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Electrostatics NEET Questions — Free MCQ Practice

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A parallel plate condenser with oil between the plates (dielectric constant of oil K = 2) has a capacitance C. If the oil is removed, then capacitance of the capacitor becomes

A.

$\sqrt{2} C$

B.

2C

C.

$\frac{C}{\sqrt{2}}$

D.

$\frac{C}{2}$

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Explanation

$C_{a i r} = \frac{C_{m e d i u m}}{K} = \frac{C}{2}$

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What is the area of the plates of a 3F parallel plate capacitor, if the separation between the plates is 5 mm ?

A.

$1 .694 \times 10^{9} m^{2}$

B.

$4 .529 \times 10^{9} m^{2}$

C.

$9 .281 \times 10^{9} m^{2}$

D.

$12 .981 \times 10^{9} m^{2}$

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Explanation

We have $C = \frac{ε_{0} A}{d}$

⇒ $A = \frac{C d}{ε_{0}} = \frac{3 \times 5 \times 10^{- 3}}{8 .85 \times 10^{- 12}}$

$= 1 .7 \times 10^{9} m^{2}$

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The energy stored in the condenser is

A.

QV

B.

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

C.

$\frac{1}{2} C$

D.

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

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Explanation

The energy stored $= \frac{1}{2} Q V$

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Two identical charged spherical drops each of capacitance C merge to form a single drop. The resultant capacitance is -

A.

Equal to 2C

B.

Greater than 2C

C.

Less than 2C but greater than C

D.

Less than C

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Explanation

$C' = n^{1 / 3} C$

⇒ $C' = 2^{1 / 3} C$

⇒ 2C > C' > C

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A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor ?

A.

1.5 × 10^–8Joule

B.

2.5 × 10^–7Joule

C.

3.5 × 10^–5Joule

D.

4.5 × 10^–2Joule

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Explanation

$U = \frac{1}{2} C V^{2} = \frac{1}{2} \times 12 \times 10^{- 12} \times {(50)}^{2}$ = 1.5 × 10^–8 J

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The capacity of a parallel plate condenser is 15 μF, when the distance between its plates is 6 cm. If the distance between the plates is reduced to 2 cm, then the capacity of this parallel plate condenser will be

A.

15 μF

B.

30 μF

C.

45 μF

D.

60 μF

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Explanation

$C \propto \frac{1}{d} \Rightarrow \frac{C_{1}}{C_{2}} = \frac{d_{2}}{d_{1}}$

⇒ $\frac{15}{C_{2}} = \frac{2}{6}$ ⇒ C₂ = 45 μF

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The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4 cm is 2 μF. The separation is reduced to half and it is filled with a dielectric substance of value 2.8. The final capacity of the capacitor is

A.

11.2 μF

B.

15.6 μF

C.

19.2 μF

D.

22.4 μF

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Explanation

$C = \frac{ε_{0} K A}{d}$ ⇒ $\frac{C_{1}}{C_{2}} = \frac{K_{1}}{K_{2}} \times \frac{d_{2}}{d_{1}}$

$\frac{2}{C_{2}} = \frac{1}{2 .8} \times \frac{(0 .4 / 2)}{(0 .4)}$ ⇒ $C_{2} = 11 .2 μ F$

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Two insulated metallic spheres of 3 μF and 5 μF capacitances are charged to 300 V and 500V respectively. The energy loss, when they are connected by a wire is

A.

0.012 J

B.

0.0218 J

C.

0.0375 J

D.

3.75 J

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Explanation

$I n i t i a l c h a r g e o n t h e c a p a c i t o r s : Q_{1} = C_{1} V_{1} Q_{2} = C_{2} V_{2} U_{i} = \frac{1}{2} \frac{{Q_{1}}^{2}}{C_{1}} + \frac{1}{2} \frac{{Q_{2}}^{2}}{C_{2}} W h e n t h e y a r e c o n n e c t e d : \frac{Q_{1}'}{Q_{2}'} = \frac{C_{1}}{C_{2}} Q_{1}' = (Q_{1} + Q_{2}) \frac{C_{1}}{C_{1} + C_{2}} Q_{2}' = (Q_{1} + Q_{2}) \frac{C_{2}}{C_{1} + C_{2}} U_{f} = \frac{1}{2} \frac{Q_{1}'^{2}}{C_{1}} + \frac{1}{2} \frac{Q_{2}'^{2}}{C_{2}} L o s s i n h e a t = U_{f} - U_{i} = \frac{1}{2} \frac{C_{1} C_{2} {(V_{2} - V_{1})}^{2}}{(C_{1} + C_{2})}$

$Δ U = \frac{1}{2} \frac{C_{1} C_{2} {(V_{2} - V_{1})}^{2}}{(C_{1} + C_{2})} = \frac{(3 \times 5) \times 10^{- 12} \times {(500 - 300)}^{2}}{(3 + 5) \times 10^{- 6}}$

$= \frac{15 \times 10^{- 12} \times 4 \times 10^{4}}{8 \times 10^{- 6}} = 0 .0375 J$

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A charge of 40 μC is given to a capacitor having capacitance C = 10 μF. The stored energy in ergs is

A.

80 × 10^–6

B.

800

C.

80

D.

8000

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Explanation

$U = \frac{Q^{2}}{2 C} = \frac{{(40 \times 10^{- 6})}^{2}}{2 \times 10^{- 6} \times 10} = \frac{16 \times 10^{- 10}}{2 \times 10^{- 5}} = 8 \times 10^{- 5} J$

$= 8 \times 10^{- 5} \times 10^{7} = 800 e r g$

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A parallel plate capacitor has plate area A and separation d. It is charged to a potential difference V₀. The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is

A.

$\frac{3 ε_{0} A V_{0}^{2}}{d}$

B.

$\frac{ε_{0} A V_{0}^{2}}{2 d}$

C.

$\frac{ε_{0} A V_{0}^{2}}{3 d}$

D.

$\frac{ε_{0} A V_{0}^{2}}{d}$

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Explanation

Work done W = U_f – U_i

$U_{i} = \frac{1}{2} C {V_{0}}^{2} {and U}_{f} = \frac{1}{2} \frac{(C)}{3} . {(3 V_{0})}^{2}$ $= 3 \times \frac{1}{2} C {V_{0}}^{2}$

So $W = \frac{ε_{0} A {V_{0}}^{2}}{d}$

Electrostatics MCQs for NEET — Physics Questions with Answers

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