Question 640: Electrostats & Capacitance

Question

To produce an instantaneous displacement current of $$2 \mathrm{~mA}$$ in the space between the parallel plates of a capacitor of capacitance $$4 ~\mu \mathrm{F}$$, the rate of change of applied variable potential difference $$\left(\frac{\mathrm{dV}}{\mathrm{dt}}\right)$$ must be :-

Accepted Answer

$$
\begin{aligned}
& \mathrm{Q}=\mathrm{CV} \\
& \begin{aligned}
\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{C} \cdot \frac{\mathrm{dV}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dV}}{\mathrm{dt}} & =\frac{\mathrm{I}}{\mathrm{C}}=\frac{2 	imes 10^{-3}}{4 	imes 10^{-6}} \
& =\frac{10^3}{2}=500 \frac{\mathrm{V}}{\mathrm{s}}
\end{aligned}
\end{aligned}
$$

Answer

$$800 \mathrm{~V} / \mathrm{s}$$

Answer

$$500 \mathrm{~V} / \mathrm{s}$$

Answer

$$200 \mathrm{~V} / \mathrm{s}$$

Answer

$$400 \mathrm{~V} / \mathrm{s}$$

Electrostats & Capacitance

To produce an instantaneous displacement current of $2 \mathrm{~mA} in the space between the parallel plates of a capacitor of capacitance 4 ~\mu \mathrm{F}, the rate of change of applied variable potential difference \left(\frac{\mathrm{dV}}{\mathrm{dt}}\right)$ must be :-

Choose the correct answer: