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Electrostatics NEET Questions — Free MCQ Practice

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Three capacitors of capacitances 3 μF, 9 μF and 18 μF are connected once in series and another time in parallel. The ratio of equivalent capacitance in the two cases $(\frac{C_{s}}{C_{p}})$ will be

A.

1 : 15

B.

15 : 1

C.

1 : 1

D.

1 : 3

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Explanation

$\frac{1}{C_{s}} = \frac{1}{3} + \frac{1}{9} + \frac{1}{18} = \frac{1}{2}$

⇒ $C_{s} = 2 μ F$

$C_{p} = 3 + 9 + 18 = 30 μ F$

⇒ $\frac{C_{s}}{C_{p}} = \frac{2}{30} = \frac{1}{15}$

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Two capacitances of capacity C₁and C₂ are connected in series and potential difference V is applied across it. Then the potential difference across C₁will be

A.

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

B.

$V \frac{C_{1} + C_{2}}{C_{1}}$

C.

$V \frac{C_{2}}{C_{1} + C_{2}}$

D.

$V \frac{C_{1}}{C_{1} + C_{2}}$

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Explanation

Charge flowing $= \frac{C_{1} C_{2}}{C_{1} + C_{2}} V$ .

So potential difference across capacitor 1,

$V_{1} = \frac{C_{1} C_{2} V}{C_{1} + C_{2}} \times \frac{1}{C_{1}} = \frac{C_{2} V}{C_{1} + C_{2}}$

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The capacities of two conductors are C₁and C₂ and their respective potentials are V₁and V₁. If they are connected by a thin wire, then the loss of energy will be given by

A.

$\frac{C_{1} C_{2} (V_{1} + V_{2})}{2 (C_{1} + C_{2})}$

B.

$\frac{C_{1} C_{2} (V_{1} - V_{2})}{2 (C_{1} + C_{2})}$

C.

$\frac{C_{1} C_{2} {(V_{1} - V_{2})}^{2}}{2 (C_{1} + C_{2})}$

D.

$\frac{(C_{1} + C_{2}) (V_{1} - V_{2})}{C_{1} C_{2}}$

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Explanation

Initial energy $U_{i} = \frac{1}{2} C_{1} V_{1}^{2} + \frac{1}{2} C_{2} V_{2}^{2}$ ,

Final energy $U_{f} = \frac{1}{2} (C_{1} + C_{2}) V^{2}$ (where $V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}}$ )

Hence energy loss $Δ U = U_{i} - U_{f} = \frac{C_{1} C_{2}}{2 (C_{1} + C_{2})} {(V_{1} - V_{2})}^{2}$

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Three condensers each of capacitance 2F are put in series. The resultant capacitance is

A.

6F

B.

$\frac{3}{2} F$

C.

$\frac{2}{3} F$

D.

5F

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Explanation

$\frac{1}{C} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2}$

⇒ $C = \frac{2}{3} F$

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2 μF capacitance has potential difference across its two terminals 200 volts. It is disconnected with battery and then another uncharged capacitance is connected in parallel to it, then P.D. becomes 20 volts. Then the capacity of another capacitance will be

A.

2 μF

B.

4 μF

C.

18 μF

D.

10 μF

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Explanation

By using, common potential $V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}}$

⇒ $20 = \frac{2 \times 200 + C_{2} \times 0}{2 + C_{2}}$ ⇒ C₂ = 18 μF

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A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of the resulting system

A.

increases by a factor of 4

B.

decreases by a factor of 2

C.

remain the same

D.

increases by a factor of 2

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Explanation

When two uncharged identical capacitors are connected in parallel, the equivalent capacitance becomes twice the individual capacitance. Since the total charge remains the same, the energy stored in the combined system decreases by a factor of 2.

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A parallel plate air capacitor of capacitance C is connected, to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?

A.

The potential difference between the plates decreases K times

B.

The energy stored in the capacitor decreases K times

C.

The change in energy stored is $\frac{1}{2}$ CV^{2 $(\frac{1}{K - 1})$}

D.

The charge on the capacitor is not conserved

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Explanation

When a parallel plate air capacitor connected to a coil of emf V, then charge stored will be

q=CV

=> V=q/C

Also, energy stored is U= $\frac{1}{2}$ CV²= $\frac{q^{2}}{2 C}$

As the battery is disconnected from the capacitor the charge will not be destroyed i.e. q'=q with the introduction of dielectric in the gap of the capacitor the new capacitance will be
C'=CK

=> V'=q/C'=q/CK

The new energy stored will be

U'= $\frac{q^{2}}{2 C K}$

ΔU=U'-U= $\frac{q^{2}}{2 c}$ $(\frac{1}{K - 1})$

= $\frac{1}{2}$ CV^{2 $(\frac{1}{K - 1})$}

So, option (a),(b),(c) is correct but (d) is incorrect

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If potential (in volts) in a region is expressed as V(x,y,z)=6xy-y+2yz, the electric field (in N/C) at point (1,1,0) is

A.

-(3 $\hat{i}$ +5 $\hat{j}$ +3 $\hat{k}$ )

B.

-(6 $\hat{i}$ +5 $\hat{j}$ +2 $\hat{k}$ )

C.

-(2 $\hat{i}$ +3 $\hat{j}$ + $\hat{k}$ )

D.

-(6 $\hat{i}$ +9 $\hat{j}$ + $\hat{k}$ )

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Explanation

$G i v e n, P o t e n t i a l i n a r e g i o n; V = 6 x y - y + 2 y z \vec{E} = - \frac{δ V}{δ x} \hat{i} - \frac{δ V}{δ y} \hat{j} - \frac{δ V}{δ z} \hat{k} \vec{E} = - \frac{δ}{δ x} (6 x y - y + 2 y z) \hat{i} - \frac{δ}{δ y} (6 x y - y + 2 y z) \hat{j} - \frac{δ}{δ z} (6 x y - y + 2 y z) \hat{k} \vec{E} = - (6 y) \hat{i} - (6 x + 2 z) \hat{j} - (2 y) \hat{k} A t (1, 1, 0), \vec{E} = - (6) \hat{i} - (6) \hat{j} - (2) \hat{k}$

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A parallel plate air capacitor has capacity C, distance of separation between plates is d and potential difference V is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is

A.

C²V²/2d

B.

CV²/2d

C.

CV²/d

D.

C²V²/2d²

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Explanation

Force between plates of parallel capacitor

F=qE=q[ $σ$ /2 $ε_{0}$ ]

∴Surface charge density $σ$ =q/A

∴F=q[q/2A $ε$ _o]

=>F=q²/2A $ε$ _o
So, net charge across a capacitor, q=CV

F=^{$\frac{C^{2} V^{2}}{2 A ε_{0}}$} _{$[C = \frac{A ε_{0}}{d}]$

=> $F = (\frac{A ε_{0}}{d})$ x $\frac{C V^{2}}{2 A ε_{0}}$ ^{$= \frac{C V^{2}}{2 d}$}}

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A conducting sphere of radius R is given a charge Q. The electric potential and field at the centre of the sphere respectively are

A.

zero and Q/4π $ε$ _oR²

B.

Q/4π $ε$ _oR and zero

C.

Q/4π $ε$ _oRand Q/4π $ε$ _oR²

D.

Both are zero

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Explanation

For a uniformly charged spherical conductor, the electric potential inside the sphere is constant and equal to Q/4Ï€Ïµâ‚€R, while the electric field inside is zero due to the cancellation of fields from different parts of the charge distribution.

Electrostatics MCQs for NEET — Physics Questions with Answers

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