Capacitors with Dielectric Concept Quiz 2

 

Advanced Worksheet: Capacitors with Dielectric

This worksheet contains advanced multiple-choice problems on Capacitors with Dielectric designed for JEE Main, JEE Advanced, and NEET preparation. The worksheet includes conceptual theory MCQs and advanced numerical problems along with answer key and detailed solutions.

Section A – Advanced Theory Concept MCQs

Q1.

A dielectric slab is inserted between the plates of an isolated charged capacitor. Which quantity definitely decreases?

A) Charge

B) Capacitance

C) Potential difference

D) Electric flux through dielectric

Q2.

For a dielectric medium, polarization vector \( \vec P \) is proportional to:

A) Electric flux

B) Potential

C) External electric field

D) Plate area

Q3.

If dielectric constant of a medium is very large, the electric field inside it becomes:

A) Larger

B) Nearly zero

C) Infinite

D) Equal to vacuum field

Q4.

A dielectric slab partially fills a capacitor. Fringing effects are neglected. The system behaves like:

A) Two capacitors in series

B) Two capacitors in parallel

C) Pure resistor

D) Inductor-capacitor combination

Q5.

When a dielectric is inserted into a capacitor connected to battery, work done by battery is used in:

A) Increasing electrostatic energy only

B) Mechanical work only

C) Increasing electrostatic energy and mechanical work

D) Producing heat only

Q6.

The energy density inside a dielectric-filled capacitor is:

A) \( \dfrac{1}{2}\epsilon_0E^2 \)

B) \( \dfrac{1}{2}KE^2 \)

C) \( \dfrac{1}{2}K\epsilon_0E^2 \)

D) \( \dfrac{1}{2}\dfrac{E^2}{K} \)

Q7.

A dielectric slab experiences force toward stronger electric field because:

A) Dielectric reduces system energy

B) Charge increases

C) Voltage increases

D) Dielectric becomes conductor

Q8.

The dielectric constant is mainly caused due to:

A) Polarization of atoms/molecules

B) Electron emission

C) Thermal conductivity

D) Mechanical compression

Q9.

A dielectric inserted obliquely between plates causes capacitance to:

A) Increase

B) Decrease

C) Remain same

D) Become zero

Q10.

In a dielectric material, bound charges appear because:

A) Free electrons move throughout material

B) Positive ions escape

C) Dipoles align under electric field

D) Neutrons polarize

Section B – Advanced Numerical MCQs

Q11.

A parallel plate capacitor has air capacitance \( 10\mu F \). A dielectric slab of dielectric constant 4 fills half the separation completely. The new capacitance is:

A) \( 16\mu F \)

B) \( 20\mu F \)

C) \( \dfrac{40}{3}\mu F \)

D) \( 30\mu F \)

Q12.

An isolated capacitor stores energy \( 18J \). A dielectric constant 6 is inserted completely. Final energy becomes:

A) \( 108J \)

B) \( 18J \)

C) \( 3J \)

D) \( 6J \)

Q13.

A capacitor of capacitance \( 5\mu F \) is charged to \( 200V \). After disconnecting battery, dielectric constant 5 is inserted. Final voltage is:

A) \( 1000V \)

B) \( 200V \)

C) \( 40V \)

D) \( 20V \)

Q14.

A dielectric slab occupies half area between capacitor plates. Its dielectric constant is 5. If original capacitance is \( C_0 \), final capacitance is:

A) \( 2C_0 \)

B) \( 3C_0 \)

C) \( \dfrac{5}{2}C_0 \)

D) \( 6C_0 \)

Q15.

A capacitor has capacitance \( 8\mu F \) and voltage \( 50V \). Battery remains connected while dielectric constant 3 is inserted. Increase in stored energy:

A) \( 0.01J \)

B) \( 0.02J \)

C) \( 0.04J \)

D) \( 0.08J \)

Q16.

Two dielectric slabs of constants 2 and 4 each occupy half separation of capacitor. Equivalent dielectric constant is:

A) \( \dfrac{8}{3} \)

B) 3

C) 4

D) 2

Q17.

A dielectric slab is inserted into a capacitor connected to battery. If electric field remains constant, electrostatic energy:

A) Decreases

B) Increases

C) Remains same

D) Becomes zero

Q18.

A dielectric of constant 9 is used in capacitor. If dielectric strength is \( 5\times10^6V/m \), maximum safe electric field inside dielectric is:

A) \( 5\times10^6V/m \)

B) \( 45\times10^6V/m \)

C) \( 5\times10^5V/m \)

D) Depends on capacitance

Q19.

A dielectric slab is pulled out from a capacitor connected to battery. External agent does:

A) Positive work

B) Negative work

C) Zero work

D) Infinite work

Q20.

A capacitor with dielectric constant 2 stores energy \( U \). If dielectric is removed after disconnecting battery, new energy becomes:

A) \( \dfrac{U}{2} \)

B) \( U \)

C) \( 2U \)

D) \( 4U \)

Answer Key

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

Detailed Step-by-Step Solutions

Q11 Solution

For dielectric filling half separation:

\[ C = \frac{K C_0}{K(1-f)+f} \]

Here \( K=4 \), \( f=\frac12 \)

\[ C=\frac{4\times10}{4(1/2)+1/2} \]

\[ C=\frac{40}{2.5} \]

\[ C=16\mu F \]

Answer: A

Q12 Solution

For isolated capacitor:

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

Capacitance increases by factor 6. Hence energy decreases by factor 6.

\[ U_f=\frac{18}{6}=3J \]

Answer: C

Q13 Solution

Battery disconnected ⇒ charge constant.

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

Capacitance becomes 5 times.

\[ V_f=\frac{200}{5}=40V \]

Answer: C

Q14 Solution

Half area dielectric means parallel combination.

\[ C_1=\frac{K}{2}C_0 \]

\[ C_2=\frac12 C_0 \]

\[ C=C_1+C_2 \]

\[ C=\frac52C_0+\frac12C_0 \]

\[ C=3C_0 \]

Answer: B

Q15 Solution

Initial energy:

\[ U_i=\frac12CV^2 \]

\[ U_i=\frac12\times8\times10^{-6}\times50^2 \]

\[ U_i=0.01J \]

Final energy:

\[ U_f=3U_i=0.03J \]

Increase:

\[ \Delta U=0.03-0.01 \]

\[ \Delta U=0.02J \]

Answer: B

Q16 Solution

Series dielectric arrangement:

\[ K_{eq}=\frac{2K_1K_2}{K_1+K_2} \]

\[ K_{eq}=\frac{2\times2\times4}{2+4} \]

\[ K_{eq}=\frac{16}{6}=\frac83 \]

Answer: A

Q17 Solution

Battery connected ⇒ voltage constant.

Energy:

\[ U=\frac12CV^2 \]

Since capacitance increases, energy increases.

Answer: B

Q18 Solution

Dielectric strength itself defines maximum safe electric field.

\[ E_{max}=5\times10^6V/m \]

Answer: A

Q19 Solution

Capacitor attracts dielectric inward. To remove it slowly, external agent must apply force outward.

Hence external work is positive.

Answer: A

Q20 Solution

Battery disconnected ⇒ charge constant.

Energy:

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

Removing dielectric halves capacitance.

Hence energy doubles.

\[ U_f=2U \]

Answer: C