Capacitor with Dielectric Concept Quiz 1

Capacitors with Dielectric Worksheet – JEE/NEET Level

This worksheet contains 20 multiple choice questions on Capacitors with Dielectric. It includes theory-based conceptual MCQs and numerical problems suitable for JEE Main, JEE Advanced, and NEET preparation. Detailed step-by-step solutions are provided at the end.

Section A – Theory Concept MCQs

Q1.

A dielectric slab completely fills the space between the plates of a capacitor. The capacitance becomes:

A) \( \dfrac{C_0}{K} \)

B) \( KC_0 \)

C) \( K^2C_0 \)

D) Unchanged

Q2.

A dielectric increases the capacitance of a capacitor because it:

A) Increases charge leakage

B) Reduces effective electric field

C) Increases plate area

D) Reduces plate separation

Q3.

The dielectric constant of vacuum is:

A) 0

B) 1

C) Infinity

D) Depends on temperature

Q4.

A capacitor remains connected to a battery while a dielectric is inserted. Which quantity remains constant?

A) Charge

B) Voltage

C) Energy

D) Electric field energy density

Q5.

When a dielectric is inserted into an isolated charged capacitor, its energy:

A) Increases

B) Decreases

C) Remains same

D) First increases then decreases

Q6.

Polar molecules in a dielectric material align mainly because of:

A) Magnetic force

B) Thermal expansion

C) External electric field

D) Gravity

Q7.

The SI unit of dielectric constant is:

A) Farad

B) Coulomb

C) Volt

D) Dimensionless

Q8.

For a capacitor partially filled with dielectric, capacitance:

A) Always decreases

B) Depends on dielectric geometry

C) Becomes zero

D) Becomes infinite

Q9.

Dielectric strength refers to:

A) Ability to conduct current

B) Maximum electric field without breakdown

C) Mechanical strength

D) Ability to store heat

Q10.

A dielectric slab is slowly removed from a charged isolated capacitor. Its potential difference:

A) Decreases

B) Becomes zero

C) Increases

D) Remains constant

Section B – Numerical MCQs

Q11.

A parallel plate capacitor has capacitance \( 4\mu F \). A dielectric of constant \( K=5 \) completely fills it. The new capacitance is:

A) \( 0.8\mu F \)

B) \( 4\mu F \)

C) \( 20\mu F \)

D) \( 25\mu F \)

Q12.

A capacitor of capacitance \( 2\mu F \) is charged to \( 100V \). A dielectric of constant 4 is inserted while battery remains connected. Final charge becomes:

A) \( 200\mu C \)

B) \( 400\mu C \)

C) \( 800\mu C \)

D) \( 1600\mu C \)

Q13.

An isolated capacitor initially stores energy \( 8J \). A dielectric of constant 2 is inserted completely. Final energy is:

A) \( 16J \)

B) \( 8J \)

C) \( 4J \)

D) \( 2J \)

Q14.

A capacitor has plate area \( 0.02m^2 \), separation \( 1mm \), dielectric constant \( 5 \). Find capacitance. Take \( \epsilon_0 = 8.85\times10^{-12} F/m \).

A) \( 885pF \)

B) \( 442pF \)

C) \( 1770pF \)

D) \( 3540pF \)

Q15.

A dielectric slab of thickness equal to half plate separation is inserted. Its dielectric constant is 3. Capacitance becomes:

A) \( \dfrac{3}{2}C_0 \)

B) \( \dfrac{6}{4}C_0 \)

C) \( \dfrac{3}{4}C_0 \)

D) \( 2C_0 \)

Q16.

An air capacitor stores \( 12\mu C \) charge at \( 6V \). If dielectric constant 3 is inserted with battery connected, new charge is:

A) \( 12\mu C \)

B) \( 18\mu C \)

C) \( 24\mu C \)

D) \( 36\mu C \)

Q17.

An isolated capacitor has potential difference \( 90V \). A dielectric constant 5 is inserted completely. New potential difference is:

A) \( 18V \)

B) \( 90V \)

C) \( 450V \)

D) \( 72V \)

Q18.

A dielectric material has dielectric strength \( 3\times10^6 V/m \). Maximum safe voltage across plates separated by \( 2mm \) is:

A) \( 300V \)

B) \( 3000V \)

C) \( 6000V \)

D) \( 9000V \)

Q19.

The capacitance of a parallel plate capacitor is doubled by:

A) Doubling separation

B) Halving area

C) Filling dielectric of constant 2

D) Removing dielectric

Q20.

A capacitor of capacitance \( 5\mu F \) is connected to \( 20V \). A dielectric of constant 4 is inserted while connected to battery. Increase in stored energy:

A) \( 1mJ \)

B) \( 3mJ \)

C) \( 6mJ \)

D) \( 12mJ \)

Answer Key

1. B
2. B
3. B
4. B
5. B
6. C
7. D
8. B
9. B
10. C
11. C
12. C
13. C
14. A
15. B
16. D
17. A
18. C
19. C
20. B

Detailed Step-by-Step Solutions

Q11 Solution

Capacitance with dielectric:

\[ C = KC_0 \]

\[ C = 5 \times 4\mu F \]

\[ C = 20\mu F \]

Answer: C

Q12 Solution

Initial capacitance:

\[ C_0 = 2\mu F \]

With dielectric:

\[ C = 4 \times 2 = 8\mu F \]

Battery connected, so voltage constant:

\[ Q = CV \]

\[ Q = 8\mu F \times 100V \]

\[ Q = 800\mu C \]

Answer: C

Q13 Solution

For isolated capacitor:

\[ U = \frac{Q^2}{2C} \]

Capacitance doubles. Therefore energy becomes half.

\[ U_f = \frac{8}{2} \]

\[ U_f = 4J \]

Answer: C

Q14 Solution

Formula:

\[ C = \frac{K\epsilon_0A}{d} \]

Substitute:

\[ C = \frac{5 \times 8.85\times10^{-12}\times0.02}{10^{-3}} \]

\[ C = 885\times10^{-12}F \]

\[ C = 885pF \]

Answer: A

Q15 Solution

Effective separation:

\[ d_{eff} = \frac{d}{2} + \frac{d/2}{3} \]

\[ d_{eff} = \frac{2d}{3} \]

Thus:

\[ C = \frac{\epsilon_0A}{2d/3} \]

\[ C = \frac{3}{2}C_0 \]

Answer: A

Q16 Solution

Initial charge:

\[ Q_0 = 12\mu C \]

Battery connected, so:

\[ Q = KQ_0 \]

\[ Q = 3 \times 12 \]

\[ Q = 36\mu C \]

Answer: D

Q17 Solution

For isolated capacitor:

\[ V = \frac{Q}{C} \]

Capacitance becomes 5 times. Hence voltage becomes:

\[ V_f = \frac{90}{5} \]

\[ V_f = 18V \]

Answer: A

Q18 Solution

Maximum voltage:

\[ V = Ed \]

\[ V = 3\times10^6 \times 2\times10^{-3} \]

\[ V = 6000V \]

Answer: C

Q19 Solution

Capacitance:

\[ C = \frac{K\epsilon_0A}{d} \]

If dielectric constant becomes 2:

\[ C = 2C_0 \]

Answer: C

Q20 Solution

Initial energy:

\[ U_i = \frac12 CV^2 \]

\[ U_i = \frac12 \times 5\times10^{-6}\times20^2 \]

\[ U_i = 1mJ \]

Final capacitance:

\[ C_f = 4\times5 = 20\mu F \]

Final energy:

\[ U_f = \frac12 \times 20\times10^{-6}\times20^2 \]

\[ U_f = 4mJ \]

Increase:

\[ \Delta U = 4 - 1 = 3mJ \]

Answer: B