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Electrostatics MCQs for NEET — Physics Questions with Answers

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A conductor with a positive charge

Is always at +ve potential

Is always at zero potential

Is always at negative potential

May be at +ve, zero or –ve potential

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Explanation

May be at positive, zero or negative potential, it is according to the way one defines the zero potential.

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Two spheres A and B of radius 4 cm and 6 cm are given charges of 80 μc and 40μc respectively. If they are connected by a fine wire, the amount of charge flowing from one to the other is -

20μc from A to B

16μc from A to B

32μc from B to A

32μc from A to B

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Explanation

Total charge $Q = 80 + 40 = 120 μ C$ . By using the formula $Q_{1} ' = Q [\frac{r_{1}}{r_{1} + r_{2}}]$ .

A new charge on sphere A is $Q_{A}^{'} = Q [\frac{r_{A}}{r_{A} + r_{B}}] = 120 [\frac{4}{4 + 6}] = 48 μ C$ .

Initially it was 80 μC i.e., (80-48)= 32 μC charge flows from A to B.

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On rotating a point charge having a charge q around a charge Q in a circle of radius r. The work done will be

$q \times 2 π r$

$\frac{q \times 2 π Q}{r}$

Zero

$\frac{Q}{2 ε_{0} r}$

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Explanation

Since charge Q moving on equipotential surface so work done is zero.

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Two charge +q and –q are situated at a certain distance. At the point exactly midway between them -

Electric field and potential both are zero

Electric field is zero but the potential is not zero

Electric field is not zero but the potential is zero

Neither electric field nor potential is zero

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Two insulated charged conducting spheres of radii 20 cm and 15 cm respectively and having an equal charge of 10 C are connected by a copper wire and then they are separated. Then -

Both the spheres will have the same charge of 10 C

Surface charge density on the 20 cm sphere will be greater than that on the 15 cm sphere

Surface charge density on the 15 cm sphere will be greater than that on the 20 cm sphere

Surface charge density on the two spheres will be equal

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Explanation

After redistribution, charges on them will be different, but they will acquire common potential

i.e. $k \frac{Q_{1}}{r_{1}} = k \frac{Q_{2}}{r_{2}}$ ⇒ $\frac{Q_{1}}{Q_{2}} = \frac{r_{1}}{r_{2}}$

As $σ = \frac{Q}{4 π r^{2}}$ ⇒ $\frac{σ_{1}}{σ_{2}} = \frac{Q_{1}}{Q_{2}} \times \frac{r_{2}^{2}}{r_{1}^{2}}$

⇒ $\frac{σ_{1}}{σ_{2}} = \frac{r_{2}}{r_{1}}$ ⇒ $σ \propto \frac{1}{r}$

i.e. surface charge density on smaller sphere will be more.

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Two equal charges q are placed at a distance of 2a and a third charge –2q is placed at the midpoint. The potential energy of the system is -

$\frac{q^{2}}{8 π ε_{0} a}$

$\frac{6 q^{2}}{8 π ε_{0} a}$

$- \frac{7 q^{2}}{8 π ε_{0} a}$

$\frac{9 q^{2}}{8 π ε_{0} a}$

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Explanation

$U_{s y s t e m} = \frac{1}{4 π ε_{0}} \frac{(q) (- 2 q)}{a} + \frac{1}{4 π ε_{0}} \frac{(- 2 q) (q)}{a} + \frac{1}{4 π ε_{0}} \frac{(q) (q)}{2 a}$

$U_{s y s t e m} = - \frac{7 q^{2}}{8 π ε_{0} a}$

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A particle of mass m and charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is -

$q E y^{2}$

$q E^{2} y$

$q E y$

$q^{2} E y$

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Explanation

Kinetic energy = Force × Displacement = qEy

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How much kinetic energy will be gained by an $α$ – particle in going from a point at 70 V to another point at 50 V ?

$40 e V$

$40 k e V$

$40 M e V$

$0 e V$

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Explanation

$K E = q (V_{1} - V_{2}) = 2 \times 1 .6 \times 10^{- 19} \times (70 - 50) = 40 e V$

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If a charged spherical conductor of radius 10 cm has potential V at a point distant 5 cm from its centre, then the potential at a point distant 15 cm from the centre will be -

$\frac{1}{3} V$

$\frac{2}{3} V$

$\frac{3}{2} V$

3 V

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Explanation

Potential inside the sphere will be same as that on its surface i.e. $V = V_{s u r f a c e} = \frac{K Q}{10} u n i t s$ , $V_{o u t} = \frac{K Q}{15} u n i t s$

∴ $\frac{V_{o u t}}{V} = \frac{2}{3}$ ⇒ $V_{o u t} = \frac{2}{3} V$

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What is the potential energy of the equal positive point charges of 1 μC each held 1 m apart in air ?

$9 \times 10^{- 3} J$

$9 \times 10^{- 3} e V$

$2 e V / m$

Zero

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Explanation

By using $U = 9 \times 10^{9} \frac{Q_{1} Q_{2}}{r}$

⇒ $U = 9 \times 10^{9} \times \frac{10^{- 6} \times 10^{- 6}}{1} = 9 \times 10^{- 3} J$

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