $V$ at the centre of the hexagon.
3. Formula: Since electric potential is a scalar quantity, $V = \sum_{i=1}^6 V_i = 6 \times \left(\frac{1}{4\pi\varepsilon_0}\frac{q}{r}\right)$.
4. Substitution:
   $V = 6 \times (9 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2) \times \frac{5 \times 10^{-6}\text{ C}}{0.1\text{ m}}$
   $V = 6 \times 9 \times 10^9 \times 5 \times 10^{-5} = 2.7 \times 10^6\text{ V}$.
5. Answer: The electric potential at the centre of the hexagon is $2.7 \times 10^6\text{ V}$ (or $2.7\text{ MV}$).

NCERT Exercise 2.3

Question: Two charges $2\ \mu\text{C}$ and $-2\ \mu\text{C}$ are placed at points A and B $6\text{ cm}$ apart. (a) Identify an equipotential surface of the system. (b) What is the direction of the electric field at every point on this surface?

1. Given:
   • $q_A = +2\ \mu\text{C}$, $q_B = -2\ \mu\text{C}$
   • Separation $AB = 6\text{ cm} = 0.06\text{ m}$
2. To Find: (a) Location and nature of the equipotential surface. (b) Direction of the electric field on this surface.
3. Analysis & Explanation:
   • Part (a): The system constitutes an electric dipole. The electric potential at any point equidistant from both equal and opposite charges is zero ($V = \frac{kq}{r} + \frac{k(-q)}{r} = 0$). Therefore, the equipotential surface is the plane perpendicular to the line segment AB and passing through its midpoint O (the equatorial plane). The potential on this entire plane is zero.
   • Part (b): Electric field lines always originate from the positive charge and terminate on the negative charge. Because electric field lines are everywhere normal to equipotential surfaces, the electric field at every point on this plane is directed perpendicular to the plane, parallel to the line segment AB, pointing from point A (positive charge) to point B (negative charge).
4. Answer: (a) The equatorial plane passing normally through the midpoint of AB is the equipotential surface ($V = 0\text{ V}$). (b) The electric field is perpendicular to the plane at every point, directed from A to B.

NCERT Exercise 2.4

Question: A spherical conductor of radius $12\text{ cm}$ has a charge of $1.6 \times 10^{-7}\text{ C}$ distributed uniformly on its surface. What is the electric field: (a) inside the sphere, (b) just outside the sphere, (c) at a point $18\text{ cm}$ from the centre of the sphere?

1. Given:
   • Radius of spherical conductor $R = 12\text{ cm} = 0.12\text{ m}$
   • Total charge $q = 1.6 \times 10^{-7}\text{ C}$
2. To Find: Electric field $E$ at (a) $r < R$, (b) $r = R$, (c) $r = 18\text{ cm} = 0.18\text{ m}$.
3. Formulas:
   • Inside conductor ($r < R$): $E = 0$
   • On/outside surface ($r \ge R$): $E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}$
4. Step-by-Step Calculation:
   • (a) Inside the sphere: For any electrostatic conductor, excess charge resides entirely on the outer surface. Hence, enclosed charge $q_{enc} = 0$, giving $E = 0\text{ N/C}$.
   • (b) Just outside the sphere ($r = 0.12\text{ m}$):
     $E = \frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{(0.12)^2} = \frac{1440}{0.0144} = 10^5\text{ N/C}$ (radially outward).
   • (c) At $r = 0.18\text{ m}$:
     $E = \frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{(0.18)^2} = \frac{1440}{0.0324} = 4.44 \times 10^4\text{ N/C}$ (radially outward).
5. Answer: (a) $E = 0$, (b) $E = 10^5\text{ N/C}$ directed radially outwards, (c) $E = 4.44 \times 10^4\text{ N/C}$ directed radially outwards.

NCERT Exercise 2.5

Question: A parallel plate capacitor with air between the plates has a capacitance of $8\text{ pF}$ ($1\text{ pF} = 10^{-12}\text{ F}$). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6$?

1. Given:
   • Initial capacitance $C_0 = \frac{\varepsilon_0 A}{d} = 8\text{ pF}$
   • New plate separation $d' = \frac{d}{2}$
   • Dielectric constant of medium $K = 6$
2. To Find: New capacitance $C'$.
3. Formula: $C' = \frac{K \varepsilon_0 A}{d'}$
4. Substitution:
   $C' = \frac{K \varepsilon_0 A}{d/2} = 2K \left(\frac{\varepsilon_0 A}{d}\right) = 2K \cdot C_0$
   $C' = 2 \times 6 \times 8\text{ pF} = 12 \times 8\text{ pF} = 96\text{ pF}$.
5. Answer: The new capacitance of the capacitor is $96\text{ pF}$.

NCERT Exercise 2.6

Question: Three capacitors each of capacitance $9\text{ pF}$ are connected in series. (a) What is the total capacitance of the combination? (b) What is the potential difference across each capacitor if the combination is connected to a $120\text{ V}$ supply?

1. Given:
   • $C_1 = C_2 = C_3 = C = 9\text{ pF}$
   • Supply voltage $V = 120\text{ V}$
2. To Find: (a) Equivalent capacitance $C_s$. (b) Potential difference across each capacitor ($V_1, V_2, V_3$).
3. Formulas:
   • Series capacitance: $\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$
   • Potential division: For identical capacitors in series, $V_i = \frac{V}{n}$.
4. Calculation:
   • (a) Total capacitance: $\frac{1}{C_s} = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9} = \frac{1}{3} \implies C_s = 3\text{ pF}$.
   • (b) Voltage across each capacitor: Since all three capacitances are equal, $V_1 = V_2 = V_3 = \frac{120\text{ V}}{3} = 40\text{ V}$.
5. Answer: (a) Total capacitance is $3\text{ pF}$. (b) The potential difference across each capacitor is $40\text{ V}$.

NCERT Exercise 2.7

Question: Three capacitors of capacitances $2\text{ pF}$, $3\text{ pF}$ and $4\text{ pF}$ are connected in parallel. (a) What is the total capacitance of the combination? (b) Determine the charge on each capacitor if the combination is connected to a $100\text{ V}$ supply.

1. Given:
   • $C_1 = 2\text{ pF} = 2 \times 10^{-12}\text{ F}$, $C_2 = 3\text{ pF} = 3 \times 10^{-12}\text{ F}$, $C_3 = 4\text{ pF} = 4 \times 10^{-12}\text{ F}$
   • Supply voltage $V = 100\text{ V}$
2. To Find: (a) Equivalent parallel capacitance $C_p$. (b) Charge on each capacitor ($Q_1, Q_2, Q_3$).
3. Formulas:
   • Parallel combination: $C_p = C_1 + C_2 + C_3$
   • Charge on individual capacitor: $Q_i = C_i V$ (potential is identical across parallel branches).
4. Substitution:
   • (a) Total capacitance: $C_p = 2 + 3 + 4 = 9\text{ pF}$.
   • (b) Charges:
     $Q_1 = C_1 V = 2 \times 10^{-12}\text{ F} \times 100\text{ V} = 200\text{ pC} = 2 \times 10^{-10}\text{ C}$.
     $Q_2 = C_2 V = 3 \times 10^{-12}\text{ F} \times 100\text{ V} = 300\text{ pC} = 3 \times 10^{-10}\text{ C}$.
     $Q_3 = C_3 V = 4 \times 10^{-12}\text{ F} \times 100\text{ V} = 400\text{ pC} = 4 \times 10^{-10}\text{ C}$.
5. Answer: (a) Total capacitance is $9\text{ pF}$. (b) The charges on the capacitors are $2 \times 10^{-10}\text{ C}$, $3 \times 10^{-10}\text{ C}$, and $4 \times 10^{-10}\text{ C}$ respectively.

NCERT Exercise 2.8

Question: 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 distance between the plates is $3\text{ mm}$. Calculate the capacitance of the capacitor. If this capacitor is connected to a $100\text{ V}$ supply, what is the charge on each plate of the capacitor?

1. Given:
   • Plate area $A = 6 \times 10^{-3}\text{ m}^2$
   • Plate separation $d = 3\text{ mm} = 3 \times 10^{-3}\text{ m}$
   • Permittivity of free space $\varepsilon_0 = 8.854 \times 10^{-12}\text{ C}^2\text{N}^{-1}\text{m}^{-2}$
   • Supply voltage $V = 100\text{ V}$
2. To Find: Capacitance $C$ and charge on each plate $Q$.
3. Formulas: $C = \frac{\varepsilon_0 A}{d}$, $Q = C V$.
4. Substitution:
   $C = \frac{8.854 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}} = 2 \times 8.854 \times 10^{-12} = 17.71 \times 10^{-12}\text{ F} = 17.71\text{ pF} \approx 18\text{ pF}$.
   $Q = C V = (17.71 \times 10^{-12}\text{ F}) \times 100\text{ V} = 1.771 \times 10^{-9}\text{ C} = 1.771\text{ nC}$.
5. Answer: The capacitance is $17.71\text{ pF}$ (or $\approx 18\text{ pF}$) and the charge on each plate is $1.77 \times 10^{-9}\text{ C}$ ($+1.77\text{ nC}$ on positive plate, $-1.77\text{ nC}$ on negative plate).

NCERT Exercise 2.9

Question: Explain what would happen if in the capacitor given in Exercise 2.8, a $3\text{ mm}$ thick mica sheet (of dielectric constant = $6$) were inserted between the plates: (a) while the voltage supply remained connected, (b) after the supply was disconnected.

1. Given:
   • Initial capacitance $C_0 = 17.71\text{ pF}$, Initial voltage $V_0 = 100\text{ V}$, Initial charge $Q_0 = 1.771 \times 10^{-9}\text{ C}$
   • Dielectric thickness $t = 3\text{ mm} = d$, Dielectric constant $K = 6$
   • New capacitance with full dielectric slab: $C' = K C_0 = 6 \times 17.71\text{ pF} = 106.26\text{ pF} \approx 106.3\text{ pF}$.
2. Detailed Analysis:
   • Part (a) While the voltage supply remains connected:
     - Potential difference: Remains constant at battery voltage, $V = V_0 = 100\text{ V}$.
     - Capacitance: Increases by factor $K$, $C' = 106.3\text{ pF}$.
     - Charge: Battery supplies extra charge. $Q' = C' V = K Q_0 = 6 \times 1.771\text{ nC} = 10.63\text{ nC} = 1.063 \times 10^{-8}\text{ C}$.
     - Electric field: $E = V/d = 100/(3 \times 10^{-3}) = 3.33 \times 10^4\text{ V/m}$ (remains unchanged).
   • Part (b) After the supply is disconnected:
     - Charge: Conservation of charge dictates $Q' = Q_0 = 1.771\text{ nC}$ (cannot change).
     - Capacitance: Increases by factor $K$, $C' = 106.3\text{ pF}$.
     - Potential difference: Decreases to $V' = \frac{Q_0}{C'} = \frac{V_0}{K} = \frac{100\text{ V}}{6} = 16.67\text{ V}$.
     - Electric field: Decreases to $E' = E_0/K = \frac{3.33 \times 10^4}{6} = 5.56 \times 10^3\text{ V/m}$.
3. Answer: (a) When connected: $V = 100\text{ V}$, $C = 106.3\text{ pF}$, $Q = 1.063 \times 10^{-8}\text{ C}$. (b) When disconnected: $Q = 1.771 \times 10^{-9}\text{ C}$, $C = 106.3\text{ pF}$, $V = 16.67\text{ V}$.

NCERT Exercise 2.10

Question: A $12\text{ pF}$ capacitor is connected to a $50\text{ V}$ battery. How much electrostatic energy is stored in the capacitor?

1. Given:
   • $C = 12\text{ pF} = 12 \times 10^{-12}\text{ F}$
   • $V = 50\text{ V}$
2. To Find: Electrostatic potential energy $U$ stored in the capacitor.
3. Formula: $U = \frac{1}{2} C V^2$
4. Substitution:
   $U = \frac{1}{2} \times (12 \times 10^{-12}\text{ F}) \times (50\text{ V})^2$
   $U = 6 \times 10^{-12} \times 2500 = 1.5 \times 10^{-8}\text{ J}$.
5. Answer: The electrostatic energy stored in the capacitor is $1.5 \times 10^{-8}\text{ J}$.

NCERT Exercise 2.11

Question: A $600\text{ pF}$ capacitor is charged by a $200\text{ V}$ supply. It is then disconnected from the supply and is connected to another uncharged $600\text{ pF}$ capacitor. How much electrostatic energy is lost in the process?

1. Given:
   • $C_1 = 600\text{ pF} = 600 \times 10^{-12}\text{ F}$, $V_1 = 200\text{ V}$
   • $C_2 = 600\text{ pF} = 600 \times 10^{-12}\text{ F}$, $V_2 = 0\text{ V}$
2. To Find: Energy lost during the redistribution of charge $\Delta U$.
3. Formulas:
   • Initial energy: $U_i = \frac{1}{2} C_1 V_1^2$
   • Common potential after connection: $V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$
   • Final energy: $U_f = \frac{1}{2} (C_1 + C_2) V^2$
   • Direct formula for energy loss: $\Delta U = \frac{C_1 C_2 (V_1 - V_2)^2}{2(C_1 + C_2)}$
4. Step-by-Step Calculation:
   • Initial stored energy:
     $U_i = \frac{1}{2} \times (600 \times 10^{-12}) \times (200)^2 = 300 \times 10^{-12} \times 40000 = 1.2 \times 10^{-5}\text{ J}$.
   • Common potential:
     $V = \frac{600 \times 10^{-12} \times 200}{600 \times 10^{-12} + 600 \times 10^{-12}} = \frac{1.2 \times 10^{-7}}{1.2 \times 10^{-9}} = 100\text{ V}$.
   • Final stored energy:
     $U_f = \frac{1}{2} \times (1200 \times 10^{-12}) \times (100)^2 = 600 \times 10^{-12} \times 10000 = 6 \times 10^{-6}\text{ J} = 0.6 \times 10^{-5}\text{ J}$.
   • Loss of electrostatic energy:
     $\Delta U = U_i - U_f = 1.2 \times 10^{-5}\text{ J} - 0.6 \times 10^{-5}\text{ J} = 0.6 \times 10^{-5}\text{ J} = 6 \times 10^{-6}\text{ J}$.
     (Note: This missing energy is dissipated as heat in connecting wires and electromagnetic radiation).
5. Answer: The electrostatic energy lost in the process is $6 \times 10^{-6}\text{ J}$ (or $0.6 \times 10^{-5}\text{ J}$).

Important Formulas and Theorems

This chapter contains essential theoretical derivations and formulas frequently asked in the CBSE Class 12 board exam:

1. Electric Potential Due to a Point Charge

The electric potential $V$ at distance $r$ from an isolated point charge $Q$ is derived as the work done in bringing a unit positive charge from infinity to $r$:

$$V(r) = -\int_{\infty}^r E\, dr = -\int_{\infty}^r \frac{1}{4\pi\varepsilon_0}\frac{Q}{r'^2}\, dr' = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$$

2. Electric Potential Due to an Electric Dipole

For an electric dipole with dipole moment $\vec{p} = q(2\vec{a})$ at distance $r \gg a$:

• General point at angle $\theta$ to dipole axis: $V = \frac{1}{4\pi\varepsilon_0}\frac{p\cos\theta}{r^2}$
• Axial point ($\theta = 0^\circ$ or $180^\circ$): $V = \pm\frac{1}{4\pi\varepsilon_0}\frac{p}{r^2}$
• Equatorial point ($\theta = 90^\circ$): $V = 0$

3. Relation Between Electric Field and Potential

Electric field is the negative gradient of electric potential:

$$E = -\frac{dV}{dr}$$

This relationship implies that electric field lines point in the direction of the steepest decrease in electrostatic potential.

4. Capacitance of a Parallel Plate Capacitor with Dielectric

For plates of area $A$ separated by distance $d$:

• Vacuum / Air: $C_0 = \frac{\varepsilon_0 A}{d}$
• Fully filled with dielectric constant $K$: $C = K C_0 = \frac{K\varepsilon_0 A}{d}$
• Partially filled with dielectric slab of thickness $t$ ($t < d$): $C = \frac{\varepsilon_0 A}{d - t(1 - 1/K)}$
• Partially filled with conducting slab of thickness $t$ ($t < d$): $C = \frac{\varepsilon_0 A}{d - t}$

5. Energy Stored and Energy Density

The work done in charging a capacitor from $0$ to $Q$ is stored as electrostatic potential energy:

$$U = \frac{1}{2}\frac{Q^2}{C} = \frac{1}{2}CV^2 = \frac{1}{2}QV$$

The electrostatic energy per unit volume (energy density) in the field region between plates is:

$$u = \frac{U}{\text{Volume}} = \frac{1}{2}\varepsilon_0 E^2$$

Important CBSE Question Bank Solutions

The following selected questions from the official CBSE Question Bank reinforce core analytical concepts and derivations for Class 12 Physics examinations:

Question 1: Total Internal Reflection

Question: What is total internal reflection? State the conditions.

Answer:

Total Internal Reflection (TIR) is the optical phenomenon in which a ray of light travelling in an optically denser medium is completely reflected back into the denser medium at the interface between the two media, without any refraction into the rarer medium.

Conditions for Total Internal Reflection:

1. Light must travel from an optically denser medium to an optically rarer medium.
2. The angle of incidence ($i$) in the denser medium must be strictly greater than the critical angle ($i_c$) for the pair of media ($\sin i_c = 1/\mu$).

Question 2: Lens Formula Calculation

Question: A concave lens of focal length $15\text{ cm}$ forms an image at $10\text{ cm}$ from the lens. Find the object distance.

Solution:

1. Given (with Cartesian sign convention): Focal length of concave lens $f = -15\text{ cm}$, Image distance $v = -10\text{ cm}$ (virtual image).
2. Formula: Lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \implies \frac{1}{u} = \frac{1}{v} - \frac{1}{f}$.
3. Substitution:
   $\frac{1}{u} = \frac{1}{-10} - \frac{1}{-15} = -\frac{1}{10} + \frac{1}{15} = \frac{-3 + 2}{30} = -\frac{1}{30}$.
4. Answer: $u = -30\text{ cm}$. The object is placed at a distance of $30\text{ cm}$ in front of the concave lens.

Question 3: Astronomical Telescope Magnification

Question: An astronomical telescope has an objective of focal length $100\text{ cm}$ and eyepiece of $5\text{ cm}$. Calculate its magnifying power when the final image is at infinity.

Solution:

1. Given: Focal length of objective $f_o = 100\text{ cm}$, Focal length of eyepiece $f_e = 5\text{ cm}$.
2. Formula (Normal Adjustment / Image at infinity): $M = \frac{f_o}{f_e}$.
3. Calculation: $M = \frac{100}{5} = 20\times$.
4. Answer: The magnifying power of the astronomical telescope in normal adjustment is $20$ (image is inverted and 20 times magnified).

Question 4: Convex Lens Ray Diagram

Question: Draw a labeled ray diagram showing image formation by a convex lens when the object is placed between $F_1$ and $2F_1$.

Answer:

When an object $AB$ is positioned between $F_1$ and $2F_1$ of a convex lens:

• Ray 1: Parallel to the principal axis from the top of the object $A$, passes through the second principal focus $F_2$ after refraction.
• Ray 2: Passing straight through the optical centre $O$ without undergoing any deviation.
• Intersection & Image Characteristics: The two refracted rays intersect beyond $2F_2$ to form image $A'B'$.
• Nature of Image: Real, inverted, magnified ($|m| > 1$), and located beyond $2F_2$.

Question 5: Lens Maker's Formula Derivation

Question: State the lens maker's formula. Derive it for a convex lens.

Answer:

Statement: The lens maker's formula relates the focal length $f$ of a lens to the refractive index of its material ($\mu = n_2/n_1$) and the radii of curvature ($R_1, R_2$) of its two refracting surfaces:

$$\frac{1}{f} = \left(\frac{n_2}{n_1} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

Derivation Outline:

1. Consider a thin convex lens of refractive index $n_2$ placed in a medium of refractive index $n_1$. For refraction at the first spherical surface with radius of curvature $R_1$:
   $\frac{n_2}{v_1} - \frac{n_1}{u} = \frac{n_2 - n_1}{R_1}$ (Equation 1, where $v_1$ is the position of intermediate image $I_1$).
2. The intermediate image $I_1$ serves as a virtual object for refraction at the second surface with radius $R_2$:
   $\frac{n_1}{v} - \frac{n_2}{v_1} = \frac{n_1 - n_2}{R_2} = -\frac{n_2 - n_1}{R_2}$ (Equation 2).
3. Adding Equation 1 and Equation 2 eliminates intermediate term $\frac{n_2}{v_1}$:
   $\frac{n_1}{v} - \frac{n_1}{u} = (n_2 - n_1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
4. Dividing by $n_1$: $\frac{1}{v} - \frac{1}{u} = \left(\frac{n_2}{n_1} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$. Since $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, we get $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.

Question 6: Mirror Formula and Magnification

Question: Derive the mirror formula for a concave mirror. Define magnification.

Answer:

Mirror Formula: Relates object distance $u$, image distance $v$, and focal length $f$ of a spherical mirror:

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

Derivation: From the ray diagram for a concave mirror forming a real image $A'B'$ of an object $AB$ placed beyond center of curvature $C$:

1. Similar triangles $\Delta A'B'F \sim \Delta MPF$ (where $MP \approx AB$ for a small aperture mirror) give: $\frac{A'B'}{AB} = \frac{B'F}{FP} = \frac{v - f}{f}$.
2. Similar triangles $\Delta A'B'P \sim \Delta ABP$ give: $\frac{A'B'}{AB} = \frac{B'P}{BP} = \frac{v}{u}$.
3. Equating both ratios: $\frac{v - f}{f} = \frac{v}{u} \implies \frac{v}{f} - 1 = \frac{v}{u}$.
4. Dividing the entire equation by $v$ yields: $\frac{1}{f} - \frac{1}{v} = \frac{1}{u} \implies \frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.

Linear Magnification ($m$): Defined as the ratio of the height of the image ($h'$) to the height of the object ($h$):

$$m = \frac{h'}{h} = -\frac{v}{u}$$

Question 7: Refraction through a Prism and Minimum Deviation

Question: What is a prism? Derive the expression for the angle of minimum deviation.

Answer:

A prism is a transparent optical medium bounded by at least two non-parallel refracting plane surfaces inclined at a specific angle called the angle of prism ($A$).

Derivation:

1. For a ray traversing a triangular prism: Angle of deviation $\delta = (i + e) - A$, where $i$ is the incident angle, $e$ is the emergent angle, and $A = r_1 + r_2$ is the prism angle.
2. At the condition of minimum deviation ($\delta = \delta_m$):
   - The refracted ray inside the prism travels parallel to the base (for equilateral/isosceles prisms).
   - $i = e$ and $r_1 = r_2 = r$.
3. From $A = r_1 + r_2 = 2r \implies r = \frac{A}{2}$.
4. From $\delta_m = 2i - A \implies i = \frac{A + \delta_m}{2}$.
5. Applying Snell's Law at the first refracting face ($n = \frac{\sin i}{\sin r}$):
   $$n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$

Common Mistakes and Tips

• Sign Errors in Electric Potential: Electric potential is a scalar quantity. Always include the algebraic sign of the charge ($+q$ or $-q$) directly in potential calculations ($V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}$). Never omit negative signs for negative charges as students often mistakenly do when calculating vector electric fields.
• Battery Connected vs. Disconnected in Capacitors:
  • Battery connected: Potential difference $V$ remains constant ($V = V_0$). Charge changes to $Q = K Q_0$.
  • Battery disconnected: Charge $Q$ is trapped and remains constant ($Q = Q_0$). Potential difference drops to $V = V_0/K$.
• Unit Conversions: Double-check unit prefixes before calculating: $1\text{ pF} = 10^{-12}\text{ F}$, $1\ \mu\text{F} = 10^{-6}\text{ F}$, $1\text{ nC} = 10^{-9}\text{ C}$, and convert distances from $\text{cm}$ or $\text{mm}$ to $\text{m}$.
• Energy Loss Misinterpretation: In Exercise 2.11, the energy loss $\Delta U = 6 \times 10^{-6}\text{ J}$ does not violate conservation of energy; the energy is converted to thermal energy in wires and high-frequency electromagnetic radiation during charge redistribution.

Board Exam Relevance

Question Type	Mark Allocation	Frequent Focus Areas
MCQ / Assertion-Reason	1 Mark	Equipotential surface properties, dielectric constant effects, energy density units.
Short Answer (SA I)	2 Marks	Work done on equipotential surfaces, capacitor energy calculation, zero-potential points.
Short Answer (SA II)	3 Marks	Derivation of dipole potential, combination of series/parallel capacitors, energy loss on sharing charge.
Long Answer / Case Study	4 / 5 Marks	Dielectric slab insertion in parallel plate capacitor, complete numericals with circuit analysis.

More NCERT Solutions and Practice

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