To produce an instantaneous displacement current of $2 ext{ mA}$ in the space between the parallel plates of

Question

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

Accepted Answer

1. **Concept:** Displacement current ($I_d$) arises due to a changing electric field (or electric flux) and is mathematically equivalent to the conduction current ($I_c$) flowing through the connecting wires of the capacitor.
2. **Formula:** The relationship between displacement current, capacitance ($C$), and the rate of change of voltage ($dV/dt$) is derived from $Q = CV$. Differentiating with respect to time gives $I = \frac{dQ}{dt} = C \frac{dV}{dt}$. Thus, $I_d = C \frac{dV}{dt}$.
3. **Given Data:**
   - Displacement Current, $I_d = 2 	ext{ mA} = 2 	imes 10^{-3} 	ext{ A}$.
   - Capacitance, $C = 4 	ext{ }\mu	ext{F} = 4 	imes 10^{-6} 	ext{ F}$.
4. **Calculation:**
   $$ \frac{dV}{dt} = \frac{I_d}{C} $$
   $$ \frac{dV}{dt} = \frac{2 	imes 10^{-3}}{4 	imes 10^{-6}} $$
   $$ \frac{dV}{dt} = 0.5 	imes 10^3 = 500 	ext{ V/s} $$

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