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(B) Displacement current is given by the formula:$i_{d} = \varepsilon_{0} \frac{d\phi_{E}}{dt}$Since electric flux $\phi_{E} = EA$,where $A$ is the ar

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$2$

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(B) Displacement current is given by the formula:$i_{d} = \varepsilon_{0} \frac{d\phi_{E}}{dt}$Since electric flux $\phi_{E} = EA$,where $A$ is the area of the plates and $E$ is the electric field,we have:$i_{d} = \varepsilon_{0} A \frac{dE}{dt}$Given that the electric field $E = \frac{V}{d}$,where $V$ is the potential difference and $d$ is the separation between the plates,we substitute this into the equation:$i_{d} = \varepsilon_{0} A \frac{d}{dt} \left( \frac{V}{d} ight) = \frac{\varepsilon_{0} A}{d} \frac{dV}{dt}$Since the capacitance $C = \frac{\varepsilon_{0} A}{d}$,the expression becomes:$i_{d} = C \frac{dV}{dt}$Given values are $C = 2\,\mu F = 2 	imes 10^{-6}\, F$ and $\frac{dV}{dt} = 10^6\, V/s$.Substituting these values:$i_{d} = (2 	imes 10^{-6}) 	imes 10^6 = 2\, A$

The potential difference between the plates of a parallel plate capacitor is changing at the rate of $10^6\, V/s$. If the capacitance is $2\,\mu F$,the displacement current in the dielectric of the capacitor will be.....$A$

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