Multiple Choice & Assertion-Reason

PHYExam Practice

This is a whole-chapter revision set: the MCQs and assertion-reason items test the cross-cutting laws of electrostatics — Coulomb's inverse-square law, dipole fields, Gauss's law, capacitor combinations and energy, equipotentials, and conductor properties. The trick is recognising which standard result a one-line question is probing and recalling the correct proportionality, factor, or sign.

Master the reusable formulas below and the conceptual traps they exploit.

Coulomb's inverse-square law

F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2}

F\propto 1/r^2

, so halving

r

gives

4F

; basis of force MCQs

Dipole axial vs equatorial field

E_{axial} = \dfrac{2p}{4\pi\varepsilon_0 r^3}, \qquad E_{eq} = \dfrac{p}{4\pi\varepsilon_0 r^3}

axial is twice equatorial at the same

r

; both

\propto 1/r^3

Gauss's law and the infinite sheet

\oint \vec{E}\cdot d\vec{A} = \dfrac{q_{enc}}{\varepsilon_0}, \qquad E_{sheet} = \dfrac{\sigma}{2\varepsilon_0}

net flux depends only on enclosed charge; sheet field is

\sigma/2\varepsilon_0

(conductor surface gives

\sigma/\varepsilon_0

)

Capacitor combination and energy

\dfrac{1}{C_s}=\sum\dfrac{1}{C_i}, \quad C_p=\sum C_i, \quad U=\dfrac{Q^2}{2C}

three equal

C

in series give

C/3

; with

Q

fixed, raising

d

lowers

C

and raises

U

Dipole energy and equilibrium

U = -pE\cos\theta

stable equilibrium at

\theta=0^{\circ}

(minimum

U

), unstable at

180^{\circ}

Work on an equipotential

W = q\,(V_B - V_A)

on one equipotential

V_B=V_A

, so

W=0

; field is everywhere perpendicular to the surface

Inverse-square scaling: any change to $r$ scales $F$ (and point-charge $E$ ) by the inverse square — halving $r$ quadruples them.
Net flux through a closed surface depends only on enclosed charge $q$ , independent of surface shape, charge position inside, or the external medium.
For a short dipole the axial field is double the equatorial field at equal distance, and both fall off as $1/r^3$ .
Three equal capacitors $C$ in series give $C/3$ ; in parallel they give $3C$ — series always reduces capacitance below the smallest member.
With the battery disconnected the charge $Q$ is fixed; pulling the plates apart lowers $C$ , so $U=Q^2/2C$ increases.
Work to move a charge between two points on the same equipotential is zero because $\Delta V=0$ ; field lines meet equipotentials at $90^{\circ}$ .
A dipole is in stable equilibrium when $\vec{p}$ is parallel to $\vec{E}$ ( $\theta=0^{\circ}$ ), where the potential energy is minimum.
Both the assertion-reason classics are true: $E=0$ inside a conductor because charge resides on the outer surface, and two capacitors at different potentials lose energy when joined as charge redistributes to a common potential.

Where the marks go

Reading 'distance halved' and answering $2F$ instead of $4F$ — the law is inverse-square, not inverse.
Saying flux depends on surface shape or the charge's position inside — it depends only on the enclosed charge $q$ .
In an assertion-reason item, marking 'R explains A' when R is merely a true statement that does not actually cause A — check the causal link, not just truth.
Confusing the sheet field $\sigma/2\varepsilon_0$ with the conductor-surface field $\sigma/\varepsilon_0$ , or forgetting that joining capacitors at different potentials dissipates energy.

How the board asks it

Assertion–Reasonconductor electrostatics: field inside and surface charge
Assertion (A): The electrostatic field inside the material of a charged conductor is zero. Reason (R): The net charge on an isolated conductor resides entirely on its outer surface. Choose the correct option: (a) Both A and R are true and R is the correct explanation of A; (b) Both A and R are true but R is not the correct explanation of A; (c) A is true, R is false; (d) A is false, R is true.
Multiple choicegauss's law: flux depends only on enclosed charge
A point charge $+q$ is placed off-centre inside a closed Gaussian surface of arbitrary shape. The net electric flux through the surface (a) increases as the charge moves nearer the wall; (b) depends on the shape of the surface; (c) equals $q/\varepsilon_0$ regardless of the position of the charge or the shape of the surface; (d) is zero. Choose the correct option and justify your choice in one line.
Multiple choicecoulomb's inverse-square law
Two point charges separated by a distance $r$ exert a force $F$ on each other. If the separation is reduced to $r/2$ with the charges unchanged, the new force is (a) $2F$ ; (b) $F/2$ ; (c) $4F$ ; (d) $F/4$ . Choose the correct option.
Multiple choicecapacitor energy at fixed charge
A parallel-plate capacitor carrying charge $Q$ is disconnected from the battery and its plates are then pulled farther apart. The stored energy $U = Q^2/2C$ (a) decreases; (b) stays the same; (c) increases; (d) becomes zero. Choose the correct option.
Multiple choiceseries combination of capacitors
Three identical capacitors, each of capacitance $C$ , are connected in series. Their equivalent capacitance is (a) $3C$ ; (b) $C$ ; (c) $C/3$ ; (d) $C/2$ . Choose the correct option.
Multiple choiceshort dipole: axial versus equatorial field
For a short electric dipole, the magnitude of the axial field at a distance $r$ compared with the equatorial field at the same distance $r$ is (a) equal; (b) half; (c) twice; (d) four times. Choose the correct option.

PreviousCapacitance, Dielectrics & Energy

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.