Vidaara.orgClass 12 · Physics
CodeVID-P12-01-CAP-01
Assignment — Capacitance & Capacitors
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
Capacitance is defined as:
- A.$QV$
- B.$Q/V$
- C.$V/Q$
- D.$Q^2 V$
2.
For a parallel-plate capacitor $C$ is proportional to:
- A.$d$
- B.$1/d$
- C.$d^2$
- D.$1/d^2$
3.
Energy stored in a capacitor is:
- A.$\frac12 CV$
- B.$\frac12 CV^2$
- C.$CV^2$
- D.$\frac12 C^2 V$
4.
In parallel, the equivalent capacitance is the:
- A.reciprocal sum
- B.sum
- C.product
- D.difference
5.
A dielectric of constant $K$ changes $C_0$ to:
- A.$C_0/K$
- B.$KC_0$
- C.$C_0$
- D.$2KC_0$
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
Define capacitance and give its SI unit.
7.
A $10\ \mu\text{F}$ capacitor is charged to 50 V. Find the charge stored.
8.
Why does inserting a dielectric increase capacitance?
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
Derive the energy stored in a charged capacitor.
10.
Two capacitors $4\ \mu\text{F}$ and $12\ \mu\text{F}$ are in series. Find the equivalent capacitance.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Derive the capacitance of a parallel-plate capacitor and explain the effect of introducing a dielectric slab that fills the gap.
Answer Key
Section A — Multiple Choice Questions
- (B) $Q/V$
- (B) $1/d$
- (B) $\frac12 CV^2$
- (B) sum
- (B) $KC_0$
Section B — Short Answer (2 marks)
- Capacitance is charge stored per unit potential difference, $C=\frac{Q}{V}$; SI unit farad (F).
- $Q=CV=10\times10^{-6}\times50=5\times10^{-4}\ \text{C}$.
- The dielectric polarises and sets up an opposing field, reducing net field and voltage for the same charge, so $C=Q/V$ rises by factor $K$.
Section C — Short Answer (3 marks)
- Work to add charge $dq$ at voltage $q/C$ is $dW=\frac{q}{C}dq$; integrating from 0 to $Q$ gives $U=\frac{Q^2}{2C}=\frac12 CV^2$.
- $\frac{1}{C_s}=\frac{1}{4}+\frac{1}{12}=\frac{4}{12}$, so $C_s=3\ \mu\text{F}$.
Section D — Long Answer (5 marks)
- Field $E=\frac{\sigma}{\epsilon_0}=\frac{Q}{\epsilon_0 A}$, voltage $V=Ed=\frac{Qd}{\epsilon_0 A}$, so $C=\frac{Q}{V}=\frac{\epsilon_0 A}{d}$. With a dielectric of constant $K$, the field reduces to $E/K$, $V$ falls, and $C=\frac{K\epsilon_0 A}{d}=KC_0$.
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