Question

If the voltage across the plates is increasing at the rate of 220 V/s, what is the displacement current in the capacitor

Answer 1

Complete Question

A parallel-plate capacitor has square plates 20 cm on a side and 0.50 cm apart. If the voltage across the plates is increasing at the rate of 220 V/s, what is the displacement current in the capacitor?

Answer:

The value is  $I = 1.416*10^{-8} \ A$

Explanation:

From the question we are told that

       The  length and breath of the square plate is  $l = b = 20 \ cm = 0.2 \ m$

        The  distance of separation between each plate is $k = 0.50 \ cm = 0.005 \ m$

       The rate of voltage  increase is  $\frac{dV}{dt} = 200 \ V/s$

Generally the charge on the plate is mathematically represented as

  $Q = CV$

Now C is the capacitance of the capacitor which is mathematically represented as

   $C = \frac{\epsilon_o * A}{k}$

Here A is the cross-sectional area which is mathematically represented as

    $A = l^2$

=>   $A = 0.2^2$

=>   $A = 0.04 \ m^2$

So  

     $C = \frac{8.85 *10^{-12} * 0.04}{0.005}$

          $C = 7.08*10^{-11} \ F$

Now the change of the charge flowing through the plates with time is mathematically represented as

        $\frac{d Q}{dt} = C \frac{dV}{dt}$

So  

       $\frac{d Q}{dt} = 7.08 *10^{-11} * 200$

       $\frac{d Q}{dt} = 1.416*10^{-8}$

Generally   $\frac{d Q}{dt} = \ current \ i.e \ I$

So

     $I = 1.416*10^{-8} \ A$