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Electrostatics MCQs for NEET — Physics Questions with Answers

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An electric dipole is in unstable equilibrium in the uniform electric field. The angle between its dipole moment and the electric field is

90 $°$

120 $°$

0 $°$

180 $°$

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Explanation

U = -PE cos $θ$

For unstable equilibrium electric dipole moment opposite to the electric field vector

i.e. $θ$ = 180 $°$

U $\to$ maximum

So $θ = π$ = 180 $°$

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A particle of mass m and charge q is at rest. If it is moved with velocity $\frac{\sqrt{3} c}{2}$ where c is speed of light in vacuum, then its charge will become/remain

$\sqrt{3} q$

$\frac{q}{2}$

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Explanation

Charged be invarient quantity.

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A charge q is to be divided on two small conducting spheres. What should be the value of charges on the spheres so that when placed at a certain distance apart, the repulsion force between them is maximum

$\frac{q}{4} a n d \frac{3 q}{4}$

$\frac{q}{2} a n d \frac{q}{2}$

$\frac{q}{3} a n d \frac{q}{3}$

$\frac{q}{4} a n d \frac{q}{4}$

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Explanation

$L e t t h e c h a r g e o n o n e s p h e r e = x T h e n, c h a r g e o n o t h e r s p h e r e = q - x F o r c e b e t w e e n t h e s p h e r e s : F = \frac{k x (q - x)}{r^{2}} = \frac{k (q x - x^{2})}{r^{2}} F o r m a x i m u m f o r c e b e t w e e n t h e s p h e r e s : \frac{d F}{d x} = 0 q - 2 x = 0 x = \frac{q}{2}$

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You are traveling in a car during a thunderstorm. In order to protect yourself from lightning, you would prefer to

Touch the nearest electric pole

Take shelter under a tree

Get out and crouch on the ground

Remain in the car

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Explanation

Due to electrostatic shielding.

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If a soap bubble is given some charge, then its radius

Increases

Decreases

Remain unchanged

May increase or decrease depending upon the give charge is positive or negative

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An electric dipole of dipole moment p is placed in an electric field of intensity E such that angle between electric field and dipole moment is $θ$ . Assuming that the potential energy of the dipole is zero when $θ = 0 °$ , the potential energy of the dipole will be

-pE cos $θ$

pE(1-cos $θ$ )

pE(cos $θ$ -1)

-2pE(cos $θ$ -1)

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Explanation

$∆ U = pE ({Cosθ}_{1} - {Cosθ}_{2}) U_{f} - U_{i} = pE ({Cosθ}_{1} - {Cosθ}_{2}) θ_{1} = 0 ° \Rightarrow U_{i} = 0 (A c c o r d i n g t o q u e s t i o n) U_{f} = pE (Cos 0 ° - Cosθ) U_{f} = pE (1 - Cosθ)$

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A point charge q is placed over a horizontal square of side L at a normal distance of L/4 from its centre. Electric flux through the square is

$ϕ = \frac{q}{6 ε_{0}}$

$ϕ < \frac{q}{6 ε_{0}}$

$\frac{q}{6 ε_{0}} < ϕ < \frac{q}{ε_{0}}$

$ϕ > \frac{q}{6 ε_{0}}$

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What is the angle between the electric dipole moment and the electric field strength due to it on the equatorial line

$0 °$

$90 °$

$180 °$

None of these

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The electrostatic field due to a charged conductor just outside the conductor is

zero and parallel to the surface at every point inside the conductor

zero and is normal to the surface at every point inside the conductor

parallel to the surface at every point and zero inside the conductor

normal to the surface at every point and zero inside the conductor

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Explanation

The electric field is zero inside a conductor. Just outside a conductor, the electric field lines are perpendicular to its surface.

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A soap bubble is given a negative charge, then its radius

Decreases

Increases

Remains unchanged

Nothing can be predicted as information is insufficient

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Explanation

Due to mutual repulsion of charges distributed on the surface of bubble.

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