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Electrostatics MCQs for NEET — Physics Questions with Answers

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In the absence of other conductors, the surface charge density

Is proportional to the charge on the conductor and its surface area

Inversely proportional to the charge and directly proportional to the surface area

Directly proportional to the charge and inversely proportional to the surface area

Inversely proportional to the charge and the surface area

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Explanation

Surface charge density $σ = \frac{q}{A}$

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Three charges 4q, Q, and q are in a straight line in the position of 0, l/2, and l respectively. The resultant force on q will be zero if Q =

– q

–2q

$- \frac{q}{2}$

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Explanation

The force between 4q and q; $F_{1} = \frac{1}{4 π ε_{0}} \cdot \frac{4 q \times q}{l^{2}}$

The force between Q and q; $F_{2} = \frac{1}{4 π ε_{0}} \cdot \frac{Q \times q}{{(l / 2)}^{2}}$

We want $F_{1} + F_{2} = 0$ or $\frac{4 q^{2}}{l^{2}} = - \frac{4 Q q}{l^{2}}$

⇒ Q = –q

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Two small spheres each having the charge +Q are suspended by insulating threads of length L from a hook. This arrangement is taken in space where there is no gravitational effect, then the angle between the two suspensions and the tension in each will be

$180^{o}, \frac{1}{4 π ε_{0}} \frac{Q^{2}}{{(2 L)}^{2}}$

$90^{o}, \frac{1}{4 π ε_{0}} \frac{Q^{2}}{L^{2}}$

$180^{o}, \frac{1}{4 π ε_{0}} \frac{Q^{2}}{2 L^{2}}$

$180^{o}, \frac{1}{4 π ε_{0}} \frac{Q^{2}}{L^{2}}$

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One metallic sphere A is given positive charge whereas another identical metallic sphere B of exactly same mass as of A is given equal amount of negative charge. Then

Mass of A and mass of B is the same

Mass of A is more

Mass of B is less

Mass of B is more

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Explanation

Negative charge means excess of electron which increases the mass of sphere B.

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Five balls numbered 1 to 5 are suspended using separate threads. Pairs (1, 2), (2, 4), and (4, 1) show electrostatic attraction, while pair (2, 3) and (4, 5) show repulsion. Therefore ball 1 must be

Positively charged

Negatively charged

Neutral

Made of metal

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Explanation

(2, 3) and (4, 5) repel each other. It is clear that 2, 3, 4, and 5 are charged bodies. Let (2, 3) have +ve charge and (4, 5) have -ve charge. (1, 2) attract each other. That means ball 1 has -ve charge. But (1, 4) also attract each other. So ball 1 should be neutral. In that case, only, ball 1 can be attracted by both 2 and 4.It means be ball one has no charge so it must negatively charge.

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Two identical conductors of copper and aluminium are placed in an identical electric field. The magnitude of induced charge in the aluminum will be

Zero

Greater than in copper

Equal to that in copper

Less than in copper

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Explanation

Since both are metals so equal amount of charge will induce on them.

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The electric field near a conducting surface having a uniform surface charge density σ is given by

$\frac{σ}{ε_{0}}$ and is parallel to the surface

$\frac{2 σ}{ε_{0}}$ and is parallel to the surface

$\frac{σ}{ε_{0}}$ and is normal to the surface

$\frac{2 σ}{ε_{0}}$ and is normal to the surface

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Explanation

Electric field near the conductor surface is given by $\frac{σ}{ε_{0}}$ and it is perpendicular to surface.

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The magnitude of electric field E in the annular region of a charged cylindrical capacitor

Is same throughout

Is higher near the outer cylinder than near the inner cylinder

Varies as 1/r, where r is the distance from the axis

Varies as 1/r², where r is the distance from the axis

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Explanation

The magnitude of electric field in the annular region of a charged cylindrical capacitor is given by $E = \frac{1}{2 π ε_{0}} \frac{λ}{r}$ where λ is the charge per unit length and r is the distance from the axis of the cylinder. Thus $E \propto \frac{1}{r}$

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A hollow insulated conducting sphere is given a positive charge of 10μC. What will be the electric field at the centre of the sphere if its radius is 2 meters

Zero

5μ Cm^–2

20μ Cm^–2

8μ Cm^–2

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Explanation

The intensity of electric field inside a hollow conducting sphere is zero.

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An electron of mass m_e initially at rest moves through a certain distance in a uniform electric field in time t₁. A proton of mass m_p also initially at rest takes time t₂ to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio of t₂/ t₁ is nearly equal to

${(m_{p} / m_{e})}^{1 / 2}$

${(m_{e} / m_{p})}^{1 / 2}$

1836

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Explanation

For electron $s = \frac{1}{2} \frac{e E}{m_{e}} \times t_{1}^{2}$ , For proton $s = \frac{1}{2} \frac{e E}{m_{p}} \times t_{2}^{2}$

∴ $\frac{t_{2}^{2}}{t_{1}^{2}} = \frac{m_{p}}{m_{e}} \Rightarrow \frac{t_{2}}{t_{1}} = \sqrt{\frac{m_{p}}{m_{e}}} = {(\frac{m_{p}}{m_{e}})}^{1 / 2}$

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