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

$$\frac{dV}{dt}=\frac{I}{C}=\frac{4}{6\times10^{-6}}=\frac{2}{3}\times10^6$$

$$\approx0.67\times10^6\text{ V/s}$$

Create a FREE account and get:

Predict your JEE Main percentile, rank & performance in seconds

Check Your Score Now

Ask our AI anything

AI can make mistakes. Please verify important information.

Download resources