Question
You are given a 2μF parallel plate capacitor. How would you establish an instantaneous displacement current of 1mA in the space between its plates?
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Answer
The capacitance of capacitor $\text{C}=2\mu\text{F}$,
Displacement current Id = 1mA
Charge in capacitor, q = CV
$\text{I}_\text{d}\text{dt}=\text{Cdv}\ \ \big[\because\ \text{q}=\text{it}\big]$
or $\text{I}_\text{d}=\text{C}\frac{\text{dV}}{\text{dt}}$
$1\times10^{-3}=2\times10^{-6}\times\frac{\text{dV}}{\text{dt}}$
or $\frac{\text{dV}}{\text{dt}}=\frac{1}{2}\times10^{3}=500\text{V/s}$
Hence by applying a varying potential difference of 500V/s, we would produce a displacement current of desired value.
Displacement current Id = 1mA
Charge in capacitor, q = CV
$\text{I}_\text{d}\text{dt}=\text{Cdv}\ \ \big[\because\ \text{q}=\text{it}\big]$
or $\text{I}_\text{d}=\text{C}\frac{\text{dV}}{\text{dt}}$
$1\times10^{-3}=2\times10^{-6}\times\frac{\text{dV}}{\text{dt}}$
or $\frac{\text{dV}}{\text{dt}}=\frac{1}{2}\times10^{3}=500\text{V/s}$
Hence by applying a varying potential difference of 500V/s, we would produce a displacement current of desired value.
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