Question

A parallel-plate capacitor in air has circular plates of radius 2.8 cm separated by 1.1 mm. Charge is flowing onto the upper plate and off the lower plate at a rate of 5 A. Find the time rate of change of the electric field between the plates.

Answer 1

Answer:

The time rate of change of the electric field between the plates is  $\frac{E }{t} = 2.29 *10^{14} \ N \cdot C \cdot s^{-1}$  

Explanation:

From the question we are told that

    The  radius is  $r = 2.8 \ cm = 0.028 \ m$

     The distance of separation is  $d = 1.1 \ mm = 0.0011 \ m$

      The  current is  $I = 5 \ A$

Generally the electric field generated is mathematically represented as

         $E = \frac{q }{ \pi * r^2 \epsilon_o }$

Where $\epsilon_o$ is the permitivity of free space with a value

          $\epsilon_o = 8.85*10^{-12 }\ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

So the time rate of change of the electric field between the plates is mathematically represented as

        $\frac{E }{t} = \frac{q}{t} * \frac{1 }{ \pi * r^2 \epsilon_o }$

But $\frac{q}{t } = I$

So  

       $\frac{E }{t} = * \frac{I }{ \pi * r^2 \epsilon_o }$

substituting values  

        $\frac{E }{t} = * \frac{5 }{3.142 * (0.028)^2 * 8.85 *10^{-12} }$

        $\frac{E }{t} = 2.29 *10^{14} \ N \cdot C \cdot s^{-1}$