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 1.5 mm apart and the voltage across them is changing at a rate of 140 V/s .

Answer 1

Step-by-step explanation:

It is known that the relation between electric field and potential difference between the plates of a parallel plate capacitor is as follows.

E =
`(V)/(D)`

So, differentiating this on both the sides with respect o time as follows.

`(dE)/(dt) = (1)/(D) (dV)/(dt)`

Hence, rate of electric field changes between the plates of parallel plate capacitor as follows.

`(dE)/(dt) = (1)/(D) (dV)/(dt)`

=
`(1)/(1.5 * 10^(-3) m) * 140 V/s`

=
`93.33 * 10^(3)` V/ms

Thus, we can conclude that the rate at which the electric field changes between the round plates of a capacitor is
`93.33 * 10^(3)` V/ms.