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A displacement current of 4.0 A can be set up in the space between two parallel plates of 6 $$\mu$$F capacitor. The rate of change of potential difference across the plates of the capacitor is nearly $$\alpha \times 10^6$$ V/s. The value of $$\alpha$$ is _______.
start with Q = C V
differentiating with respect to time:
dQ/dt = C dV/dt
since dQ/dt = I, we get:
I = C dV/dt
given:
I = 4 A
C = 6 μF = 6 × 10⁻⁶ F
$$\frac{dV}{dt}=\frac{I}{C}=\frac{4}{6\times10^{-6}}=\frac{2}{3}\times10^6$$
$$\approx0.67\times10^6\text{ V/s}$$
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