Vidaara.orgClass 12 · Physics
CodeVID-P12-01-GLP-01
Assignment — Gauss's Law & Electric Potential
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.
The SI unit of electric potential is the:
- A.coulomb
- B.volt
- C.newton
- D.joule
2.
Flux through a closed surface depends on:
- A.surface shape
- B.enclosed charge only
- C.external charges
- D.surface area
3.
The field of an infinite line charge varies as:
- A.$1/r$
- B.$1/r^2$
- C.$1/r^3$
- D.constant
4.
Work done in moving a charge along an equipotential surface is:
- A.maximum
- B.zero
- C.negative
- D.infinite
5.
The relation between field and potential is:
- A.$E=V/r$
- B.$E=-\frac{dV}{dr}$
- C.$E=Vr$
- D.$E=\frac{dr}{dV}$
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
State Gauss\u2019s law.
7.
Why is the field perpendicular to an equipotential surface?
8.
Find the potential 0.3 m from a $+6\ \mu\text{C}$ charge.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
Using Gauss\u2019s law, derive the field of an infinite charged sheet.
10.
Two charges $+2\ \mu\text{C}$ and $+2\ \mu\text{C}$ are 0.4 m apart. Find the potential energy.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Using Gauss\u2019s law, derive the electric field due to a uniformly charged thin spherical shell at points outside and inside it.
Answer Key
Section A — Multiple Choice Questions
- (B) volt
- (B) enclosed charge only
- (A) $1/r$
- (B) zero
- (B) $E=-\frac{dV}{dr}$
Section B — Short Answer (2 marks)
- The total electric flux through a closed surface equals $\frac{1}{\epsilon_0}$ times the net charge enclosed: $\oint\vec{E}\cdot d\vec{A}=\frac{q_{enc}}{\epsilon_0}$.
- If it had a tangential component, work would be done moving a charge along the surface, contradicting constant $V$; hence $E$ is normal to it.
- $V=9\times10^{9}\times\frac{6\times10^{-6}}{0.3}=1.8\times10^{5}\ \text{V}$.
Section C — Short Answer (3 marks)
- Take a cylindrical pillbox of area $A$ through the sheet; flux $=2EA=\frac{\sigma A}{\epsilon_0}$, giving $E=\frac{\sigma}{2\epsilon_0}$.
- $U=9\times10^{9}\times\frac{(2\times10^{-6})^2}{0.4}=0.09\ \text{J}$.
Section D — Long Answer (5 marks)
- Outside ($r>R$): Gaussian sphere encloses $Q$, $E\cdot4\pi r^2=\frac{Q}{\epsilon_0}$ so $E=\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}$. Inside ($r
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