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Electrostatics MCQs for NEET — Physics Questions with Answers

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A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in -

Reduction of charge on the plates and increase of potential difference across the plates

Increase in the potential difference across the plate, reduction in stored energy, but no change in the charge on the plates

Decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates

None of the above

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Explanation

Battery is disconnected so Q will be constant as C ∝ K. So with introduction of dielectric slab capacitance will increase using Q = CV, V will decrease and using $U = \frac{Q^{2}}{2 C}$ , energy will decrease.

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The capacity of a condenser is 4 × 10^–6 farad and its potential is 100 volts. The energy released on discharging it fully will be -

0.02 Joule

0.04 Joule

0.025 Joule

0.05 Joule

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Explanation

$The energy released = the energy stored in the capacitor . Energy stored in the capacitor - U = \frac{1}{2} C V^{2} = \frac{1}{2} \times 4 \times 10^{- 6} \times {(100)}^{2} = 0 .02 J$

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The insulated spheres of radii R₁ and R₂ having charges Q₁and Q₂ respectively are connected to each other. There is -

No change in the energy of the system

An increase in the energy of the system

Always a decrease in the energy of the system

A decrease in the energy of the system unless Q₁R₂ = Q₂R₁

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Explanation

When $\frac{Q_{1}}{R_{1}} \neq \frac{Q_{2}}{R_{2}};$ current will flow in connecting wire so that energy decreases in the form of heat through the connecting wire.

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The energy stored in a condenser of capacity C which has been raised to a potential V is given by -

$\frac{1}{2} C V$

$\frac{1}{2} C V^{2}$

$\frac{1}{2 V C}$

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Explanation

$U = \int_{0}^{V} C V d V = \frac{1}{2} C V^{2}$

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If two conducting spheres are separately charged and then brought in contact -

The total energy of the two spheres is conserved

The total charge on the two spheres is conserved

Both the total energy and charge are conserved

The final potential is always the mean of the original potentials of the two spheres

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Explanation

Law of conservation of charge.

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Eight drops of mercury of equal radii possessing equal charges combine to form a big drop. Then the capacitance of bigger drop compared to each individual small drop is

8 times

4 times

2 times

32 times

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Explanation

Volume of 8 small drops = Volume of the big drop

$8 \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} \Rightarrow R = 2 r As capacity is 4 {πε}_{0} r, hence capacity becomes 2 times .$

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A condenser of capacity 50 μF is charged to 10 volts. Its energy is equal to

2.5 × 10^–3Joule

2.5 × 10^–4Joule

5 × 10^–2Joule

1.2 × 10^–8Joule

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Explanation

$U = \frac{1}{2} C V^{2} = \frac{1}{2} \times 50 \times 10^{- 6} \times {(10)}^{2} = 2 .5 \times 10^{- 3} J$

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A parallel plate condenser has a capacitance 50 μF in air and 110 μF when immersed in an oil. The dielectric constant ‘k’ of the oil is

0.45

0.55

1.10

2.20

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Explanation

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

⇒ $K = \frac{C_{m e d i u m}}{C_{a i r}} = \frac{110}{50} = 2 .20$

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Separation between the plates of a parallel plate capacitor is d and the area of each plate is A. When a slab of material of dielectric constant k and thickness t(t < d) is introduced between the plates, its capacitance becomes -

$\frac{ε_{0} A}{d + t (1 - \frac{1}{k})}$

$\frac{ε_{0} A}{d + t (1 + \frac{1}{k})}$

$\frac{ε_{0} A}{d - t (1 - \frac{1}{k})}$

$\frac{ε_{0} A}{d - t (1 + \frac{1}{k})}$

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The capacity of parallel plate condenser depends on

The type of metal used

The thickness of plates

The potential applied across the plates

The separation between the plates

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Explanation

$C = \frac{K ε_{0} A}{d}$

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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.