Vidaara.orgClass 12 · Physics
CodeVID-P12-01-CLF-01
Assignment — Electric Charge, Coulomb's Law & Field
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 charge is the:
- A.newton
- B.coulomb
- C.volt
- D.farad
2.
The value of $\frac{1}{4\pi\epsilon_0}$ is approximately:
- A.$9\times10^{9}$
- B.$8.85\times10^{-12}$
- C.$1.6\times10^{-19}$
- D.$6.25\times10^{18}$
3.
Like charges:
- A.attract
- B.repel
- C.neither
- D.first attract then repel
4.
The dipole moment points from:
- A.$+q$ to $-q$
- B.$-q$ to $+q$
- C.either way
- D.it has no direction
5.
Electric flux is a:
- A.vector
- B.scalar
- C.tensor
- D.pseudovector
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
State the principle of superposition of electric forces.
7.
Define electric field and give its SI unit.
8.
Two charges $+2\ \mu\text{C}$ and $+2\ \mu\text{C}$ are 10 cm apart. Find the force.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
Derive the expression for the torque on a dipole in a uniform field.
10.
Find the field at 0.5 m from a $+10\ \mu\text{C}$ charge.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Define electric dipole. Derive the field on its axial line for $r\gg a$ and state how it differs from the equatorial field.
Answer Key
Section A — Multiple Choice Questions
- (B) coulomb
- (A) $9\times10^{9}$
- (B) repel
- (B) $-q$ to $+q$
- (B) scalar
Section B — Short Answer (2 marks)
- The net force on a charge is the vector sum of the forces exerted on it independently by every other charge.
- Field is force per unit positive test charge, $\vec{E}=\frac{\vec{F}}{q}$; SI unit $\text{N/C}$ or $\text{V/m}$.
- $F=9\times10^{9}\times\frac{(2\times10^{-6})^2}{(0.1)^2}=3.6\ \text{N}$, repulsive.
Section C — Short Answer (3 marks)
- Forces $+qE$ and $-qE$ act at separation $2a$; the couple gives $\tau=(qE)(2a\sin\theta)=pE\sin\theta$, i.e. $\vec{\tau}=\vec{p}\times\vec{E}$.
- $E=9\times10^{9}\times\frac{10\times10^{-6}}{0.25}=3.6\times10^{5}\ \text{N/C}$, radially outward.
Section D — Long Answer (5 marks)
- A dipole is $\pm q$ separated by $2a$, $p=q(2a)$. Axial: $E=\frac{1}{4\pi\epsilon_0}\frac{2p}{r^3}$; equatorial $E=\frac{1}{4\pi\epsilon_0}\frac{p}{r^3}$; axial is twice the equatorial and points along $p$, while equatorial points opposite to $p$.
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