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Electrostatics MCQs for NEET — Physics Questions with Answers

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The capacity of a parallel plate condenser is C. Its capacity when the separation between the plates is halved will be

4 C

2 C

$\frac{C}{2}$

$\frac{C}{4}$

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Explanation

$C = \frac{ε_{0} A}{d} . C' = \frac{ε_{0} A}{d / 2}$

⇒ C’ = 2C

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Eight small drops, each of radius r and having same charge q are combined to form a big drop. The ratio between the potentials of the bigger drop and the smaller drop is

8 : 1

4 : 1

2 : 1

1 : 8

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Explanation

$V o l u m e o f l a r g e d r o p = V o l u m e o f n s m a l l d r o p s \Rightarrow \frac{4}{3} {πR}^{3} = n \times \frac{4}{3} {πr}^{3} \Rightarrow R = n^{\frac{1}{3}} r Charge on large drop, Q = nq Potential at the centre of small drop, V_{small} = \frac{kq}{r} Potential at the centre of large drop, V_{Big} = \frac{KQ}{R} = \frac{nkq}{n^{\frac{1}{3}} r} = n^{\frac{2}{3}} \frac{kq}{r} = n^{\frac{2}{3}} V_{small} Put n = 8 - \frac{V_{B i g}}{V_{s m a l l}} = {(8)}^{2 / 3} = \frac{4}{1}$

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1000 small water drops each of radius r and charge q coalesce together to form one spherical drop. The potential of the big drop is larger than that of the smaller drop by a factor of

1000

100

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Explanation

$Potential of small drop : ‐ V = \frac{kq}{r} Charge on larger drop : ‐ Q_{net} = 1000 q Let radius of larger drop = R . \Rightarrow \frac{4}{3} {πR}^{3} = 1000 \times \frac{4}{3} {πr}^{3} R = 10 r Then potential of larger drop : ‐ V_{net} = \frac{kQ}{R} = \frac{K \times 1000 q}{10 r} V_{net} = \frac{kq}{r} \times 100 V_{net} = 100 V$

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A parallel plate condenser is immersed in an oil of dielectric constant 2. The field between the plates is

Increased proportional to 2

Decreased proportional to $\frac{1}{2}$

Increased proportional to $\sqrt{2}$

Decreased proportional to $\frac{1}{\sqrt{2}}$

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Explanation

$E_{medium} = \frac{E_{air}}{K} = \frac{E}{2}$

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If the dielectric constant and dielectric strength be denoted by k and x respectively, then a material suitable for use as a dielectric in a capacitor must have

High k and high x

High k and low x

Low k and low x

Low k and high x

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Explanation

High K means good insulating property and high x means able to withstand electric field to a higher value.

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When air in a capacitor is replaced by a medium of dielectric constant K, the capacity -

Decreases K times

Increases K times

Increases K² times

Remains constant

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Explanation

$C_{m e d i u m} = K \times C_{a i r}$

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64 drops each having the capacity C and potential V are combined to form a big drop. If the charge on the small drop is q, then the charge on the big drop will be

16q

64q

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Explanation

By using Q = nq ⇒ Q = 64q

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The capacity of a parallel plate capacitor increases with the

Decrease of its area

Increase of its distance

Increase of its area

None of the above

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Explanation

Capacity of parallel plate capacitor $C = \frac{ε_{0} A}{d}$

⇒ C ∝ A

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The radii of two metallic spheres P and Q are r₁ and r₂ respectively. They are given the same charge. If r₁ > r₂ , then on connecting them with a thin wire, the charge will flow

From P to Q

From Q to P

Neither the charge will flow from P to Q nor from Q to P

The information is incomplete

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Explanation

Since charge flows from high potential to lower potential.

If positive charge is given, then V₁ < V₂ as r₁ > r₂

So positive charge flows from Q → P

If negative charge is given, then V₁ > V₂

So negative charge flows from P → Q.

Since it is not given that whether the charge given is positive or negative, hence the information is incomplete.

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Between the plates of a parallel plate condenser, a plate of thickness t₁ and dielectric constant k₁ is placed. In the rest of the space, there is another plate of thickness t₂ and dielectric constant k₂. The potential difference across the condenser will be

$\frac{Q}{A ε_{0}} (\frac{t_{1}}{k_{1}} + \frac{t_{2}}{k_{2}})$

$\frac{ε_{0} Q}{A} (\frac{t_{1}}{k_{1}} + \frac{t_{2}}{k_{2}})$

$\frac{Q}{A ε_{0}} (\frac{k_{1}}{t_{1}} + \frac{k_{2}}{t_{2}})$

$\frac{ε_{0} Q}{A} (k_{1} t_{1} + k_{2} t_{2})$

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Explanation

Potential difference across the condenser

$V = V_{1} + V_{2} = E_{1} t_{1} + E_{2} t_{2} = \frac{σ}{K_{1} ε_{0}} t_{1} + \frac{σ}{K_{2} ε_{0}} t_{2}$

⇒ $V = \frac{σ}{ε_{0}} (\frac{t_{1}}{K_{1}} + \frac{t_{2}}{K_{2}}) = \frac{Q}{A ε_{0}} (\frac{t_{1}}{K_{1}} + \frac{t_{2}}{K_{2}})$

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