To produce an instantaneous displacement current of 2 mA in the space between the parallel plates of a capacitor of capacitance 4 , the rate of change of applied variable potential difference must be [2023]

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Solution

Q.
To produce an instantaneous displacement current of 2 mA in the space between the parallel plates of a capacitor of capacitance 4 $μ F$ , the rate of change of applied variable potential difference $(\frac{d V}{d t})$ must be [2023]

1 800 V/s

2 500 V/s

3 200 V/s

4 400 V/s

Ans.
(2)

Displacement current, $I_{D} = ε_{0} \frac{d ϕ_{E}}{d t} = ε_{0} \frac{d}{d t} (E A)$

$= ε_{0} A \frac{d}{d t} (\frac{V}{d}) or I_{D} = \frac{ε_{0} A}{d} \frac{d V}{d t}$

$\Rightarrow I_{D} = \frac{C d V}{d t} or \frac{d V}{d t} = \frac{I_{D}}{C}$

Given, $I_{D} = 2 mA = 2 \times 10^{- 3} A;$ $C = 4 \times 10^{- 6} F$

So, $\frac{d V}{d t} = \frac{2 \times 10^{- 3}}{4 \times 10^{- 6}} = 500 V/s$

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