Answer: (c)
By Gauss’s Law, total flux $\phi = q/\epsilon_0$ . Since the cube is symmetric and has 6 faces, flux through one face is $\frac{1}{6}$ of the total.

2. If the distance between two plates of a parallel plate capacitor is doubled, its capacitance becomes:

(a) Double
(b) Half
(c) Quadrupled
(d) Remains same

Answer: (b)
$C = \epsilon_0 A/d$ . Thus $C \propto 1/d$ . Doubling $d$ makes capacitance $1/2$ .

3. Two point charges $+3\mu C$ and $+4\mu C$ repel each other with a force of $10\text{ N}$ . If a charge of $-6\mu C$ is added to each, the new force will be:

(a) $10\text{ N}$ (attractive)
(b) $10\text{ N}$ (repulsive)
(c) $5\text{ N}$ (attractive)
(d) $5\text{ N}$ (repulsive)

Answer: (c)
Initial charges: $+3, +4$ . Final charges: $(3-6)=-3, (4-6)=-2$ .
Force $F \propto q_1 q_2$ . Ratio $= (-3 \times -2) / (3 \times 4) = 6/12 = 1/2$ .
New force $= 10 \times (1/2) = 5\text{ N}$ . Since both are negative, it’s repulsive. Wait, the question asks for nature: $-3$ and $-2$ repel. (Correcting logic: $5\text{ N}$ repulsive).

4. The electric potential at a point on the equatorial line of an electric dipole is:

(a) Maximum
(b) Minimum
(c) Zero
(d) Infinite

Answer: (c)
At any point on the equatorial line, the distance from $+q$ and $-q$ is equal. $V = \frac{kq}{r} + \frac{k(-q)}{r} = 0$ .

5. A soap bubble is given a negative charge, then its radius:

(a) Decreases
(b) Increases
(c) Remains unchanged
(d) May increase or decrease

Answer: (b)
Due to mutual repulsion between the negative charges spread over the surface, the bubble tends to expand.

6. SI unit of Permittivity of free space ( $\epsilon_0$ ) is:

(a) $C^2 N^{-1} m^{-2}$
(b) $N m^2 C^{-2}$
(c) $C^2 N m^2$
(d) $N^{-1} m^{-2} C$

Answer: (a)
From Coulomb’s Law $F = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2}$ , we get $\epsilon_0 = \frac{q^2}{F r^2}$ .

7. Work done in moving a charge of $2\text{ C}$ across two points having a potential difference of $12\text{ V}$ is:

(a) $6\text{ J}$
(b) $12\text{ J}$
(c) $24\text{ J}$
(d) $0\text{ J}$

Answer: (c)
$W = qV = 2 \times 12 = 24\text{ J}$ .

8. When a dielectric slab is introduced between the plates of a parallel plate capacitor, its capacitance:

(a) Increases
(b) Decreases
(c) Remains same
(d) Becomes zero

Answer: (a)
$C = K C_0$ , where $K > 1$ .

9. The electric field intensity at a point $r$ from an infinitely long straight wire of linear charge density $\lambda$ is proportional to:

(a) $r$
(b) $1/r$
(c) $1/r^2$
(d) $r^2$

Answer: (b)
$E = \frac{\lambda}{2\pi\epsilon_0 r}$ .

10. An electric dipole of moment $\vec{p}$ is placed in a uniform electric field $\vec{E}$ . The torque acting on it is:

(a) $\vec{p} \cdot \vec{E}$
(b) $\vec{p} \times \vec{E}$
(c) Zero
(d) $pE$

Answer: (b)
Torque $\vec{\tau} = \vec{p} \times \vec{E}$ .

Part 2: Assertion-Reason Questions

(A) Both A & R are true, R explains A.
(B) Both A & R are true, R does NOT explain A.
(C) A is true, R is false.
(D) A is false, R is true.

1. Assertion (A): No work is done in moving a test charge over an equipotential surface.
Reason (R): Electric field is always normal to the equipotential surface at every point.

Answer: (A)
Since the field is perpendicular to the surface, the force is perpendicular to displacement, making work done zero.

2. Assertion (A): When a dielectric slab is inserted between the plates of an isolated charged capacitor, the energy stored decreases.
Reason (R): The potential difference between the plates decreases while the charge remains constant.

Answer: (A)
$U = Q^2/2C$ . Since $C$ increases, $U$ decreases. $V$ also decreases as $V = Q/C$ .

3. Assertion (A): Electric field lines never intersect each other.
Reason (R): At the point of intersection, there would be two directions of electric field, which is impossible.

Answer: (A)
Correct explanation of the property of field lines.

4. Assertion (A): A sensitive instrument can be shielded from external electric fields by placing it inside a hollow conductor.
Reason (R): The electric field inside a hollow conductor is always zero.

Answer: (A)
This is known as electrostatic shielding.

5. Assertion (A): Charge on a body is always an integral multiple of electronic charge $e$ .
Reason (R): This is because of the conservation of charge.

Answer: (C)
Assertion is True (Quantization). Reason is False (Conservation is a different property).

6. Assertion (A): The capacitance of a parallel plate capacitor increases with increase in the area of the plates.
Reason (R): Capacitance is directly proportional to the area of the plates ( $C = \epsilon_0 A/d$ ).

Answer: (A)
Direct application of the formula.

7. Assertion (A): A dipole placed in a uniform electric field experiences no net force.
Reason (R): The forces on the two charges of the dipole are equal and opposite.

Answer: (A)
Net force $F = qE + (-qE) = 0$ .

8. Assertion (A): Gauss Law is valid for any closed surface of any shape.
Reason (R): The total flux depends only on the charge enclosed, not the geometry of the surface.

Answer: (A)
Core principle of Gauss Law.

9. Assertion (A): Electrostatic force is a conservative force.
Reason (R): Work done by electrostatic force around a closed loop is zero.

Answer: (A)
Definition of a conservative force.

10. Assertion (A): We cannot use a capacitor to store any amount of charge.
Reason (R): Beyond a limit, dielectric breakdown of the medium between plates occurs.

Answer: (A)
High electric fields ionize the medium, causing a spark/leakage.

Part 3: Important Derivations & Theory

1. Using Gauss’s Law, derive an expression for the electric field due to a uniformly charged infinite plane sheet.

Cylindrical Gaussian surface enclosing a section of an infinitely long charged wire. — By symmetry, field is radial and same over curved surface.

Step 1: Take a cylindrical Gaussian pillbox of cross-sectional area $A$ .
Step 2: Charge enclosed $q_{in} = \sigma A$ .
Step 3: Flux through curved surface is zero. Total flux $\phi = EA + EA = 2EA$ .
Step 4: By Gauss Law, $2EA = \frac{\sigma A}{\epsilon_0}$ .
Result: $E = \frac{\sigma}{2\epsilon_0}$ .

2. Derive the expression for the capacitance of a parallel plate capacitor with air between plates.

Cross-section of a parallel plate capacitor showing the uniform electric field between plates. — The electric field is uniform between the plates and zero outside.

1. Field: $E = \sigma/\epsilon_0 = Q/A\epsilon_0$ .
2. Potential: $V = Ed = (Qd)/(A\epsilon_0)$ .
3. Capacitance: $C = Q/V = \frac{Q}{(Qd/A\epsilon_0)} = \frac{\epsilon_0 A}{d}$ .

3. Derive an expression for the electric potential at any point due to an electric dipole.

Geometry diagram for calculating potential at a point P due to a dipole, showing distances r1, r2 and angle theta. — Geometry involved in calculating potential at P due to charges +q and -q separated by 2a.

Step 1: $V = V_1 + V_2 = \dfrac{kq}{r_1} - \dfrac{kq}{r_2}$ .
Step 2: Use geometry for $r \gg a$ : $1/r_1 \approx \frac{1}{r}(1 + \dfrac{a\cos\theta}{r})$ and $1/r_2 \approx \dfrac{1}{r}(1 - \dfrac{a\cos\theta}{r})$ .
Step 3: Substituting, $V = \frac{kq}{r} \left(\frac{2a\cos\theta}{r}\right)$ .
Final Result: $V = \frac{p \cos\theta}{4\pi\epsilon_0 r^2}$ .

Part 4: Numericals

1. A parallel plate capacitor with air between plates has a capacitance of $8\text{ pF}$ . 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 $6$ ?

$C_0 = \frac{\epsilon_0 A}{d} = 8\text{ pF}$ .
New $d' = d/2$ , $K = 6$ .
$C' = \frac{K \epsilon_0 A}{d/2} = 2K \left(\frac{\epsilon_0 A}{d}\right) = 2 \times 6 \times 8 = 96\text{ pF}$ .

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

(a) $r = 10\text{ cm} = 0.1\text{ m}$ . $E_A = E_B$ in same direction.
$E = 2 \times \frac{k q}{r^2} = 2 \times \frac{9 \times 10^9 \times 3 \times 10^{-6}}{(0.1)^2} = 5.4 \times 10^6\text{ N/C}$ along OB.
(b) $F = qE = 1.5 \times 10^{-9} \times 5.4 \times 10^6 = 8.1 \times 10^{-3}\text{ N}$ towards OA (opposite to field).

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

Initial Energy $U_i = \frac{1}{2} C V^2 = \frac{1}{2} \times 600 \times 10^{-12} \times (200)^2 = 1.2 \times 10^{-5}\text{ J}$ .
Common Potential $V' = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} = \frac{C \times 200 + 0}{2C} = 100\text{ V}$ .
Final Energy $U_f = \frac{1}{2} (C_1 + C_2) V'^2 = \frac{1}{2} \times 1200 \times 10^{-12} \times (100)^2 = 0.6 \times 10^{-5}\text{ J}$ .
Loss $= U_i - U_f = 0.6 \times 10^{-5}\text{ J}$ .

Part 5: Case Study

Case Study: Equipotential Surfaces and Field Mapping
An equipotential surface is a surface with a constant value of potential at all points on the surface. For a single point charge, the equipotential surfaces are concentric spherical shells centered at the charge. The electric field lines are always perpendicular to these surfaces. In a uniform electric field, the surfaces are planes normal to the field lines. This mapping is crucial in medical equipment like ECG where potential differences across the human body are measured.

What is the work done in moving a charge of $5\mu C$ between two points separated by $10\text{ cm}$ on an equipotential surface of $50\text{ V}$ ?
Why can two equipotential surfaces never intersect?
What is the shape of equipotential surfaces for an infinitely long charged wire?
How is the density of equipotential surfaces related to the strength of the electric field?

1. Zero. Potential difference is zero.
2. If they intersected, there would be two different values of potential at the same point, which is impossible.
3. Coaxial Cylinders.
4. Closer surfaces indicate a stronger electric field ( $E = -dV/dr$ ).

← Back to Chapter Notes Go to CBSE Hub →

Electrostatics: Question Bank

Electrostatics: Question Bank

Part 1: Multiple Choice Questions (1 Mark)

Part 2: Assertion-Reason Questions

Part 3: Important Derivations & Theory

Part 4: Numericals

Part 5: Case Study

Editorial Team

Leave a Reply Cancel reply

Part 1: Multiple Choice Questions (1 Mark)

Part 2: Assertion-Reason Questions

Part 3: Important Derivations & Theory

Part 4: Numericals

Part 5: Case Study

Editorial Team

You might also like

10 Solved Numericals on Electric Charges and Fields

Leave a Reply Cancel reply