How would you establish an instantaneous current 4A in the space between the two parallel plates of $2 μF$ capacitor?

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By changing the potential difference across the parallel plates of capacitor at the rate of $1 \times 10^{6} V / s$

By changing the potential difference across the parallel plates of capacitor at the rate of $4 \times 10^{6} V / s$

By changing the potential difference across the parallel plates of capacitor at the rate of $8 \times 10^{6} V / s$

By changing the potential difference across the parallel plates of capacitor at the rate of $2 \times 10^{6} V / s$

answer is D.

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Detailed Solution

Displacement current $I_{D} = \in_{0} \frac{d ϕ_{E}}{d t}$
$\Rightarrow I_{D} = \in_{0} A \frac{d E}{d t}$
$\Rightarrow I_{D} = \in_{0} A \frac{d \frac{V}{d}}{d t} \Rightarrow I_{D} = \in_{o} \frac{A}{d} \frac{d V}{d t} = C \frac{d V}{d t}$
$\Rightarrow \frac{d V}{d t} = \frac{I_{D}}{C} = \frac{4}{2 \times 10^{- 6}} = 2 \times 10^{6} \frac{V}{s}$