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Electrostatics MCQs for NEET — Physics Questions with Answers

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The plates of a capacitor are charged to a potential difference of 320 volts and are then connected across a resistor. The potential difference across the capacitor decays exponentially with time. After 1 second the potential difference between the plates of the capacitor is 240 volts, then after 2 and 3 seconds the potential difference between the plates will be

200 and 180 V

180 and 135 V

160 and 80 V

140 and 20 V

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Explanation

$V = V_{0} (e^{- λ t})$

After 1 seconds

$V_{1} = 320 (e^{- λ}) \Rightarrow 240 = 320 (e^{- λ}) \Rightarrow e^{- λ} = \frac{3}{4}$

After 2 seconds

$V_{2} = 320 {(e^{- λ})}^{2} = 320 \times {(\frac{3}{4})}^{2} = 180 v o l t$

After 3 seconds

$V_{3} = 320 {(e^{- λ})}^{3} = 320 \times {(\frac{3}{4})}^{3} = 135 v o l t$

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The plates of a parallel plate condenser are pulled apart with a velocity v. If at any instant their mutual distance of separation is d, then the magnitude of the time of rate of change of capacity depends on d as follows

1/d

1/d²

d²

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Explanation

$C = \frac{ε_{0} A}{x};$

∴ $\frac{d C}{d t} = ε_{0} A \frac{d}{d t} (\frac{1}{x}) = \frac{- ε_{0} A}{x^{2}} (\frac{d x}{d t}) = \frac{- ε_{0} A}{d^{2}} (\frac{d x}{d t})$

⇒ $|\frac{d C}{d t}| = \frac{ε_{0} A}{d^{2}} v$

i.e. $|\frac{d C}{d t}| \propto \frac{1}{d^{2}}$

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A fully charged capacitor has a capacitance ‘C’. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity ‘s’ and mass ‘m’. If the temperature of the block is raised by ‘ΔT’, the potential difference ‘V’ across the capacitance is

$\frac{m s Δ T}{C}$

$\sqrt{\frac{2 m s Δ T}{C}}$

$\sqrt{\frac{2 m C Δ T}{s}}$

$\frac{m C Δ T}{s}$

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Explanation

$\frac{1}{2} C V^{2} = m . s . Δ T$

⇒ $V = \sqrt{\frac{2 m s Δ T}{C}}$

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A 4 μF capacitor, a resistance of 2.5 MΩ is in series with 12 V battery. Find the time after which the potential difference across the capacitor is 3 times the potential difference across the resistor. [Given ln(2)= 0.693]

13.86 s

6.93 s

7 s

14 s

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Explanation

$V_{R} = \frac{V_{0}}{4}$ = $V_{0} e^{- \frac{t}{R C}}$

⇒ $\frac{1}{4}$ = $e^{- \frac{t}{10}}$

⇒ $4 = e^{\frac{t}{10}}$

⇒ $\log_{e} 4 = \frac{t}{10}$

⇒ t = 10 log 4 = 13.86 s (RC = 2.5 × 10⁶ × 4 × 10⁻⁶ = 10)

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A point charge Q is placed at a distance d from the centre of an uncharged conducting sphere of radius R. The potential of the sphere is (d > R) –

$\frac{1}{4 π \in_{0}} . \frac{Q}{(d - R)}$

$\frac{1}{4 π \in_{0}} . \frac{Q}{d}$

$\frac{1}{4 π \in_{0}} . \frac{Q}{R}$

zero

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The plates of a parallel plate capacitor are separated by d cm. A plate of thickness t cm with dielectric constant $k_{1}$ is inserted and the remaining space is field with a plate of dielectric constant $k_{2}$ . If Q is the charge on the capacitor and area of plates is $A {cm}^{2}$ each, then potential difference between the plates is –

$\frac{Q}{ε_{0} A} (\frac{t}{k_{1}} + \frac{d - t}{k_{2}})$

$\frac{4 πQ}{A} (\frac{t}{k_{1}} + \frac{d - t}{k_{2}})$

$\frac{4 πQ}{A} (\frac{k_{1}}{t} + \frac{k_{2}}{d - t})$

$\frac{Q}{ε_{0} A} (\frac{k_{1}}{t} + \frac{d - t}{k_{2}})$

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Explanation

Potential difference across the plates,

$v = v_{1} + v_{2} = E_{1} t_{1} + E_{2} t_{2} = \frac{σ t}{K_{1} \in_{0}} + \frac{σ}{K_{2} \in_{0}} (d - t) = \frac{Q}{A \in_{0}} [\frac{t}{K_{1}} + \frac{d - t}{K_{2}}]$

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Electrical potential ‘v’ in space as a function of coordinates is given by, $v = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ . Then the electric field intensity at (1, 1, 1) is given by –

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

$\hat{i} + \hat{j} + \hat{k}$

zero

$\frac{1}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k})$

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Explanation

$2 . \vec{E} = - \nabla V = + \frac{1}{x^{2}} \hat{i} + \frac{1}{y^{2}} \hat{j} + \frac{1}{z^{2}} \hat{k} = \hat{i} + \hat{j} + \hat{k}$

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Two concentric, thin metallic spheres of radii $R_{1}$ and $R_{2}$ $(R_{1} > R_{2})$ bear changes $Q_{1}$ and $Q_{2}$ respectively. Then the potential at distance r between $R_{1}$ and $R_{2}$ will be $(k = \frac{1}{4 {πε}_{0}})$

$k (\frac{Q_{1} + Q_{2}}{r})$

$k (\frac{Q_{1}}{r} + \frac{Q_{2}}{R_{2}})$

$k (\frac{Q_{2}}{r} + \frac{Q_{1}}{R_{1}})$

$k (\frac{Q_{1}}{R_{1}} + \frac{Q_{1}}{R_{2}})$

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A parallel plate capacitor with air between the plates is charged to a potential difference of 500V and then insulated. A plastic plate is inserted between the plates filling the whole gap. The potential difference between the plates now becomes 75V. The dielectric constant of plastic is –

10/3

20/3

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Explanation

$3 . V = \frac{q}{C} \Rightarrow \frac{V_{1}}{V_{2}} = \frac{C_{2}}{C_{1}} \Rightarrow \frac{C_{2}}{C_{1}} = \frac{500}{75} = \frac{20}{3} By definition, C_{2} = {kC}_{1} \Rightarrow k = \frac{20}{3}$

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A capacitor of $1 µ F$ withstands a maximum voltage of 6 kilovolt while another capacitor of $2 µ F$ withstands a maximum voltage of 4 kilovolt. If the two capacitors are connected in series, the system will withstand a maximum voltage of –

2 kV

4 kV

6 kV

9 kV

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Explanation

4. For series combination $C_{s} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} \Rightarrow C_{s} = \frac{2}{3} μF$

When connected in series the maximum charge that can flow through the combination equals

the lower value of charge accommodated by the first capacitor i.e. $6000 µ C$

$∴ Q_{1} = 6000 μC and 8000 μC V_{s} = \frac{Q_{1}}{C_{s}} = \frac{6000 μC}{(2 / 3) μC} \Rightarrow V_{s} = 9 kV$

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