Question

A $1.0\mu F$ capacitor is discharged, starting at $t=0$. The displacement current through the plates is $I_{disp}=\left(10A\right)exp(-t/2.0\mu s)$. What was the capacitor’s initial voltage ${(∆V_c)}_o$?

Short answer

The initial voltage of capacitor is${(∆V_c)}_o=20V$

Step-by-step solution

 : Given information

We have given that

$C=$$1.0\mu F$ at$t=0$

$I_{disp}=\left(10A\right)exp(-t/2.0\mu s)$

Simplification

As we know that displacement current $I_{disp}$flow in capacitor.

so, $\frac{dQ}{dt}=I_{disp}$

$dQ=I_{disp}dt$

$\int dQ=\int I_{disp}dt$ integrate both side.

$Q=\int \left(10\right)e^{(-t/2\mu s)}dt$= $\frac{\left(10\right)e^{(-t/2\mu s)}}{-1/2\mu s}$

$Q_{(t=0)=\frac{\left(10\right)e^{(-0/2\mu s)}}{-1/2\mu s}}$

$Q_{(t=0)}=-20\left({10}^{-6}\right)C$ since $\mu ={10}^{-6}$

$∆Q=20\left({10}^{-6}\right)C$

$∆Q=20\mu C$

We know that

$∆Q=C∆V$

put value of $∆Q$$=20\mu C$and C$=1.0\mu F$

$20\mu C=1\mu F(∆V)$

${\left(V_c\right)}_o=20V$