Question

Determine the rate at which the electric field changes between the round plates of a capacitor, 8.0 cm in diameter, if the plates are spaced 1.4 mm apart and the voltage across them is changing at a rate of 110 V/s .

Answer 1

Solution:

The relation between the potential difference and the electric field between the plates of the parallel plate capacitor is given by :

$E=\frac{V}{D}$

Differentiating on both the sides with respect to time, we get

$\frac{dE}{dt}=\frac{1}{D}\frac{dV}{dt}$

Therefore, the rate of the electric field changes between the plates of the parallel plate capacitor is given by :

$\frac{dE}{dt}=\frac{1}{D}\frac{dV}{dt}$

      $=\frac{1}{1.4 \times 10^{-3}} \times 110$

      $=7.85 \times 10^4$  V/m-s