Displacement Current and Rate of Potential Change in Capacitor

To find the value of $\alpha$, we use the relation between the displacement current ($I_d$) and the rate of change of potential difference across a parallel plate capacitor.

The displacement current $I_d$ between the plates of a capacitor is given by: $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$

Since the electric field $E$ inside a parallel plate capacitor is $E = \frac{V}{d}$ and the electric flux is $\Phi_E = E A = \frac{V A}{d}$, we have: $I_d = \epsilon_0 \frac{d}{dt}\left(\frac{V A}{d}\right) = \left(\frac{\epsilon_0 A}{d}\right) \frac{dV}{dt}$

Since $C = \frac{\epsilon_0 A}{d}$, the formula simplifies to: $I_d = C \frac{dV}{dt}$

Given values:

• Displacement current, $I_d = 4.0\text{ A}$
• Capacitance, $C = 6\text{ }\mu\text{F} = 6 \times 10^{-6}\text{ F}$
• Rate of change of potential difference, $\frac{dV}{dt} = \alpha \times 10^6\text{ V/s}$

Substituting these values into the equation: $4.0 = 6 \times 10^{-6} \times \frac{dV}{dt}$

Rearranging for $\frac{dV}{dt}$: $\frac{dV}{dt} = \frac{4.0}{6 \times 10^{-6}} = \frac{2}{3} \times 10^6\text{ V/s} \approx 0.67 \times 10^6\text{ V/s}$

Comparing this with $\frac{dV}{dt} = \alpha \times 10^6\text{ V/s}$: $\alpha \approx 0.67$

Thus, the correct option is B.