Blog Log in Get Started Free

Electrostatics MCQs for NEET — Physics Questions with Answers

Practice free Electrostatics (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

Chapter Difficulty Show only questions I answered incorrectly Clear Register free for difficulty & keyword filters

● ● ● ● ●

An oil drop having charge 2e is kept stationary between two parallel horizontal plates 2.0 cm apart when a potential difference of 12000 volts is applied between them. If the density of oil is 900 kg/m³, the radius of the drop will be -

$2 .0 \times 10^{- 6} m$

$1 .7 \times 10^{- 6} m$

$1 .4 \times 10^{- 6} m$

$1 .1 \times 10^{- 6} m$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

In equilibrium QE = mg ⇒ $Q . \frac{V}{d} = m g = (\frac{4}{3} π r^{3} ρ) g$

⇒ $2 \times 1 .6 \times 10^{- 19} \times \frac{12000}{2 \times 10^{- 2}} = \frac{4}{3} π r^{3} \times 900 \times 10$

⇒ r = 1.7 × 10^–6 m

● ● ● ● ●

The ratio of momenta of an electron and an $α$ -particle which are accelerated from rest by a potential difference of 100 volt is

$\sqrt{\frac{2 m_{e}}{m_{α}}}$

$\sqrt{\frac{m_{e}}{m_{α}}}$

$\sqrt{\frac{m_{e}}{2 m_{α}}}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Momentum $p = \sqrt{2 m K};$ where K = kinetic energy = Q.V

⇒ $p = \sqrt{2 m Q V}$ ⇒ $p \propto \sqrt{m Q}$

$\Rightarrow \frac{p_{e}}{p_{α}} = \sqrt{\frac{m_{e} Q_{e}}{m_{α} Q_{α}}} = \sqrt{\frac{m_{e}}{2 m_{α}}}$

● ● ● ● ●

When a proton is accelerated through 1V, then its kinetic energy will be -

1840 eV

13.6 eV

1 eV

0.54 eV

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$Δ K E = q V = e V = e \times 1 = 1 e V$

● ● ● ● ●

Ten electrons are equally spaced and fixed around a circle of radius R. Relative to V = 0 at infinity, the electrostatic potential V and the electric field E at the centre C are

$V \neq 0$ and $\vec{E} \neq 0$

$V \neq 0$ and $\vec{E} = 0$

V = 0 and $\vec{E} = 0$

V = 0 and $\vec{E} \neq 0$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

At centre E = 0, V ≠ 0 because the electric field is a vector quantity but the electric potential is not a vector quantity.

● ● ● ● ●

The displacement of a charge Q in the electric field $E = e_{1} \hat{i} + e_{2} \hat{j} + e_{3} \hat{k}$ is $\hat{r} = a \hat{i} + b \hat{j}$ . The work done is

$Q (a e_{1} + b e_{2})$

$Q \sqrt{{(a e_{1})}^{2} + {(b e_{2})}^{2}}$

$Q (e_{1} + e_{2}) \sqrt{a^{2} + b^{2}}$

$Q (\sqrt{e_{1}^{2} + e_{2}^{2})} (a + b)$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

By using $W = Q (\vec{E} . Δ \vec{r})$

⇒ $W = Q [(e_{1} \hat{i} + e_{2} \hat{j} + e_{3} \hat{k}) . (a \hat{i} + b \hat{j})] = Q (e_{1} a + e_{2} b)$

● ● ● ● ●

A cube of a metal is given a positive charge Q. For the above system, which of the following statements is true

Electric potential at the surface of the cube is zero

Electric potential within the cube is zero

Electric field is normal to the surface of the cube

Electric field varies within the cube

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Electric lines of force are always normal to metallic body.

● ● ● ● ●

Electric potential at any point is $V = - 5 x + 3 y + \sqrt{15} z$ , then the magnitude of the electric field is

$3 \sqrt{2}$

$4 \sqrt{2}$

$5 \sqrt{2}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$E_{x} = - \frac{d V}{d x} = - (- 5) = 5;$ $E_{y} = - \frac{d V}{d y} = - 3$

and $E_{z} = - \frac{d V}{d z} = - \sqrt{15}$

$E_{n e t} = \sqrt{E_{x}^{2} + E_{y}^{2} + E_{z}^{2}} = \sqrt{{(5)}^{2} + {(- 3)}^{2} + {(- \sqrt{15})}^{2}} = 7$

● ● ● ● ●

Kinetic energy of an electron accelerated in a potential difference of 100 V is

1.6 × 10^–17 J

1.6 × 10²¹ J

1.6 × 10^–29 J

1.6 × 10^–34 J

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

By using KE = QV ⇒ KE = 1.6 × 10^–19 × 100

= 1.6 × 10^–17 J

● ● ● ● ●

If identical charges (–q) are placed at each corner of a cube of side b, then electric potential energy of charge (+q) which is placed at centre of the cube will be -

$\frac{8 \sqrt{2} q^{2}}{4 π ε_{0} b}$

$\frac{- 8 \sqrt{2} q^{2}}{π ε_{0} b}$

$\frac{- 4 \sqrt{2} q^{2}}{π ε_{0} b}$

$\frac{- 4 q^{2}}{\sqrt{3} π ε_{0} b}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Length of the diagonal of a cube having each side b is $\sqrt{3} b .$ So distance of centre of cube from each vertex is $\frac{\sqrt{3} b}{2} .$

Hence potential energy of the given system of charge is

$U = 8 \times \{\frac{1}{4 π ε_{0}} . \frac{(- q) (q)}{\sqrt{3} b / 2}\} = \frac{- 4 q^{2}}{\sqrt{3} π ε_{0} b}$

● ● ● ● ●

A proton is about 1840 times heavier than an electron. When it is accelerated by a potential difference of 1 kV, its kinetic energy will be -

1840 keV

1/1840 keV

1 keV

920 keV

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

KE = QV = e × 10³ V = 1KeV.

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every Electrostatics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.