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Electrostatics MCQs for NEET — Physics Questions with Answers

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A thin spherical conducting shell of radius R has a charge q. Another charge Q is placed at the centre of the shell. The electrostatic potential at a point p a distance $\frac{R}{2}$ from the centre of the shell is

$\frac{(q + Q)}{4 π ε_{0}} \frac{2}{R}$

$\frac{2 Q}{4 π ε_{0} R}$

$\frac{2 Q}{4 π ε_{0} R} - \frac{2 q}{4 π ε_{0} R}$

$\frac{2 Q}{4 π ε_{0} R} + \frac{q}{4 π ε_{0} R}$

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Explanation

For a conducting spherical shell, the electric field and potential inside the shell are due solely to the charge Q at the center. Outside the shell, the potential is the sum of the potentials due to Q and the charge q on the shell. At a distance R/2 from the center, the potential contributions from Q and q add up to give the given expression.

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A sphere of 4 cm radius is suspended within a hollow sphere of 6 cm radius. The inner sphere is charged to potential 3 e.s.u. and the outer sphere is earthed. The charge on the inner sphere is

54 e.s.u.

$\frac{1}{4}$ e.s.u.

30 e.s.u.

36 e.s.u.

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Four charges $+ Q, - Q, + Q, - Q$ are placed at the corners of a square taken in order. At the centre of the square

$E = 0, V = 0$

$E = 0, V \neq 0$

$E \neq 0, V = 0$

$E = 0, V \neq 0$

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Point charge q₁ = 2 μC and q₂ = –1 μC are kept at points x = 0 and x = 6 respectively. Electrical potential will be zero at points

x = 2 and x = 9

x = 1 and x = 5

x = 4 and x = 12

x = –2 and x = 2

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Equipotential surfaces associated with an electric field which is increasing in magnitude along the x-direction are

Planes parallel to yz-plane

Planes parallel to xy-plane

Planes parallel to xz-plane

Coaxial cylinders of increasing radii around the x-axis

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A bullet of mass 2 gm is having a charge of 2 μC. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of 10 m/s ?

5 kV

50 kV

5 V

50 V

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Explanation

By using $\frac{1}{2} m v^{2} = Q V$

⇒ $\frac{1}{2} \times 2 \times 10^{- 6} \times {(10)}^{2} = 2 \times 10^{- 6} V$ ⇒ V = 50 kV

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In a certain charge distribution, all points having zero potential can be joined by a circle S. Points inside S have positive potential, and points outside S have a negative potential. A positive charge, which is free to move, is placed inside S .

It will remain in equilibrium

It can move inside S, but it cannot cross S

It must cross S at some time

It may move, but will ultimately return to its starting point

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Explanation

A free positive charge move from higher (positive) potential to lower (negative) potential. Hence, it must cross S at some time.

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A square of side ‘a’ has charge Q at its centre and charge ‘q’ at one of the corners. The work required to be done in moving the charge ‘q’ from the corner to the diagonally opposite corner is -

Zero

$\frac{Q q}{4 π \in_{0} a}$

$\frac{Q q \sqrt{2}}{4 π \in_{0} a}$

$\frac{Q q}{2 π \in_{0} a}$

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Two thin wire rings each having a radius R are placed at a distance d apart with their axes coinciding. The charges on the two rings are +q and –q. The potential difference between the centres of the two rings is -

Zero

$\frac{Q}{4 π ε_{0}} [\frac{1}{R} - \frac{1}{\sqrt{R^{2} + d^{2}}}]$

$Q R / 4 π ε_{0} d^{2}$

$\frac{Q}{2 π ε_{0}} [\frac{1}{R} - \frac{1}{\sqrt{R^{2} + d^{2}}}]$

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Explanation

The potential difference between the centers of two thin wire rings of radius R, placed at a distance d apart with charges +q and -q, is given by (q/2Ï€Îµâ‚€)[1/R - 1/âˆš(RÂ² + dÂ²)]. This is obtained by considering the potential due to each ring at the center of the other ring and taking their difference.

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A hollow metallic sphere of radius R is given a charge Q. Then the potential at the centre is -

Zero

$\frac{1}{4 π ε_{0}} . \frac{Q}{R}$

$\frac{1}{4 π ε_{0}} . \frac{2 Q}{R}$

$\frac{1}{4 π ε_{0}} . \frac{Q}{2 R}$

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Explanation

Potential V any where inside the hollow sphere, including the centre is $V = \frac{1}{4 π ε_{0}} \frac{Q}{r}$ .

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