The potential difference between the two plat | vedclass

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(C) The displacement current $I_d$ in a capacitor is given by the formula: $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$.Since the electric flux $\Phi_E = E \

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$2$

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(C) The displacement current $I_d$ in a capacitor is given by the formula: $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$.Since the electric flux $\Phi_E = E \cdot A = \frac{V}{d} \cdot A = \frac{V \cdot A}{d}$, and the capacitance $C = \frac{\epsilon_0 A}{d}$, we can write the displacement current as $I_d = C \frac{dV}{dt}$.Given values are $C = 2 \; \mu F = 2 	imes 10^{-6} \; F$ and $\frac{dV}{dt} = 10^6 \; V/s$.Substituting these values into the formula:$I_d = (2 	imes 10^{-6} \; F) 	imes (10^6 \; V/s) = 2 \; A$.Therefore, the displacement current is $2 \; A$.

The potential difference between the two plates of a parallel plate capacitor changes at a rate of $10^6 \; V/s$. If the capacitance of the capacitor is $2 \; \mu F$, what is the displacement current in the dielectric of the capacitor?

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