NEETPhysicsCapacitorMCQ+4 / −1
To produce an instantaneous displacement current of in the space between the parallel plates of a capacitor of capacitance , the rate of change of applied variable potential difference must be :-
- A
- B
- C
- D
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Correct answer: B
$$ \begin{aligned} & \mathrm{Q}=\mathrm{CV} \\\\ & \begin{aligned} \frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{C} \cdot \frac{\mathrm{dV}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dV}}{\mathrm{dt}} & =\frac{\mathrm{I}}{\mathrm{C}}=\frac{2 \times 10^{-3}}{4 \times 10^{-6}} \\ & =\frac{10^3}{2}=500 \frac{\mathrm{V}}{\mathrm{s}} \end{aligned} \end{aligned} $$
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