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Electrostatics MCQs for NEET — Physics Questions with Answers

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The electric field between the plates of a parallel plate capacitor when connected to a certain battery is E₀. If the space between the plates of the capacitor is filled by introducing a material of dielectric constant K without disturbing the battery connections, the field between the plates shall be

K E₀

E₀

$\frac{E_{0}}{K}$

None of the above

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Explanation

In the presence of battery potential difference remains constant. Also $E = \frac{V}{d},$ so E remains same.

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If the distance between parallel plates of a capacitor is halved and dielectric constant is doubled then the capacitance

Decreases two times

Increases two times

Increases four times

Remain the same

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Explanation

Capacitance with dielectric $C_{m e d i u m} = \frac{K ε_{0} A}{d}$

⇒ $C_{m e d i u m} \propto \frac{K}{d}$

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If there are n capacitors each of capacitance C in parallel connected to V volt source, then the energy stored is equal to

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

CV²

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

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Explanation

$U = \frac{1}{2} C_{e q} V^{2} = \frac{1}{2} (n C) V^{2}$

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A capacitor of capacitance 6 μF is charged upto 100 volt. The energy stored in the capacitor is

0.6 Joule

0.06 Joule

0.03 Joule

0.3 Joule

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Explanation

$U = \frac{1}{2} C V^{2} = \frac{1}{2} \times 6 \times 10^{- 6} {(100)}^{2} = 0 .03 J$

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The unit of electric permittivity is

Volt/m²

Joule/coulomb

Farad/m

Henry/m

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Explanation

$C = \frac{ε_{0} A}{d}$ ⇒ $ε_{0} = \frac{C d}{A}$ ⇒ $ε_{0}$ → $\frac{F a r a d \times m}{m^{2}} \to \frac{F}{m}$

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The work done in placing a charge of 8 × 10^–18 coulomb on a condenser of capacity 100 micro-farad is

32× 10^–32Joule

16 × 10^–32Joule

3.1 × 10^–26Joule

4 × 10^–10Joule

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Explanation

$W = \frac{Q^{2}}{2 C} = \frac{{(8 \times 10^{- 18})}^{2}}{2 \times 100 \times 10^{- 6}} = 32 \times 10^{- 32} J$

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A parallel plate capacitor of capacity C₀ is charged to a potential V₀

(i) The energy stored in the capacitor when the battery is disconnected and the separation is doubled E₁

(ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is E₂. Then E₁ / E₂ value is

3/2

1/2

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Explanation

Let $E = \frac{1}{2} C_{0} {V_{0}}^{2} then E_{1} = 2 E$ and $E_{2} = \frac{E}{2}$

So $\frac{E_{1}}{E_{2}} = \frac{4}{1}$

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Two identical capacitors are joined in parallel, charged to a potential V and then separated and then connected in series i.e. the positive plate of one is connected to negative of the other

The charges on the free plates connected together are destroyed

The charges on the free plates are enhanced

The energy stored in the system increases

The potential difference in the free plates becomes 2V

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Explanation

Q₁ = CV and Q₂ = CV

Applying charge conservation $C V_{1} + C V_{2} = Q_{1} + Q_{2}$

$C V_{1} + C V_{2} = 2 C V$ ⇒ $V_{1} + V_{2} = 2 V$

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A parallel plate capacitor is made by stacking n equally spaced plates connected alternately. If the capacitance between any two plates is C then the resultant capacitance is

(n – 1)C

(n + 1)C

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Explanation

The given arrangement becomes an arrangement of (n – 1) capacitors connected in parallel. So C_R = (n – 1)C

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n identical condensers are joined in parallel and are charged to potential V. Now they are separated and joined in series. Then the total energy and potential difference of the combination will be

Energy and potential difference remain the same

Energy remains the same and the potential difference is nV

Energy increases n times and potential difference is nV

Energy increases n times and potential difference remains the same

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Explanation

According to energy conservation, energy remains the same

⇒ $U_{p a r a l l e l} = U_{s e r i e s}$

⇒ $\frac{1}{2} (n C) V^{2} = \frac{1}{2} (\frac{C}{n}) V'^{2}$

⇒ V' = nV

(V’ = potential difference across series combination)

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