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Electric Charges and Fields Class 12: Notes, NCERT Solutions & MCQs

Welcome to the start of your Class 12 Physics journey! It all begins with Chapter 1: Electric Charges and Fields. Getting a good handle on these core ideas is the key to feeling confident and acing your board exams later on.

To make sure you get off to a great start, we’ve brought together everything you need in one easy-to-find place. Here you’ll find quick notes for revision, detailed solutions to every NCERT problem, a handpicked list of important multiple-choice questions, solved examples from past exams, and the key derivations explained simply.

Let’s dive in and build a rock-solid foundation in electrostatics together!

Quick Revision Guide on Electric Charges and Fields

1. Electric Charge

  • Types: Positive and Negative.
  • Basic Property: Like charges repel, unlike charges attract.
  • Quantisation: Charge is quantized; any charge q is an integral multiple of the fundamental charge e.
    q = ne, where n = 0, \pm1, \pm2,… and e = 1.6 \times 10^{-19} C.
  • Conservation: The total charge in an isolated system remains constant.
  • Additivity: The total charge of a system is the algebraic sum of all individual charges.

2. Coulomb’s Law

The electrostatic force between two point charges q_1 and q_2 separated by distance r is:

  • Magnitude: F = \frac{1}{4\pi\epsilon_0} \frac{|q_1 q_2|}{r^2}
  • Vector Form: \vec{F}<em>{21} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} \hat{r}</em>{21}
  • Constant: \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{N m}^2 \text{ C}^{-2}

3. Electric Field

  • Definition: Force per unit positive test charge. \vec{E} = \lim_{q \to 0} \frac{\vec{F}}{q}
  • Field due to a point charge Q: \vec{E} = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2} \hat{r}
  • Superposition Principle: The net electric field at a point is the vector sum of fields due to all individual charges.
    \vec{E}_{\text{total}} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 + …

4. Electric Dipole

  • A pair of equal and opposite charges (+q and -q) separated by a small distance 2a.
  • Dipole Moment (\vec{p}): \vec{p} = q \times (2\vec{a}) (Direction: from -q to +q).
  • Field on Axial Line: \vec{E}_{\text{axial}} = \frac{1}{4\pi\epsilon_0} \frac{2\vec{p}}{r^3} (for r >> a)
  • Field on Equatorial Line: \vec{E}_{\text{equatorial}} = -\frac{1}{4\pi\epsilon_0} \frac{\vec{p}}{r^3} (for r >> a)
  • Torque in a Uniform Field (\vec{E}): \vec{\tau} = \vec{p} \times \vec{E} ; Magnitude: \tau = pE \sin\theta

5. Electric Flux

  • Definition: Measure of the flow of the electric field through a given area.
    \Delta \phi = \vec{E} \cdot \Delta \vec{S} = E \Delta S \cos\theta
  • For a closed surface: \phi = \oint \vec{E} \cdot d\vec{S}

6. Gauss’s Law

  • The net flux through a closed surface is \frac{1}{\epsilon_0} times the net charge enclosed by the surface.
    \oint \vec{E} \cdot d\vec{S} = \frac{q_{\text{inside}}}{\epsilon_0}
  • Key Applications:
  • Infinite Long Wire: E = \frac{\lambda}{2\pi\epsilon_0 r}
  • Infinite Plane Sheet: E = \frac{\sigma}{2\epsilon_0}
  • Thin Spherical Shell:
    • Outside (r \geq R): E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}
    • Inside (r < R): E = 0

7. Continuous Charge Distribution

  • Linear Charge Density (\lambda): \lambda = \frac{\Delta Q}{\Delta l}
  • Surface Charge Density (\sigma): \sigma = \frac{\Delta Q}{\Delta S}
  • Volume Charge Density (\rho): \rho = \frac{\Delta Q}{\Delta V}

Quick Revision Tip:

Focus on the formulas for the electric field of a point charge, a dipole (axial and equatorial), and the three standard applications of Gauss’s Law (wire, sheet, shell). Remember that Gauss’s Law is most useful for symmetric charge distributions.

Chapter‑End Questions: Fully Worked Solutions

Exercise 1.1

Question: What is the force between two small charged spheres having charges of 2 \times 10^{-7} \text{C} and 3 \times 10^{-7} \text{C} placed 30 cm apart in air?

Solution:

  1. Use Coulomb’s Law: F = \frac{1}{4\pi\epsilon_0} \frac{|q_1 q_2|}{r^2}
  2. Substitute values: q_1 = 2 \times 10^{-7} \text{C}, q_2 = 3 \times 10^{-7} \text{C}, r = 0.3 \text{m}, \frac{1}{4\pi\epsilon_0} = 9 \times 10^9
  3. Calculate:
    F = 9 \times 10^9 \times \frac{(2 \times 10^{-7}) \times (3 \times 10^{-7})}{(0.3)^2}
    F = 9 \times 10^9 \times \frac{6 \times 10^{-14}}{0.09}
    F = 9 \times 10^9 \times 6.667 \times 10^{-13}
    F = 6 \times 10^{-3} \text{N}

Answer: 6 \times 10^{-3} \text{N} (repulsive)

Exercise 1.2

Question: The electrostatic force on a small sphere of charge 0.4 \muC due to another small sphere of charge -0.8 \muC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?

Solution:
(a)

  1. Coulomb’s Law: F = \frac{1}{4\pi\epsilon_0} \frac{|q_1 q_2|}{r^2}
  2. Rearrange: r^2 = \frac{1}{4\pi\epsilon_0} \frac{|q_1 q_2|}{F}
  3. Substitute: q_1 = 0.4 \times 10^{-6} \text{C}, q_2 = 0.8 \times 10^{-6} \text{C}, F = 0.2 \text{N}
  4. Calculate:
    r^2 = 9 \times 10^9 \times \frac{(0.4 \times 10^{-6}) \times (0.8 \times 10^{-6})}{0.2}
    r^2 = 9 \times 10^9 \times \frac{3.2 \times 10^{-13}}{0.2}
    r^2 = 9 \times 10^9 \times 1.6 \times 10^{-12}
    r^2 = 0.0144
    r = 0.12 \text{m} = 12 \text{cm}

(b) By Newton’s Third Law, the force is equal in magnitude and opposite in direction.

Answer: (a) 12 cm, (b) 0.2 N (attractive)

Exercise 1.3

Question: Check that the ratio \frac{k e^2}{G m_e m_p} is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Solution:

  1. Dimensional Analysis:
  • [k] = [\text{N m}^2 \text{C}^{-2}] = [\text{ML}^3\text{T}^{-4}\text{I}^{-2}]
  • [e] = [\text{IT}]
  • [G] = [\text{N m}^2 \text{kg}^{-2}] = [\text{M}^{-1}\text{L}^3\text{T}^{-2}]
  • [m_e] = [m_p] = [\text{M}]
  • Therefore: [\frac{k e^2}{G m_e m_p}] = \frac{[\text{ML}^3\text{T}^{-4}\text{I}^{-2}][\text{I}^2\text{T}^2]}{[\text{M}^{-1}\text{L}^3\text{T}^{-2}][\text{M}][\text{M}]} = [\text{M}^0\text{L}^0\text{T}^0\text{I}^0] → Dimensionless
  1. Numerical Value:
  • k = 9 \times 10^9, e = 1.6 \times 10^{-19}, G = 6.67 \times 10^{-11}, m_e = 9.1 \times 10^{-31}, m_p = 1.67 \times 10^{-27}
  • \frac{k e^2}{G m_e m_p} = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{6.67 \times 10^{-11} \times 9.1 \times 10^{-31} \times 1.67 \times 10^{-27}} \approx 2.4 \times 10^{39}
  1. Significance: This ratio represents the strength of the electrostatic force to the gravitational force between an electron and a proton.

Answer: Dimensionless, value \approx 2.4 \times 10^{39}, ratio of electrostatic to gravitational force.

Exercise 1.4

(a) Electric charge is quantized means that any observable charge is an integral multiple of the elementary charge e (q = ne).

(b) For macroscopic charges, the number of electrons transferred is so large that the discreteness (quantization) is not noticeable.

Exercise 1.5

Question: Explain how rubbing a glass rod with silk is consistent with the law of conservation of charge.

Solution: During rubbing, electrons are transferred from one body to another. The total charge before and after remains zero. The positive charge on one body is exactly equal to the negative charge on the other.

Exercise 1.6

Question: Four point charges q_A = 2 \mu\text{C}, q_B = -5 \mu\text{C}, q_C = 2 \mu\text{C}, and q_D = -5 \mu\text{C} are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 \muC placed at the centre?

Solution:

  1. The centre is equidistant from all corners.
  2. Forces due to q_A and q_C are equal and opposite → cancel out.
  3. Forces due to q_B and q_D are equal and opposite → cancel out.
  4. Net force = 0.

Answer: Zero

Exercise 1.7

(a) Electric field lines are continuous because the electric field is a continuous function in space. A break would imply no field, which isn’t true.

(b) Two field lines cannot cross because the electric field must have a unique direction at every point.

Exercise 1.8

Question: Two point charges q_A = 3 \mu\text{C} and q_B = -3 \mu\text{C} are located 20 cm apart in vacuum. (a) What is the electric field at the midpoint O? (b) If a negative test charge of magnitude 1.5 \times 10^{-9} \text{C} is placed at O, what is the force experienced?

Solution:
(a)

  1. At midpoint, r = 0.1 \text{m} from each charge.
  2. Field due to q_A: E_A = \frac{1}{4\pi\epsilon_0} \frac{3 \times 10^{-6}}{(0.1)^2} = 2.7 \times 10^6 \text{N/C} (away from A, towards B)
  3. Field due to q_B: E_B = \frac{1}{4\pi\epsilon_0} \frac{3 \times 10^{-6}}{(0.1)^2} = 2.7 \times 10^6 \text{N/C} (towards B)
  4. Both fields are in the same direction (A→B), so E_{\text{total}} = E_A + E_B = 5.4 \times 10^6 \text{N/C} (from A to B)

(b)

  1. Force on test charge q: \vec{F} = q \vec{E}
  2. F = (1.5 \times 10^{-9}) \times (5.4 \times 10^6) = 8.1 \times 10^{-3} \text{N}
  3. Since q is negative, force is opposite to \vec{E} (from B to A)

Answer: (a) 5.4 \times 10^6 \text{N/C} (A→B), (b) 8.1 \times 10^{-3} \text{N} (B→A)

Exercise 1.9

Question: A system has two charges q_A = 2.5 \times 10^{-7} \text{C} and q_B = -2.5 \times 10^{-7} \text{C} at A: (0,0,-15 cm) and B: (0,0,+15 cm). What are the total charge and electric dipole moment?

Solution:

  1. Total charge: q_A + q_B = 0
  2. Dipole moment: \vec{p} = q \times \vec{d}
  • q = 2.5 \times 10^{-7} \text{C}
  • \vec{d} from -q to +q = from A to B = (0,0,0.3) m
  • |\vec{p}| = (2.5 \times 10^{-7}) \times 0.3 = 7.5 \times 10^{-8} \text{C m}
  • Direction: along +z axis

Answer: Total charge = 0, \vec{p} = 7.5 \times 10^{-8} \hat{k} \text{C m}

Exercise 1.10

Question: An electric dipole with dipole moment 4 \times 10^{-9} \text{C m} is aligned at 30^\circ with a uniform electric field of magnitude 5 \times 10^4 \text{N C}^{-1}. Calculate the torque.

Solution:

  1. Torque: \tau = p E \sin\theta
  2. Substitute: p = 4 \times 10^{-9}, E = 5 \times 10^4, \theta = 30^\circ
  3. \tau = (4 \times 10^{-9}) \times (5 \times 10^4) \times \sin 30^\circ
    \tau = (2 \times 10^{-4}) \times 0.5 = 1 \times 10^{-4} \text{N m}

Answer: 1 \times 10^{-4} \text{N m}

Exercise 1.11

Question: A polythene piece rubbed with wool has a negative charge of 3 \times 10^{-7} \text{C}. (a) Estimate the number of electrons transferred. (b) Is there a transfer of mass?

Solution:
(a)

  1. n = \frac{\text{Total charge}}{e} = \frac{3 \times 10^{-7}}{1.6 \times 10^{-19}} = 1.875 \times 10^{12} electrons
  2. Electrons are transferred from wool to polythene.

(b) Yes, mass of electrons transferred:
m = n \times m_e = (1.875 \times 10^{12}) \times (9.1 \times 10^{-31}) = 1.71 \times 10^{-18} \text{kg}

Answer: (a) 1.875 \times 10^{12} electrons (wool to polythene), (b) 1.71 \times 10^{-18} \text{kg}

Exercise 1.12

(a)

  1. F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}
  2. F = 9 \times 10^9 \times \frac{(6.5 \times 10^{-7})^2}{(0.5)^2} = 1.52 \times 10^{-2} \text{N}

(b) If charge doubled and distance halved:
F_{\text{new}} = \frac{(2q)(2q)}{(r/2)^2} = 16 \times F = 16 \times 1.52 \times 10^{-2} = 0.243 \text{N}

Answer: (a) 1.52 \times 10^{-2} \text{N}, (b) 0.243 \text{N}

Exercise 1.13

Question: Figure 1.30 shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Solution:

  • Positive charges deflect with the field, negative charges against it.
  • The particle with the sharpest curvature has the highest charge-to-mass ratio (q/m).

(Without the figure, a general answer is provided)

Exercise 1.14

Question: Consider \vec{E} = 3 \times 10^3 \hat{i} \text{N/C}. (a) Flux through a square of side 10 cm parallel to the yz plane? (b) Flux if normal makes 60^\circ with x-axis?

Solution:
(a)

  1. Area S = (0.1)^2 = 0.01 \text{m}^2
  2. Flux \phi = \vec{E} \cdot \vec{S} = E S \cos 0^\circ = (3 \times 10^3) \times 0.01 = 30 \text{N m}^2/\text{C}

(b)

  1. \phi = E S \cos 60^\circ = (3 \times 10^3) \times 0.01 \times 0.5 = 15 \text{N m}^2/\text{C}

Answer: (a) 30 \text{N m}^2/\text{C}, (b) 15 \text{N m}^2/\text{C}

Exercise 1.15

Question: Net flux of the same field through a cube of side 20 cm with faces parallel to coordinate planes?

Solution:

  • In a uniform field, the net flux through any closed surface is zero.
  • Reason: Equal flux enters and exits the cube.

Answer: 0

Exercise 1.16

Question: Net outward flux through a black box is 8.0 \times 10^3 \text{N m}^2/\text{C}. (a) Net charge inside? (b) If flux were zero, could you conclude no charges inside?

Solution:
(a)

  1. Gauss’s Law: \phi = \frac{q}{\epsilon_0}
  2. q = \phi \epsilon_0 = (8.0 \times 10^3) \times (8.854 \times 10^{-12}) = 7.08 \times 10^{-8} \text{C}

(b) No, there could be equal positive and negative charges inside (net charge zero).

Answer: (a) 7.08 \times 10^{-8} \text{C}, (b) No

Exercise 1.17

Question: A point charge +10 \muC is 5 cm above the centre of a square of side 10 cm. Find flux through the square.

Solution:

  1. Imagine the square as one face of a cube of side 10 cm with the charge at the centre.
  2. By symmetry, flux through each face is equal.
  3. Total flux through cube: \phi_{\text{cube}} = \frac{q}{\epsilon_0} = \frac{10 \times 10^{-6}}{8.854 \times 10^{-12}} = 1.13 \times 10^6 \text{N m}^2/\text{C}
  4. Flux through one face: \phi_{\text{square}} = \frac{1}{6} \phi_{\text{cube}} \approx 1.88 \times 10^5 \text{N m}^2/\text{C}

Answer: 1.88 \times 10^5 \text{N m}^2/\text{C}

Exercise 1.18

Question: A point charge of 2.0 \muC at the centre of a cubic Gaussian surface 9.0 cm on edge. Net flux?

Solution:

  1. Gauss’s Law: \phi = \frac{q}{\epsilon_0}
  2. \phi = \frac{2.0 \times 10^{-6}}{8.854 \times 10^{-12}} = 2.26 \times 10^5 \text{N m}^2/\text{C}

Answer: 2.26 \times 10^5 \text{N m}^2/\text{C}

Exercise 1.19

Question: A point charge causes a flux of -1.0 \times 10^3 \text{N m}^2/\text{C} through a spherical Gaussian surface of 10.0 cm radius. (a) Flux if radius doubled? (b) Value of charge?

Solution:
(a) Flux depends only on enclosed charge, not radius. So, flux remains -1.0 \times 10^3 \text{N m}^2/\text{C}.

(b)

  1. q = \phi \epsilon_0 = (-1.0 \times 10^3) \times (8.854 \times 10^{-12}) = -8.854 \times 10^{-9} \text{C}

Answer: (a) -1.0 \times 10^3 \text{N m}^2/\text{C}, (b) -8.85 \text{nC}

Exercise 1.20

Question: A conducting sphere of radius 10 cm has an unknown charge. Electric field 20 cm from centre is 1.5 \times 10^3 \text{N/C} inward. Net charge on sphere?

Solution:

  1. For a conducting sphere: E = \frac{1}{4\pi\epsilon_0} \frac{|q|}{r^2}
  2. 1.5 \times 10^3 = 9 \times 10^9 \times \frac{|q|}{(0.2)^2}
  3. |q| = \frac{1.5 \times 10^3 \times 0.04}{9 \times 10^9} = 6.67 \times 10^{-9} \text{C}
  4. Since field is inward, charge is negative.

Answer: -6.67 \text{nC}

Exercise 1.21

Question: A uniformly charged conducting sphere of 2.4 m diameter has \sigma = 80.0 \mu\text{C/m}^2. (a) Charge on sphere? (b) Total flux leaving the surface?

Solution:
(a)

  1. Radius R = 1.2 \text{m}
  2. Surface area = 4\pi R^2 = 4\pi (1.2)^2 = 18.1 \text{m}^2
  3. Charge q = \sigma \times \text{Area} = (80.0 \times 10^{-6}) \times 18.1 = 1.45 \times 10^{-3} \text{C}

(b)

  1. \phi = \frac{q}{\epsilon_0} = \frac{1.45 \times 10^{-3}}{8.854 \times 10^{-12}} = 1.64 \times 10^8 \text{N m}^2/\text{C}

Answer: (a) 1.45 \text{mC}, (b) 1.64 \times 10^8 \text{N m}^2/\text{C}

Exercise 1.22

Question: An infinite line charge produces E = 9 \times 10^4 \text{N/C} at 2 cm. Calculate \lambda.

Solution:

  1. For infinite line charge: E = \frac{\lambda}{2\pi\epsilon_0 r}
  2. \lambda = E \cdot 2\pi\epsilon_0 r
  3. \lambda = (9 \times 10^4) \times 2\pi \times (8.854 \times 10^{-12}) \times 0.02
    \lambda \approx 1.0 \times 10^{-7} \text{C/m} = 0.1 \mu\text{C/m}

Answer: 0.1 \mu\text{C/m}

Exercise 1.23

Question: Two large parallel plates have surface charge densities +17.0 \times 10^{-22} \text{C/m}^2 and -17.0 \times 10^{-22} \text{C/m}^2 on inner faces. Find E in (a) outer region of first plate, (b) outer region of second plate, (c) between plates.

Solution:

  • For two parallel plates with equal and opposite charges:
  • Field outside the plates = 0
  • Field between plates: E = \frac{\sigma}{\epsilon_0}

(a) E = 0
(b) E = 0
(c) E = \frac{17.0 \times 10^{-22}}{8.854 \times 10^{-12}} = 1.92 \times 10^{-10} \text{N/C}

Answer: (a) 0, (b) 0, (c) 1.92 \times 10^{-10} \text{N/C}

20 MCQs with Answers: Electric Charges and Fields

JEE‑oriented MCQs from NCERT Class 12 Physics Chapter 1 (Electric Charges and Fields).

  1. Quantisation check
    A metal sphere carries charge q = 7.2\times 10^{-8},\text{C}. Which statement is most accurate?
    A. q/e may be any real number
    B. q must be an integer multiple of 2e
    C. q/e is an integer, but q can appear continuous at macroscopic scale
    D. Quantisation holds only for insulators
  2. Conservation in rubbing
    When a glass rod is rubbed with silk, which process best describes what happens?
    A. Protons flow from silk to glass; total charge increases
    B. Electrons flow from glass to silk; total charge is conserved
    C. Protons flow from glass to silk; total charge is conserved
    D. Electrons flow from silk to glass; total charge decreases
  3. Coulomb vector form
    Charges q_1 at \mathbf r_1 and q_2 at \mathbf r_2. Force on q_2 due to q_1 is
    A. \displaystyle \mathbf F_{21}=\frac{1}{4\pi\varepsilon_0},q_1q_2,\frac{\mathbf r_2-\mathbf r_1}{\lVert \mathbf r_2-\mathbf r_1\rVert^3}
    B. \displaystyle \mathbf F_{21}=\frac{1}{4\pi\varepsilon_0},q_1q_2,\frac{\mathbf r_1-\mathbf r_2}{\lVert \mathbf r_1-\mathbf r_2\rVert^2}
    C. \displaystyle \mathbf F_{21}=k(q_1+q_2)\frac{\mathbf r_2-\mathbf r_1}{\lVert \cdot\rVert^3}
    D. \displaystyle \mathbf F_{21}=\frac{1}{4\pi\varepsilon_0},\frac{q_1q_2}{r^2},\hat{\mathbf r}<em>{12} with \hat{\mathbf r}</em>{12} from 2 to 1
  4. Superposition
    Which statement correctly expresses electrostatic superposition?
    A. Forces add scalarly; direction is irrelevant
    B. Each pairwise Coulomb force is unaffected by other charges; vector sum gives net force
    C. A genuine three-body force appears for three charges
    D. Net force depends only on total enclosed charge regardless of positions
  5. Force ratio scale
    For an electron–proton pair separated by r, the ratio |F_e|/|F_g| equals
    A. \dfrac{k e^2}{G m_em_p} (independent of r)
    B. \dfrac{k e^2}{G m_em_p r^2}
    C. \dfrac{k e^2 r^2}{G m_em_p}
    D. Depends only on \varepsilon_0
  6. Field definition
    Best operational definition of electric field at a point is
    A. \mathbf E=\mathbf F/q for any test charge q
    B. \mathbf E=\lim_{q\to 0}\mathbf F/q to avoid disturbing sources
    C. \mathbf E=\nabla V by definition for all cases
    D. \mathbf E equals force per unit mass
  7. Field lines
    Which property of electrostatic field lines is correct?
    A. They can intersect where fields superpose
    B. They form closed loops in charge-free space
    C. They never cross; local density indicates |\mathbf E|
    D. They start on negative and end on positive charges
  8. Flux element
    Electric flux through a small oriented area d\mathbf S with outward normal is
    A. d\Phi=E,dS,\sin\theta
    B. d\Phi=\mathbf E\cdot d\mathbf S=E,dS,\cos\theta
    C. d\Phi=E,dS for all orientations
    D. d\Phi=0 for closed surfaces always
  9. Dipole far-field (axial)
    At a distant axial point (r\gg a) of a dipole with moment p, the field magnitude is
    A. \dfrac{p}{4\pi\varepsilon_0 r^2}
    B. \dfrac{2p}{4\pi\varepsilon_0 r^3}
    C. \dfrac{p}{2\pi\varepsilon_0 r}
    D. \dfrac{2p}{4\pi\varepsilon_0 r^2}
  10. Dipole in uniform field
    A rigid electric dipole \mathbf p in a uniform \mathbf E experiences
    A. Net force pE and zero torque
    B. Zero force and zero torque for any orientation
    C. Zero net force and torque \boldsymbol\tau=\mathbf p\times\mathbf E
    D. Net force toward higher |\mathbf E| if aligned
  11. Gauss’s law (global)
    Total electric flux through any closed surface equals
    A. (q_{\text{inside}}+q_{\text{outside}})/\varepsilon_0
    B. q_{\text{encl}}/\varepsilon_0
    C. Zero if the conductor is neutral
    D. Depends on shape for the same enclosed charge
  12. Infinite line charge
    For an infinitely long straight wire with uniform linear density \lambda, the field at radial distance r is
    A. E=\dfrac{\lambda}{4\pi\varepsilon_0 r^2}
    B. E=\dfrac{\lambda}{2\pi\varepsilon_0 r}
    C. E=\dfrac{2\lambda}{4\pi\varepsilon_0 r^2}
    D. E=\dfrac{\lambda r}{2\pi\varepsilon_0}
  13. Infinite plane sheet
    A uniformly charged infinite plane sheet with surface density \sigma produces field magnitude
    A. |E|=\sigma/(2\varepsilon_0) on each side, normal to the sheet
    B. |E|=\sigma/\varepsilon_0 decreasing with distance
    C. |E|=2\sigma/\varepsilon_0 only near edges
    D. |E|=0 inside a conductor only
  14. Spherical shell
    For a thin uniformly charged spherical shell (total charge q, radius R), the field is
    A. E=0 for r>R and E=\dfrac{q}{4\pi\varepsilon_0 r^2} for r<R
    B. E=\dfrac{q}{4\pi\varepsilon_0 r^2} for r>R and E=0 for r<R
    C. Constant everywhere
    D. Non-zero only at the centre
  15. Continuous distribution integral
    For a volume charge density \rho(\mathbf r'), the field at \mathbf r is
    A. \mathbf E(\mathbf r)=\dfrac{1}{4\pi\varepsilon_0}\displaystyle\int \rho(\mathbf r),\dfrac{\mathbf r-\mathbf r'}{\lVert\cdot\rVert^2},d\tau'
    B. \mathbf E(\mathbf r)=\dfrac{1}{4\pi\varepsilon_0}\displaystyle\int \rho(\mathbf r'),\dfrac{\mathbf r-\mathbf r'}{\lVert\mathbf r-\mathbf r'\rVert^3},d\tau'
    C. \mathbf E(\mathbf r)=\displaystyle\int \rho(\mathbf r'),\dfrac{\mathbf r'-\mathbf r}{\lVert\cdot\rVert^3},d\tau' (no constant)
    D. \mathbf E(\mathbf r)=-\nabla\cdot \rho(\mathbf r')
  16. Flux in non-uniform field
    A cube of side a occupies 0\le x,y,z\le a. If \mathbf E=(\alpha x^2),\hat{\mathbf i}, the net outward flux through the cube is
    A. 0
    B. \alpha a^3
    C. 2\alpha a^3/2
    D. \alpha a^4
  17. Atom-like model
    A point charge +Ze at the centre of a sphere of radius R is embedded in a uniform negative charge -Ze spread throughout the sphere. For r<R, the radial field magnitude E_r is
    A. 0
    B. \dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze}{r^2}
    C. \dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze,r}{R^3}
    D. \dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze,R}{r^2}
  18. Dipole in gradient field
    A permanent dipole aligned with \nabla|\mathbf E| placed in a non-uniform field experiences
    A. Zero net force and nonzero torque
    B. Force toward weaker field, torque zero
    C. Force toward stronger field and torque 0 when aligned
    D. Neither force nor torque
  19. Sharing + distance change
    Identical insulated spheres A and B initially carry +Q each and are far apart. A is touched by an identical uncharged sphere C and B by an identical uncharged sphere D. After removing C and D, A and B are brought to half the original separation. The force between A and B becomes
    A. Four times the original
    B. Unchanged
    C. One fourth of the original
    D. Doubles
  20. Flux by symmetry
    A point charge +q is located a distance a above the centre of a square of side a. The electric flux through the square is
    A. q/\varepsilon_0
    B. q/(6\varepsilon_0)
    C. q/(4\varepsilon_0)
    D. Depends on a in a non-trivial way

Answer Key with Explanations

  1. C. Charge is quantised: q=ne with integer n. For macroscopic n, q appears continuous, but quantisation remains valid.
  2. B. Electrons transfer from glass to silk; glass loses electrons and becomes positive. Net charge of the isolated system is conserved.
  3. A. The correct vector form uses \mathbf r_2-\mathbf r_1 divided by its cube; it points along the line joining the charges with the correct inverse-square magnitude.
  4. B. In electrostatics, each pairwise Coulomb force is independent of other charges; the net is the vector sum (no three-body term).
  5. A. Both forces vary as 1/r^2, hence the ratio \dfrac{k e^2}{G m_em_p} is independent of r and extremely large.
  6. B. Using \mathbf E=\lim_{q\to 0}\mathbf F/q ensures the test charge does not disturb the source configuration.
  7. C. Field lines never cross (unique direction). Their density indicates field strength qualitatively.
  8. B. By definition, d\Phi=\mathbf E\cdot d\mathbf S=E,dS,\cos\theta.
  9. B. On the axis and for r\gg a, E\simeq \dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{r^3}.
  10. C. In a uniform field, a dipole experiences zero net force but a torque \boldsymbol\tau=\mathbf p\times\mathbf E.
  11. B. Gauss’s law: total flux through a closed surface equals q_{\text{encl}}/\varepsilon_0.
  12. B. From Gauss’s law with cylindrical symmetry: E=\lambda/(2\pi\varepsilon_0 r), radial.
  13. A. For an infinite sheet, |E|=\sigma/(2\varepsilon_0) on each side, uniform and normal to the sheet.
  14. B. A thin charged shell behaves as a point charge externally; internally E=0 (for r<R).
  15. B. Coulomb integral for a continuous volume charge: \mathbf E(\mathbf r)=\dfrac{1}{4\pi\varepsilon_0}\int \rho(\mathbf r')\dfrac{\mathbf r-\mathbf r'}{\lVert \mathbf r-\mathbf r'\rVert^3},d\tau'.
  16. D. Only the x-faces contribute: \Phi=\big[\alpha x^2\big]_{0}^{a}a^2=\alpha a^4.
  17. C. Enclosed negative charge at radius r is -Ze(r^3/R^3); net enclosed =Ze[1-r^3/R^3] gives E_r=(1/4\pi\varepsilon_0)(Ze,r/R^3).
  18. C. In a non-uniform field, a dipole aligned with \mathbf E is pulled toward stronger field regions; torque is zero when aligned or anti-aligned.
  19. B. After sharing, A and B each have Q/2. Halving separation multiplies F by 4, but (Q/2)^2=Q^2/4 cancels it, leaving the force unchanged.
  20. B. By symmetry (imagining a cube with the charge at its centre), each of the six faces receives flux q/(6\varepsilon_0).

CBSE PYQs: Electric Charges and Fields

Question 1: Charge Quantization (MCQ)

Source : CBSE Class 12, 2022-25, Paper Code 55(B), Question 1 (Section A), 1 Mark

Question: A charge of -1 \mu \text{C} on a body represents (A) loss of 6.25 \times 10^{12} electrons by the body. (B) gain of 6.25 \times 10^{12} electrons by the body. (C) gain of 1.6 \times 10^{13} electrons by the body. (D) loss of 1.6 \times 10^{13} electrons by the body.

Solution:

  1. Identify the given charge and sign. The charge is Q = -1 \mu \text{C}. Since the charge is negative, the body has gained electrons.
  2. Convert the charge magnitude to Coulombs:

        \[Q = 1 \mu \text{C} = 1 \times 10^{-6} \text{ C}\]

  3. Use the principle of quantisation of charge, Q = n |e|, where n is the number of electrons and |e| is the magnitude of elementary charge (1.6 \times 10^{-19} \text{ C}).

        \[n = \frac{Q}{|e|}\]

  4. Substitute the values:

        \[n = \frac{1 \times 10^{-6} \text{ C}}{1.6 \times 10^{-19} \text{ C}}\]

        \[n = \frac{1}{1.6} \times 10^{13} = 0.625 \times 10^{13}\]

        \[n = 6.25 \times 10^{12} \text{ electrons}\]

  5. Since the charge is negative, this represents a gain of electrons.

Final answer: (B) gain of 6.25 \times 10^{12} electrons by the body.

Question 2: Electric Dipole Work

Source: CBSE Class 12, Paper Code 2155/4/3, 2022-25, Question 1 (Section A), 1 Mark

Question: An electric dipole of dipole moment \vec{p} is kept in a uniform electric field \vec{E}. The amount of work done to rotate it from the position of stable equilibrium to that of unstable equilibrium will be (A) 2 \text{ pE} (B) -2 \text{ pE} (C) \text{pE} (D) zero

Solution:

  1. State the formula for the potential energy (U) of a dipole \vec{p} in a uniform electric field \vec{E}:

        \[U = - \vec{p} \cdot \vec{E} = - \text{pE} \cos \theta\]

  2. Determine the initial position (stable equilibrium). This occurs when the dipole moment is aligned with the field (\theta_i = 0^\circ).

        \[U_i = - \text{pE} \cos(0^\circ) = - \text{pE}\]

  3. Determine the final position (unstable equilibrium). This occurs when the dipole moment is anti-parallel to the field (\theta_f = 180^\circ).

        \[U_f = - \text{pE} \cos(180^\circ) = - \text{pE} (-1) = + \text{pE}\]

  4. Calculate the work done (W) to rotate the dipole:

        \[W = U_f - U_i\]

        \[W = (\text{pE}) - (-\text{pE})\]

        \[W = \text{pE} + \text{pE} = 2 \text{ pE}\]

Final answer: (A) 2 \text{ pE}

Question 3: Force Due to Infinite Line Charge

Source: CBSE Class 12, Paper Code 2155/4/3, 2022-25, Question 2 (Section A), 1 Mark

Question: An infinite long straight wire having a charge density \lambda is kept along y'y axis in x\text{-}y plane. The Coulomb force on a point charge q at a point P(x, 0) will be (A) attractive and \frac{q\lambda}{2\pi\epsilon_0 x} (B) repulsive and \frac{q\lambda}{2\pi\epsilon_0 x} (C) attractive and \frac{q\lambda}{\pi\epsilon_0 x} (D) repulsive and \frac{q\lambda}{\pi\epsilon_0 x}

Solution:

  1. State the electric field (E) produced by an infinitely long straight wire with linear charge density \lambda at a perpendicular distance x (using Gauss’s Law):

        \[E = \frac{\lambda}{2\pi\epsilon_0 x}\]

  2. The force (F) exerted on a point charge q placed in this field is given by F = qE.

        \[F = q \left( \frac{\lambda}{2\pi\epsilon_0 x} \right) = \frac{q\lambda}{2\pi\epsilon_0 x}\]

  3. Determine the nature/direction of the force. The charge density \lambda is assumed positive (standard assumption unless otherwise specified, leading to an outward field), and the charge q is also usually assumed positive in the general case where direction is not provided, or simply relying on the magnitude calculation. Since the force formula is always proportional to q \lambda, if q and \lambda have the same sign (both positive or both negative), the force will be repulsive (away from the wire, along the positive x-axis).
  4. Since the field formula matches Option (B), and repulsion is the case for charges of the same sign:

Final answer: (B) repulsive and \frac{q\lambda}{2\pi\epsilon_0 x}

Question 4: Electric Field Calculation (Superposition Principle)

Source: CBSE Class 12, Paper Code 2255/5/2, 2022-25, Question 17 (b) OR (Section B), 2 Marks

Question: Three point charges, 1 \text{ pC} each, are kept at the vertices of an equilateral triangle of side 10 \text{ cm}. Find the net electric field at the centroid of triangle.

Solution:

  1. Define the system: Three identical charges (q) placed at the vertices (A, B, C) of an equilateral triangle with side a = 10 \text{ cm}. We need the net electric field at the centroid (G).

        \[q_A = q_B = q_C = q = 1 \text{ pC}\]

  2. Determine the distance (r) from each vertex to the centroid (G). For an equilateral triangle, r is the same for all vertices due to symmetry.

        \[r = \frac{a}{\sqrt{3}}\]

  3. Calculate the magnitude of the electric field (E_0) produced by each individual charge at the centroid G:

        \[E_A = E_B = E_C = E_0 = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}\]

  4. Determine the direction of the fields. Since all charges are positive, the field due to each charge points radially outwards, away from the charge, towards the opposite side of the triangle.
  5. Apply the superposition principle. At the centroid G, the three electric field vectors (\vec{E}_A, \vec{E}_B, \vec{E}_C) have equal magnitude (E_0) and are directed outwards, making 120^\circ angles with each other.

        \[|\vec{E}_A| = |\vec{E}_B| = |\vec{E}_C| = E_0\]

  6. The vector sum of three equal magnitude vectors separated by 120^\circ is zero:

        \[\vec{E}_{\text{net}} = \vec{E}_A + \vec{E}_B + \vec{E}_C = 0\]

Final answer: The net electric field at the centroid of the triangle is zero.

Question 5: Electric Field Calculation (Superposition Principle)

Source: CBSE Class 12, Paper Code 2255/5/1, 2022-25, Question 17 (a) (Section B), 2 Marks

Question: Four point charges of 1 \mu\text{C}, -2 \mu\text{C}, 1 \mu\text{C} and -2 \mu\text{C} are placed at the corners \text{A}, \text{B}, \text{C} and \text{D} respectively, of a square of side 30 \text{ cm}. Find the net force acting on a charge of 4 \mu\text{C} placed at the centre of the square.

Solution:

  1. Define the given charges and geometry: q_A = 1 \mu\text{C} q_B = -2 \mu\text{C} q_C = 1 \mu\text{C} q_D = -2 \mu\text{C} q_O = 4 \mu\text{C} (Charge at the centre O) Side of square a = 30 \text{ cm} = 0.30 \text{ m}.
  2. Determine the distance (r) from each corner to the centre (O). This distance is half the diagonal (d).

        \[d = a \sqrt{2}\]

        \[r = \frac{d}{2} = \frac{a \sqrt{2}}{2} = \frac{a}{\sqrt{2}}\]

        \[r = \frac{0.30 \text{ m}}{\sqrt{2}} \approx 0.212 \text{ m}\]

  3. Apply the principle of superposition. The net force \vec{F}<em>{\text{net}} on q_O is the vector sum of forces from A, B, C, and D.

        \[\vec{F}</em>{\text{net}} = \vec{F}<em>{OA} + \vec{F}</em>{OB} + \vec{F}<em>{OC} + \vec{F}</em>{OD}\]

  4. Compare forces exerted by opposite pairs (A and C, B and D):
    • Force due to A (q_A = 1 \mu\text{C}) and C (q_C = 1 \mu\text{C}): Since q_A = q_C and the distance r is the same, the magnitudes of force F_{OA} and F_{OC} are equal. Since both q_A and q_C are positive, they both repel the central positive charge q_O. F_{OA} points away from A (towards C), and F_{OC} points away from C (towards A). Since they are equal in magnitude and opposite in direction:

          \[\vec{F}<em>{AC} = \vec{F}</em>{OA} + \vec{F}_{OC} = 0\]

    • Force due to B (q_B = -2 \mu\text{C}) and D (q_D = -2 \mu\text{C}): Since q_B = q_D and the distance r is the same, the magnitudes of force F_{OB} and F_{OD} are equal. Since q_B and q_D are negative, they both attract the central positive charge q_O. F_{OB} points towards B, and F_{OD} points towards D. Since they are equal in magnitude and opposite in direction:

          \[\vec{F}<em>{BD} = \vec{F}</em>{OB} + \vec{F}_{OD} = 0\]

  5. Calculate the net force:

        \[\vec{F}<em>{\text{net}} = \vec{F}</em>{AC} + \vec{F}_{BD} = 0 + 0 = 0\]

Final answer: The net force acting on the charge placed at the centre of the square is zero.

(Note: Questions 18 in 55/5/2 and 17(a) in 55/5/2 are identical to Question 15, confirming the zero net force due to charge and geometric symmetry.)

Question 6: Electric Field/Force (Work Done)

Source: CBSE Class 12, Paper Code 55/2/2, 2022-25, Question 16 (Assertion/Reason) (Section A), 1 Mark

Question: Assertion (A) : Work done in moving a charge around a closed path, in an electric field is always zero. Reason (R) : Electrostatic force is a conservative force.

Solution:

  1. Analyze Assertion (A): Work done W in moving a charge q between two points A and B is defined by the negative change in potential energy, W = - \Delta U. For a closed path, the final point is the initial point, so \Delta U = 0. Hence, W=0. Assertion (A) is true.
  2. Analyze Reason (R): A conservative force is one where the work done depends only on the initial and final positions, and is zero around a closed path. Electrostatic force is, by definition, a conservative force. Reason (R) is true.
  3. Evaluate the relationship: The assertion (work done is zero in a closed path) is a direct consequence of the electrostatic force being conservative.

Final answer: Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Question 7: Gauss’s Law (Electric Flux through a Cube Face)

Source: CBSE Class 12, Paper Code 55/5/2, 2022-25, Question 1 (Section A), 1 Mark

Question: A charge Q is placed at the centre of a cube. The electric flux through one if its face is (A) \frac{Q}{\epsilon_0} (B) \frac{Q}{6\epsilon_0} (C) \frac{Q}{8\epsilon_0} (D) \frac{Q}{3\epsilon_0}

Solution:

  1. State Gauss’s Law: The total electric flux (\Phi_{\text{total}}) passing through a closed surface (Gaussian surface) is equal to \frac{1}{\epsilon_0} times the net charge (Q) enclosed within that surface.

        \[\Phi_{\text{total}} = \frac{Q_{\text{enclosed}}}{\epsilon_0}\]

  2. For a charge Q placed exactly at the centre of a cube, the cube acts as the closed Gaussian surface. Thus, the total flux through the entire cube is:

        \[\Phi_{\text{cube}} = \frac{Q}{\epsilon_0}\]

  3. Due to the perfect symmetry of the cube, the total flux must be equally distributed across its six identical faces.

        \[\Phi_{\text{one face}} = \frac{\Phi_{\text{cube}}}{6}\]

  4. Substitute the total flux:

        \[\Phi_{\text{one face}} = \frac{Q}{6\epsilon_0}\]

Final answer: (B) \frac{Q}{6\epsilon_0}

Question 8: Gauss’s Law Application (Infinite Wire)

Source: CBSE Class 12, Paper Code 55/2/1, 2022-25, Question 31 (a)(i) OR (Section D), 5 Marks

Question: Use Gauss’ law to obtain an expression for the electric field due to an infinitely long thin straight wire with uniform linear charge density \lambda.

Solution:

  1. Choose Gaussian Surface: Consider an infinitely long, thin straight wire having a uniform linear charge density \lambda. By symmetry, the electric field \vec{E} must be radial (perpendicular to the wire) and constant in magnitude at any fixed distance r from the wire. We choose a co-axial cylindrical Gaussian surface of radius r and arbitrary length l.
  2. Apply Gauss’s Law: Gauss’s Law states:

        \[\Phi = \oint \vec{E} \cdot d\vec{S} = \frac{Q_{\text{enclosed}}}{\epsilon_0}\]

  3. Calculate Total Flux (\Phi): The cylindrical surface has three parts: the curved surface (S_1) and the two flat circular end caps (S_2 and S_3).

        \[\Phi = \oint_{S_1} \vec{E} \cdot d\vec{S} + \oint_{S_2} \vec{E} \cdot d\vec{S} + \oint_{S_3} \vec{E} \cdot d\vec{S}\]

    • For the end caps (S_2, S_3): \vec{E} is perpendicular to d\vec{S} (parallel to the surface), so \theta = 90^\circ. \cos(90^\circ) = 0. The flux through the end caps is zero.
    • For the curved surface (S_1): \vec{E} is parallel to d\vec{S} (radially outwards), so \theta = 0^\circ. \cos(0^\circ) = 1. Since E is constant on the curved surface:

          \[\Phi = E \oint_{S_1} dS = E (2\pi r l)\]

  4. Calculate Enclosed Charge (Q_{\text{enclosed}}): The charge enclosed within the length l of the wire is:

        \[Q_{\text{enclosed}} = \lambda l\]

  5. Equate Flux and Charge: Equating the expressions from steps 2, 3, and 4:

        \[E (2\pi r l) = \frac{\lambda l}{\epsilon_0}\]

  6. Solve for E: Rearranging the equation to find the electric field E:

        \[E = \frac{\lambda l}{2\pi \epsilon_0 r l}\]

        \[E = \frac{\lambda}{2\pi \epsilon_0 r}\]

    The electric field \vec{E} is directed radially outward (if \lambda is positive).

Final answer: The expression for the electric field is E = \frac{\lambda}{2\pi \epsilon_0 r}.

Question 9: Electric Force and Kinematics (Circular Motion)

Source: CBSE Class 12, Paper Code 55/2/1, 2022-25, Question 31 (a)(ii) (Section D), 5 Marks

Question: An infinitely long positively charged straight wire has a linear charge density \lambda. An electron is revolving in a circle with a constant speed v such that the wire passes through the centre, and is perpendicular to the plane, of the circle. Find the kinetic energy of the electron in terms of magnitudes of its charge and linear charge density \lambda on the wire.

Solution:

  1. Identify the forces: The electron (charge e) revolves around the wire at a distance r. The centripetal force required for circular motion is provided by the electrostatic force (F_e) exerted by the charged wire on the electron. Since the wire is positively charged (\lambda is positive) and the electron is negative (e), the force is attractive and directed radially towards the wire (i.e., acting as the centripetal force).
  2. State the Electric Field (E): From Gauss’s law (Q8), the magnitude of the electric field at distance r from the wire is:

        \[E = \frac{\lambda}{2\pi \epsilon_0 r}\]

  3. Equate Forces: The electrostatic force F_e = eE must equal the centripetal force F_c = \frac{m_e v^2}{r}:

        \[F_e = F_c\]

        \[e E = \frac{m_e v^2}{r}\]

  4. Substitute E and Solve for Kinetic Energy (K): Kinetic Energy is K = \frac{1}{2} m_e v^2. We substitute the expression for E into the force equation:

        \[e \left( \frac{\lambda}{2\pi \epsilon_0 r} \right) = \frac{m_e v^2}{r}\]

  5. Rearrange the equation to isolate m_e v^2:

        \[\frac{e \lambda}{2\pi \epsilon_0 r} = \frac{m_e v^2}{r}\]

        \[m_e v^2 = \frac{e \lambda}{2\pi \epsilon_0}\]

  6. Find the kinetic energy K:

        \[K = \frac{1}{2} m_e v^2 = \frac{1}{2} \left( \frac{e \lambda}{2\pi \epsilon_0} \right)\]

        \[K = \frac{e \lambda}{4\pi \epsilon_0}\]

Final answer: The kinetic energy of the electron is K = \frac{e \lambda}{4\pi \epsilon_0}.

Question 10: Work Done in Non-Uniform Electric Field

Source: CBSE Class 12, Paper Code 2155/4/1, 2022-25, Question 23 (Section C), 3 Marks

Question: The electric field in a region is given by

    \[\vec{E} = (10x + 4) \hat{i}\]

where x is in m and E is in N/C. Calculate the amount of work done in taking a unit charge from (i) (5 \text{ m}, 0) to (10 \text{ m}, 0) (ii) (5 \text{ m}, 0) to (5 \text{ m}, 10 \text{ m})

Solution: The work done W in moving a charge q along a path L in an electric field \vec{E} is given by:

    \[W = - q \int_L \vec{E} \cdot d\vec{l}\]

Here, we are moving a unit charge, so q = 1 \text{ C}. The displacement vector d\vec{l} is given by d\vec{l} = dx \hat{i} + dy \hat{j}. Since \vec{E} = (10x + 4) \hat{i}, the dot product is:

    \[\vec{E} \cdot d\vec{l} = (10x + 4) \hat{i} \cdot (dx \hat{i} + dy \hat{j}) = (10x + 4) dx\]

(i) Work done from (5 \text{ m}, 0) to (10 \text{ m}, 0):
  1. Set up the integral for the work done W:

        \[W = - (1 \text{ C}) \int_{x=5}^{x=10} (10x + 4) dx\]

  2. Integrate the expression:

        \[W = - \left[ \frac{10x^2}{2} + 4x \right]<em>{5}^{10} = - \left[ 5x^2 + 4x \right]</em>{5}^{10}\]

  3. Substitute the limits:

        \[W = - \left[ (5(10)^2 + 4(10)) - (5(5)^2 + 4(5)) \right]\]

        \[W = - \left[ (500 + 40) - (125 + 20) \right]\]

        \[W = - \left[ 540 - 145 \right]\]

        \[W = - 395 \text{ J}\]

Final answer (i): -395 \text{ J}

(ii) Work done from (5 \text{ m}, 0) to (5 \text{ m}, 10 \text{ m}):
  1. Analyze the path: The initial position is (x_i, y_i) = (5 \text{ m}, 0) and the final position is (x_f, y_f) = (5 \text{ m}, 10 \text{ m}). Along this path, the displacement is purely vertical (dy \neq 0), meaning dx = 0.
  2. Calculate the dot product \vec{E} \cdot d\vec{l}:

        \[\vec{E} \cdot d\vec{l} = (10x + 4) dx\]

    Since dx = 0:

        \[\vec{E} \cdot d\vec{l} = 0\]

  3. Calculate the work done:

        \[W = - q \int \vec{E} \cdot d\vec{l} = 0 \text{ J}\]

Final answer (ii): 0 \text{ J}

Question 11: Electric Flux and Net Enclosed Charge in a Non-Uniform Field

Source: CBSE Class 12, Paper Code 2155/4/1, 2022-25, Question 32 (b)(ii) (Section E), 5 Marks

Question: Electric field \vec{E} in a region is given by

    \[\vec{E} = (5x^2 + 2) \hat{i}\]

where \text{E} is in \text{N/C} and x is in meters. A cube of side 10 \text{ cm} is placed in the region as shown in figure. Calculate (1) the electric flux through the cube, and (2) the net charge enclosed by the cube.

Figure missing: (The figure describes a cube placed in the first octant, with one corner at the origin (0, 0, 0) and edges aligned along the x, y, and z axes. Side length a = 10 \text{ cm}.)

Solution: Given a = 10 \text{ cm} = 0.1 \text{ m}. \vec{E} = (5x^2 + 2) \hat{i}. The electric field is uniform only in the sense that it only varies with x.

(1) Calculate the electric flux through the cube (\Phi_{\text{net}})
  1. Analyze flux components: Since the electric field \vec{E} is along the x-axis (\hat{i} direction), flux passes only through the two faces perpendicular to the x-axis (the ‘Left’ face at x=0 and the ‘Right’ face at x=a). The flux through the four remaining faces (parallel to the x-axis) is zero because \vec{E} \cdot d\vec{S} = 0.
  2. Calculate Flux through Left Face (x=0): The area vector \vec{A}<em>{\text{Left}} points along -\hat{i}.

        \[E</em>{\text{Left}} = (5(0)^2 + 2) \hat{i} = 2 \hat{i} \text{ N/C}\]

        \[A = a^2 = (0.1)^2 = 0.01 \text{ m}^2\]

        \[\Phi_{\text{Left}} = \vec{E}<em>{\text{Left}} \cdot \vec{A}</em>{\text{Left}} = (2 \hat{i}) \cdot (-(0.01) \hat{i})\]

        \[\Phi_{\text{Left}} = -0.02 \text{ N m}^2/\text{C}\]

  3. Calculate Flux through Right Face (x=a=0.1 \text{ m}): The area vector \vec{A}<em>{\text{Right}} points along +\hat{i}.

        \[E</em>{\text{Right}} = (5(0.1)^2 + 2) \hat{i} \text{ N/C}\]

        \[E_{\text{Right}} = (5(0.01) + 2) \hat{i} = 2.05 \hat{i} \text{ N/C}\]

        \[\Phi_{\text{Right}} = \vec{E}<em>{\text{Right}} \cdot \vec{A}</em>{\text{Right}} = (2.05 \hat{i}) \cdot ((0.01) \hat{i})\]

        \[\Phi_{\text{Right}} = +0.0205 \text{ N m}^2/\text{C}\]

  4. Calculate Net Flux (\Phi_{\text{net}}):

        \[\Phi_{\text{net}} = \Phi_{\text{Left}} + \Phi_{\text{Right}}\]

        \[\Phi_{\text{net}} = -0.02 \text{ N m}^2/\text{C} + 0.0205 \text{ N m}^2/\text{C}\]

        \[\Phi_{\text{net}} = 0.0005 \text{ N m}^2/\text{C}\]

Final answer (1): The electric flux through the cube is 0.0005 \text{ N m}^2/\text{C}.

(2) Calculate the net charge enclosed by the cube (Q_{\text{enclosed}})
  1. Use Gauss’s Law to relate net flux and enclosed charge:

        \[\Phi_{\text{net}} = \frac{Q_{\text{enclosed}}}{\epsilon_0}\]

        \[Q_{\text{enclosed}} = \Phi_{\text{net}} \epsilon_0\]

  2. Substitute the values (\epsilon_0 = 8.854 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2}):

        \[Q_{\text{enclosed}} = (0.0005 \text{ N m}^2/\text{C}) \times (8.854 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2})\]

        \[Q_{\text{enclosed}} = 5 \times 10^{-4} \times 8.854 \times 10^{-12} \text{ C}\]

        \[Q_{\text{enclosed}} = 44.27 \times 10^{-16} \text{ C}\]

        \[Q_{\text{enclosed}} \approx 4.427 \times 10^{-15} \text{ C}\]

Final answer (2): The net charge enclosed by the cube is approximately 4.427 \times 10^{-15} \text{ C}.

Question 12: Gauss’s Law (Flux through Concentric Shells)

Source: CBSE Class 12, Paper Code 55/2/1, 2022-25, Question 32 (b)(iii) OR (Section E), 5 Marks

Question: A small spherical shell \text{S}_1 has point charges q_1 = -3 \mu\text{C}, q_2 = -2 \mu\text{C} and q_3 = 9 \mu\text{C} inside it. This shell is enclosed by another big spherical shell \text{S}_2. A point charge Q is placed in between the two surfaces \text{S}_1 and \text{S}_2. If the electric flux through the surface \text{S}_2 is four times the flux through surface \text{S}_1, find charge Q.

Solution:

  1. Calculate charge enclosed by \text{S}<em>1 (Q</em>{\text{encl}, 1}):

        \[Q_{\text{encl}, 1} = q_1 + q_2 + q_3\]

        \[Q_{\text{encl}, 1} = (-3 \mu\text{C}) + (-2 \mu\text{C}) + (9 \mu\text{C})\]

        \[Q_{\text{encl}, 1} = 4 \mu\text{C}\]

  2. Calculate flux through \text{S}_1 (\Phi_1): Using Gauss’s Law:

        \[\Phi_1 = \frac{Q_{\text{encl}, 1}}{\epsilon_0} = \frac{4 \mu\text{C}}{\epsilon_0}\]

  3. Calculate charge enclosed by \text{S}<em>2 (Q</em>{\text{encl}, 2}): \text{S}_2 encloses q_1, q_2, q_3, and the charge Q placed between \text{S}<em>1 and \text{S}2.

        \[Q{\text{encl}, 2} = q_1 + q_2 + q_3 + Q\]

        \[Q</em>{\text{encl}, 2} = 4 \mu\text{C} + Q\]

  4. Calculate flux through \text{S}_2 (\Phi_2):

        \[\Phi_2 = \frac{Q_{\text{encl}, 2}}{\epsilon_0} = \frac{4 \mu\text{C} + Q}{\epsilon_0}\]

  5. Use the given condition: The electric flux through \text{S}_2 is four times the flux through \text{S}_1:

        \[\Phi_2 = 4 \Phi_1\]

  6. Solve for Q:

        \[\frac{4 \mu\text{C} + Q}{\epsilon_0} = 4 \left( \frac{4 \mu\text{C}}{\epsilon_0} \right)\]

        \[4 \mu\text{C} + Q = 16 \mu\text{C}\]

        \[Q = 16 \mu\text{C} - 4 \mu\text{C}\]

        \[Q = 12 \mu\text{C}\]

Final answer: The charge Q is 12 \mu\text{C}.

Question 13: Electric Field due to Point Charges (Conceptual/MCQ)

Source: CBSE Class 12, Paper Code 2255/5/2, 2022-25, Question 15 (Assertion/Reason) (Section A), 1 Mark

Question: Assertion (A) : Equal amount of positive and negative charges are distributed uniformly on two halves of a thin circular ring as shown in figure. The resultant electric field at the centre O of the ring is along OC. Reason (R) : It is so because the net potential at O is not zero.

Figure missing: (The figure shows a circular ring centered at O. The upper semi-circle (say A-B-D, moving clockwise) is labeled with positive charges, and the lower semi-circle (say A-C-D) is labeled with negative charges. C is at the bottom, and the line segment OC points downwards).

Solution:

  1. Analyze Assertion (A) (Electric Field): Due to symmetry, the horizontal components of the electric field (E_x) created by the charges cancel out. Both the positive charge on the top half and the negative charge on the bottom half create a net electric field component pointing downwards (along the direction OC).
    • Positive charges repel a test charge down.
    • Negative charges attract a test charge down.
    • Therefore, the resultant electric field at the centre O is along OC (downwards). Assertion (A) is true.
  2. Analyze Reason (R) (Electric Potential): Electric potential (V) is a scalar quantity. Potential at O is the sum of potentials due to positive charges (V_{pos}) and negative charges (V_{neg}). Since the positive charge magnitude (+Q) equals the negative charge magnitude (-Q), and all charges are equidistant (R) from the centre O:

        \[V_{\text{net}} = V_{\text{pos}} + V_{\text{neg}} = \frac{k(+Q)}{R} + \frac{k(-Q)}{R} = 0\]

    The net potential at O is zero. Reason (R) states the net potential is not zero. Reason (R) is false.
  3. Conclusion: Assertion (A) is true, but Reason (R) is false.

Final answer: (C) Assertion (A) is true, but Reason (R) is false.

Question 14: Force/Distance Relation (Coulomb’s Law Graph)

Source: CBSE Class 12, Paper Code 2155/4/2, 2022-25, Question 2 (Section A), 1 Mark

Question: The Coulomb force (F) versus (1/r^2) graphs for two pairs of point charges (q_1 \text{ and } q_2) and (q_2 \text{ and } q_3) are shown in figure. The charge q_2 is positive and has least magnitude. Then (A) q_1 > q_2 > q_3 (B) q_3 > q_2 > q_1 (C) q_3 > q_2 > q_1 (Repeat of B, likely error in source options) (D) q_3 > q_1 > q_2

Figure missing: (The figure shows a graph of Force F (y-axis) vs 1/r^2 (x-axis). A straight line labeled (q_1, q_2) has a positive slope, and a straight line labeled (q_2, q_3) has a negative slope.)

Solution:

  1. Relate F to 1/r^2: Coulomb’s Law magnitude is F = \frac{1}{4\pi\epsilon_0} \frac{|q_a q_b|}{r^2}. This gives a linear relationship between F and 1/r^2.

        \[F = \text{Constant} \times \frac{1}{r^2}\]

    The slope of the F vs 1/r^2 graph is proportional to the magnitude of the product of the charges: \text{Slope} \propto |q_a q_b|.
  2. Analyze the Signs (Attraction/Repulsion):
    • The line for (q_1, q_2) has a positive slope (F increases as 1/r^2 increases, and F is plotted as positive/magnitude). Since q_2 is positive, for the force to be plotted in the positive F region, we assume the graph shows the magnitude of the force. We cannot definitively determine the sign relationship from the visual information provided without specific labeling conventions, but for magnitude comparison, we rely on the slope magnitude.
    • Reinterpreting typical convention: If F represents repulsive force (positive y) and attractive force (negative y). Let’s assume the question implies comparison of magnitudes of slopes \text{Slope} = \frac{1}{4\pi\epsilon_0} |q_a q_b|.
  3. Analyze Slope Magnitudes: Let S_{12} be the slope for (q_1, q_2) and S_{23} be the slope for (q_2, q_3). Based on the visual description (Line (q_2, q_3) appears to have a steeper magnitude of slope than (q_1, q_2) if the graph lines are plotted on the same axes, implying |q_2 q_3| > |q_1 q_2|). Assuming the visual evidence (if present) confirms |S_{23}| > |S_{12}|:

        \[|q_2 q_3| > |q_1 q_2|\]

  4. Use Given Information: We are given that q_2 is positive and has the least magnitude:

        \[|q_2| < |q_1| \text{ and } |q_2| < |q_3|\]

  5. Compare q_1 and q_3: If |q_2 q_3| > |q_1 q_2|, dividing by |q_2| (which is the smallest charge magnitude) yields:

        \[|q_3| > |q_1|\]

  6. Order the charges by magnitude: Combining the constraints:

        \[|q_3| > |q_1| > |q_2|\]

    Therefore, q_3 has the greatest magnitude, followed by q_1, and then q_2.

Final answer: (D) q_3 > q_1 > q_2 (assuming q_i represents the magnitude |q_i|, consistent with the context that q_2 has the least magnitude).

Question 15: Potential Energy of a Dipole (Work Done)

Source: CBSE Class 12, Paper Code 2155/5/2, 2022-25, Question 32 (b)(iii) (in Hindi, refers to Dipole Moment 10^{-30} \text{ Cm} and E = 10^5 \text{ V/m})

Question (Reconstructed from English source 59, Q31(a)(i) OR): The dipole moment of a molecule is 10^{-30} \text{ Cm}. It is placed in an electric field \vec{E} of 10^5 \text{ V/m} such that its axis is along the electric field. The direction of \vec{E} is suddenly changed by 60^\circ at an instant. Find the change in the potential energy of the dipole, at that instant.

Solution:

  1. State the formula for the potential energy (U) of a dipole:

        \[U = - \text{pE} \cos \theta\]

  2. Identify the initial conditions (p, E, \theta_i):

        \[p = 10^{-30} \text{ C m}\]

        \[E = 10^5 \text{ V/m}\]

    Initial angle \theta_i = 0^\circ (axis along the field, stable equilibrium).
  3. Calculate Initial Potential Energy (U_i):

        \[U_i = - \text{pE} \cos(0^\circ) = - \text{pE}\]

        \[U_i = - (10^{-30} \text{ C m}) \times (10^5 \text{ V/m}) = -10^{-25} \text{ J}\]

  4. Identify the final condition (\theta_f): The direction of \vec{E} is suddenly changed by 60^\circ, while the dipole orientation (\vec{p}) remains instantaneously fixed (due to inertia).

        \[\theta_f = 60^\circ\]

  5. Calculate Final Potential Energy (U_f):

        \[U_f = - \text{pE} \cos(60^\circ)\]

        \[U_f = - (10^{-30} \text{ C m}) \times (10^5 \text{ V/m}) \times (0.5)\]

        \[U_f = - 0.5 \times 10^{-25} \text{ J}\]

  6. Calculate the Change in Potential Energy (\Delta U):

        \[\Delta U = U_f - U_i\]

        \[\Delta U = (- 0.5 \times 10^{-25} \text{ J}) - (- 1.0 \times 10^{-25} \text{ J})\]

        \[\Delta U = (- 0.5 + 1.0) \times 10^{-25} \text{ J} = 0.5 \times 10^{-25} \text{ J}\]

Final answer: The change in the potential energy of the dipole is 0.5 \times 10^{-25} \text{ J}.

Question 16: Electric Field (Potential Calculation)

Source: CBSE Class 12, Paper Code 2155/4/2, 2022-25, Question 1 (Section A), 1 Mark

Question: Three point charges, each of charge q are placed on vertices of a triangle \text{ABC}, with \text{AB} = \text{AC} = 5\text{L}, \text{BC} = 6\text{L}. The electrostatic potential at midpoint M of side BC will be (A) \frac{11 q}{48 \pi \epsilon_0 \text{L}} (B) \frac{8 q}{36 \pi \epsilon_0 \text{L}} (C) \frac{5 q}{24 \pi \epsilon_0 \text{L}} (D) \frac{1}{16} \frac{q}{\pi \epsilon_0 \text{L}}

Solution:

  1. Determine Distances to M: Let M be the midpoint of BC. The triangle is isosceles (\text{AB} = \text{AC} = 5\text{L}).
    • Distance from B to M (r_B): \text{BM} = \text{BC}/2 = 6\text{L}/2 = 3\text{L}.
    • Distance from C to M (r_C): \text{CM} = 3\text{L}.
    • Distance from A to M (r_A): AM is the altitude. Using Pythagoras theorem on triangle ABM:

          \[\text{AM}^2 = \text{AB}^2 - \text{BM}^2\]

          \[r_A^2 = (5\text{L})^2 - (3\text{L})^2 = 25\text{L}^2 - 9\text{L}^2 = 16\text{L}^2\]

          \[r_A = 4\text{L}\]

  2. Calculate Potential (V) at M: The potential due to a system of point charges is the algebraic sum of the individual potentials V = \sum V_i.

        \[V_M = V_A + V_B + V_C = \frac{1}{4\pi\epsilon_0} \left( \frac{q_A}{r_A} + \frac{q_B}{r_B} + \frac{q_C}{r_C} \right)\]

    (Given that each charge is q: q_A = q_B = q_C = q)

        \[V_M = \frac{q}{4\pi\epsilon_0} \left( \frac{1}{4\text{L}} + \frac{1}{3\text{L}} + \frac{1}{3\text{L}} \right)\]

  3. Simplify the expression:

        \[V_M = \frac{q}{4\pi\epsilon_0 \text{L}} \left( \frac{1}{4} + \frac{2}{3} \right)\]

    Find the common denominator (12):

        \[\frac{1}{4} + \frac{2}{3} = \frac{3}{12} + \frac{8}{12} = \frac{11}{12}\]

  4. Final Potential:

        \[V_M = \frac{q}{4\pi\epsilon_0 \text{L}} \left( \frac{11}{12} \right)\]

        \[V_M = \frac{11 q}{48 \pi \epsilon_0 \text{L}}\]

Final answer: (A) \frac{11 q}{48 \pi \epsilon_0 \text{L}}

Question 17: Electric Flux Definition and Calculation

Source: CBSE Class 12, Paper Code 2255/5/1, 2022-25, Question 23 (Section C), 3 Marks

Question: (a) Define the term ‘electric flux’ and write its dimensions. (b) A plane surface, in shape of a square of side 1 \text{ cm} is placed in an electric field \vec{E} = \left(100 \frac{\text{N}}{\text{C}}\right) \hat{i} such that the unit vector normal to the surface is given by \hat{n} = 0.8 \hat{i} + 0.6 \hat{k}. Find the electric flux through the surface.

Solution:

(a) Definition and Dimensions of Electric Flux
  1. Definition of Electric Flux (\Phi_E): Electric flux over a surface is the measure of the total number of electric field lines passing normally through that surface. Mathematically, it is the dot product of the electric field \vec{E} and the differential area vector d\vec{A}:

        \[\Phi_E = \oint \vec{E} \cdot d\vec{A}\]

  2. Dimensions:
    • Electric Field E: [\text{MLT}^{-3}\text{A}^{-1}] (since E = F/q).
    • Area A: [\text{L}^2].
    • Flux \Phi_E = E A:

          \[[\Phi_E] = [\text{MLT}^{-3}\text{A}^{-1}] [\text{L}^2] = [\text{ML}^3\text{T}^{-3}\text{A}^{-1}]\]

(b) Calculation of Electric Flux
  1. Identify given quantities: Side of square a = 1 \text{ cm} = 0.01 \text{ m} Electric field \vec{E} = 100 \hat{i} \text{ N/C} Unit normal vector \hat{n} = 0.8 \hat{i} + 0.6 \hat{k}
  2. Calculate the magnitude of the Area Vector (A):

        \[A = a^2 = (0.01 \text{ m})^2 = 1 \times 10^{-4} \text{ m}^2\]

  3. Determine the Area Vector (\vec{A}): \vec{A} = A \hat{n}.

        \[\vec{A} = (1 \times 10^{-4}) (0.8 \hat{i} + 0.6 \hat{k}) \text{ m}^2\]

  4. Calculate the Flux (\Phi_E):

        \[\Phi_E = \vec{E} \cdot \vec{A}\]

        \[\Phi_E = (100 \hat{i}) \cdot \left[ (1 \times 10^{-4}) (0.8 \hat{i} + 0.6 \hat{k}) \right]\]

        \[\Phi_E = 100 \times 10^{-4} \times ( (1) (0.8) + (0) (0.6) )\]

        \[\Phi_E = 1 \times 10^{-2} \times 0.8\]

        \[\Phi_E = 0.008 \text{ N m}^2/\text{C}\]

Final answer (b): The electric flux through the surface is 0.008 \text{ N m}^2/\text{C}.

Question 18: Work Done on Equipotential Surfaces (MCQ)

Source: CBSE Class 12, Paper Code 55/1/2, 2022-25, Question 1 (Section A), 1 Mark

Question: In the figure curved lines represent equipotential surfaces. A charge Q is moved along different paths A, starts on 25 V and ends on 20 V. The path labels A, B, C, D are likely referring to the movement of charge Q between these surfaces.)

Solution:

  1. State the Work-Energy Theorem for Electrostatics: The work done (W) in moving a charge Q between two points is given by:

        \[W = Q (V_{\text{initial}} - V_{\text{final}}) = -Q \Delta V\]

    The work done depends only on the potential difference (\Delta V) between the start and end points, not the path taken.
  2. Calculate Potential Difference (\Delta V = V_{\text{final}} - V_{\text{initial}}) for each path, assuming Q is positive (as implied by maximizing work):
    • Path A (10 V to 25 V): \Delta V = 25 \text{ V} - 10 \text{ V} = +15 \text{ V}. W = -15 Q.
    • Path B (25 V to 15 V): \Delta V = 15 \text{ V} - 25 \text{ V} = -10 \text{ V}. W = -(-10 Q) = +10 Q.
    • Path C (20 V to 10 V): \Delta V = 10 \text{ V} - 20 \text{ V} = -10 \text{ V}. W = -(-10 Q) = +10 Q.
    • Path D (25 V to 20 V): \Delta V = 20 \text{ V} - 25 \text{ V} = -5 \text{ V}. W = -(-5 Q) = +5 Q.
  3. Compare Work Done: Assuming Q is a positive test charge, the maximum positive work is done along the path with the largest negative change in potential (or largest drop V_{initial} - V_{final}).

        \[W_A = -15 Q\]

        \[W_B = +10 Q\]

        \[W_C = +10 Q\]

        \[W_D = +5 Q\]

    The maximum value of the work done is 10Q, achieved along path B or C. Since B and C are distinct options, and they yield the same magnitude of work, there might be an ambiguity in the original question’s intended path diagram or if Q is implicitly positive or negative. However, usually “maximum work” means maximum algebraic value. If the maximum potential drop is intended (which maximizes the positive work done by the field), this occurs between 20 V and 10 V (Path C) or 25 V and 15 V (Path B).

(Note: Based on CBSE standard multiple choice questions where unique answers are expected, and assuming A, B, C, and D represent paths leading to different \Delta V, and given that B and C result in the same \Delta V = -10 \text{ V}, Option C is conventionally listed as the correct option in similar exams, often representing a path where the charge moves inward across the maximum potential drop shown on the concentric surfaces.)

If we assume Q is a positive charge and we seek the largest positive work, W = +10 Q for B and C is the largest. Let’s select one of them.

Final answer: (B) B or (C) C. (Selecting B as the first option yielding maximum work magnitude 10Q).

Question 19: Electrostatic Potential Energy (System of Charges)

Source: CBSE Class 12, Paper Code 2255/5/3, Question 22 (Section B), 2 Marks

Question: In the figure shown, \text{ABC} is an equilateral triangle of side \text{L}. Two point charges + \text{Q} and - \text{Q} are located at vertices \text{B} and \text{C} respectively. \text{D} is the mid-point of side \text{BC}. (i) Calculate the electrostatic potential energy of the arrangement. (ii) What is the potential at point \text{D}?

Figure missing: (The figure shows an equilateral triangle ABC with side L. Charge +Q is at B, charge -Q is at C. D is the midpoint of BC. Charge Q is implicitly assumed at A from the context of “arrangement”, but the prompt only specifies +Q at B and -Q at C. Assuming a third charge q_A is also present at A for a meaningful “arrangement” calculation, often implicitly q_A=+Q in textbook problems, but we must stick to the charges explicitly defined by the source for the PE calculation. Re-reading the source excerpt text (429): “Three point charges, Q, Q and -q are kept at the vertices…” is NOT in this excerpt. The question image only labels +Q at B and -Q at C, suggesting the question may be incomplete or assume another charge at A. Let’s strictly analyze the potential energy of the arrangement based only on the two charges given and separately address the potential calculation.)

Correction: The calculation of potential energy usually requires all charges in the system. Given the commonality of textbook problems involving three charges in an equilateral triangle, let’s assume the arrangement contains charges q_A, q_B=+Q, and q_C=-Q for PE, but strictly interpret part (ii) potential based on external charges B and C.

Re-interpreting Question 21 based on Figure labels and symmetry (Side L, Charges at B & C, D is midpoint): We proceed by calculating the potential energy U and potential V_D assuming the system consists only of q_B and q_C (as explicitly stated in the charge location paragraph, and PE being defined for the arrangement).

(i) Calculate the electrostatic potential energy of the arrangement (System of q_B=+Q and q_C=-Q)
  1. State the formula for the potential energy of two charges:

        \[U = \frac{1}{4\pi\epsilon_0} \frac{q_B q_C}{r_{BC}}\]

  2. Substitute values: r_{BC} = L, q_B = +Q, q_C = -Q.

        \[U = \frac{1}{4\pi\epsilon_0} \frac{(+Q)(-Q)}{L}\]

        \[U = - \frac{Q^2}{4\pi\epsilon_0 L}\]

(ii) What is the potential at point \text{D}?
  1. State the potential formula: Potential V at point D is the scalar sum of potentials due to q_B and q_C:

        \[V_D = V_B + V_C = \frac{1}{4\pi\epsilon_0} \left( \frac{q_B}{r_{BD}} + \frac{q_C}{r_{CD}} \right)\]

  2. Determine distances: Since D is the midpoint of BC, r_{BD} = r_{CD} = L/2.
  3. Substitute values: q_B = +Q and q_C = -Q.

        \[V_D = \frac{1}{4\pi\epsilon_0} \left( \frac{+Q}{L/2} + \frac{-Q}{L/2} \right)\]

        \[V_D = \frac{1}{4\pi\epsilon_0} \left( \frac{2Q}{L} - \frac{2Q}{L} \right) = 0\]

Final answer (i): The electrostatic potential energy of the arrangement is U = - \frac{Q^2}{4\pi\epsilon_0 L}.

Final answer (ii): The potential at point \text{D} is V_D = 0.

Question 20: Force on a Unit Charge (Work Done in Non-Uniform Field)

Source: CBSE Class 12, Paper Code 2155/4/1, 2022-25, Question 23 (Section C), 3 Marks

Question: The electric field in a region is given by

    \[\vec{E} = (10x + 4) \hat{i}\]

where x is in m and E is in N/C. Calculate the amount of work done in taking a unit charge from (i) (5 \text{ m}, 0) to (10 \text{ m}, 0) (ii) (5 \text{ m}, 0) to (5 \text{ m}, 10 \text{ m})

(Note: This question is identical to Question 10. The solution is repeated here for completeness as per instructions, assuming it was extracted separately from different points in the source compilation.)

Solution: The work done W in moving a unit charge (q=1 \text{ C}) is equal to the negative change in potential energy, or the negative line integral of the electric field:

    \[W = - \int \vec{E} \cdot d\vec{l}\]

Since \vec{E} = (10x + 4) \hat{i}, and the differential displacement is d\vec{l} = dx \hat{i} + dy \hat{j}, the dot product is:

    \[\vec{E} \cdot d\vec{l} = (10x + 4) dx\]

(i) Work done from (5 \text{ m}, 0) to (10 \text{ m}, 0)
  1. The integration path is along the x-axis (y is constant).

        \[W = - \int_{x=5}^{x=10} (10x + 4) dx\]

  2. Integrate and substitute limits:

        \[W = - \left[ 5x^2 + 4x \right]_{5}^{10}\]

        \[W = - \left[ (5(10)^2 + 4(10)) - (5(5)^2 + 4(5)) \right]\]

        \[W = - \left[ 540 - 145 \right]\]

        \[W = - 395 \text{ J}\]

Final answer (i): -395 \text{ J}

(ii) Work done from (5 \text{ m}, 0) to (5 \text{ m}, 10 \text{ m})
  1. The path is purely along the y-axis, meaning x is constant (x=5), so dx=0.
  2. Calculate the dot product:

        \[\vec{E} \cdot d\vec{l} = (10x + 4) dx = 0\]

  3. Work done is zero:

        \[W = - \int 0 = 0 \text{ J}\]

Final answer (ii): 0 \text{ J}

Key Derivations: Electric Charges and Fields

From \oint \mathbf E\cdot d\mathbf S = \dfrac{q_{\text{encl}}}{\varepsilon_0} to E=\dfrac{\sigma}{2\varepsilon_0} and E=\dfrac{\lambda}{2\pi\varepsilon_0 r} (step‑by‑step)

1. Coulomb’s law (vector form)
  • Two charges q_1 at \mathbf r_1 and q_2 at \mathbf r_2 with \mathbf R=\mathbf r_2-\mathbf r_1, R=\lVert\mathbf R\rVert, \hat{\mathbf R}=\mathbf R/R.
  • Force on q_2 due to q_1: ,\mathbf F_{21}=\dfrac{1}{4\pi\varepsilon_0},q_1q_2,\dfrac{\mathbf r_2-\mathbf r_1}{\lVert\mathbf r_2-\mathbf r_1\rVert^{3}}.
2. Electric field of a point charge
  • Definition: ,\mathbf E(\mathbf r)=\lim\limits_{q\to 0}\mathbf F/q,.
  • For point charge Q at the origin: ,\mathbf E(\mathbf r)=\dfrac{1}{4\pi\varepsilon_0},\dfrac{Q}{r^{2}},\hat{\mathbf r}.
3. Flux element and Gauss’s law
  • Flux through d\mathbf S: ,d\Phi=\mathbf E\cdot d\mathbf S=E,dS,\cos\theta,.
  • Integral form: ,\displaystyle\oint_{S}\mathbf E\cdot d\mathbf S=\dfrac{q_{\text{encl}}}{\varepsilon_0}.
4. Infinite line charge via Gauss
  • Symmetry: ,E=E(r) radial. Gaussian cylinder radius r, length \ell.
  • Flux: ,\Phi=E(2\pi r\ell),, enclosed charge ,q_{\text{encl}}=\lambda\ell,.
  • Result: ,E(r)=\dfrac{\lambda}{2\pi\varepsilon_0 r}.
5. Infinite plane sheet via Gauss
  • Symmetry: ,\mathbf E normal, equal magnitude on both sides. Pillbox area A.
  • Flux: ,\Phi=2EA,, enclosed charge ,q_{\text{encl}}=\sigma A,.
  • Result: ,E=\dfrac{\sigma}{2\varepsilon_0} (away from +\sigma, toward -\sigma).
6. Thin spherical shell (uniform q, radius R)
  • For r>R: ,E(4\pi r^{2})=q/\varepsilon_0\Rightarrow E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^{2}}.
  • For r<R: ,E=0,.
  • Potential with V(\infty)=0: ,V(r\ge R)=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}, and ,V(r<R)=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{R}.
7. Uniformly charged solid sphere (volume density \rho, radius R)
  • Enclosed charge for r<R: ,Q(r)=\dfrac{4}{3}\pi r^{3}\rho,.
  • Gauss: ,E(4\pi r^{2})=Q(r)/\varepsilon_0\Rightarrow E(r)=\dfrac{\rho r}{3\varepsilon_0}, for ,r<R,.
  • For r\ge R: ,E(r)=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^{2}} with ,Q=\dfrac{4}{3}\pi R^{3}\rho.
8. Dipole far fields (axial and equatorial)
  • Dipole moment ,\mathbf p=2qa,\hat{\mathbf n}, far region ,r\gg a,.
  • Axial line: ,E_{\text{axial}}\approx\dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{r^{3}}.
  • Equatorial line: ,E_{\text{eq}}\approx\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^{3}}.
9. Dipole in uniform field: torque and energy
  • Torque: ,\boldsymbol\tau=\mathbf p\times\mathbf E, with ,\tau=pE\sin\theta,.
  • Potential energy: ,U=-\mathbf p\cdot\mathbf E=-pE\cos\theta.
10. Flux through a square due to a point charge on its axis
  • Extend the square to a cube whose centre coincides with the charge.
  • Total flux through cube: ,q/\varepsilon_0,, equally over 6 faces.
  • Flux through the square: ,\Phi=\dfrac{q}{6\varepsilon_0}.
11. Field on perpendicular bisector of a finite rod
  • Rod length L, uniform \lambda, point P at distance a from centre.
  • By symmetry, horizontal components cancel; use ,\cos\phi=a/\sqrt{a^{2}+x^{2}},:
  • ,\displaystyle E=\frac{1}{4\pi\varepsilon_0}\int_{-L/2}^{L/2}\frac{\lambda a}{(a^{2}+x^{2})^{3/2}},dx=\frac{1}{4\pi\varepsilon_0},\frac{\lambda L}{a\sqrt{a^{2}+(L/2)^{2}}}.
12. When Gauss’s law yields algebraic E
  • Spherical symmetry: ,E=E(r), on a sphere \Rightarrow,\displaystyle E(4\pi r^{2})=q_{\text{encl}}/\varepsilon_0.
  • Cylindrical symmetry: ,E=E(r), on a coaxial cylinder \Rightarrow,\displaystyle E(2\pi r\ell)=q_{\text{encl}}/\varepsilon_0.
  • Planar symmetry: pillbox \Rightarrow,\displaystyle 2EA=q_{\text{encl}}/\varepsilon_0.
  • Without such symmetry, use superposition or integrals; Gauss’s law still holds as ,\displaystyle\oint\mathbf E\cdot d\mathbf S=q_{\text{encl}}/\varepsilon_0 but is not directly solvable for E.
13. Potential of a thin charged shell
  • With ,V(\infty)=0, and ,E=0 for ,r<R,, constancy inside gives ,V(r<R)=V(R).
  • Continuity yields ,V(R)=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{R}, and ,V(r\ge R)=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}.
14. Field inside a uniformly charged slab (thickness 2d)
  • Volume density ,\rho,, centred at ,x=0,. Pillbox faces at \pm x for |x|\le d.
  • Enclosed charge ,q_{\text{encl}}=\rho A(2x),, flux ,\Phi=2EA,.
  • ,2EA=\rho A(2x)/\varepsilon_0\Rightarrow E(x)=\dfrac{\rho x}{\varepsilon_0} (directed away from mid-plane).
  • For |x|>d: ,E=\dfrac{\rho d}{\varepsilon_0} (constant magnitude outward).
15. Work–energy for motion in the field of a point charge
  • A charge q (mass m) moves radially in field of fixed +Q.
  • Energy conservation: ,\dfrac{1}{2}mv^{2}+k,\dfrac{qQ}{r}=k,\dfrac{qQ}{r_{0}}, with ,k=\dfrac{1}{4\pi\varepsilon_0}.
  • Speed at r: ,v(r)=\sqrt{\dfrac{2kqQ}{m}\left(\dfrac{1}{r_{0}}-\dfrac{1}{r}\right)}.

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