Question

The displacement current density between the plates of a parallel-plate capacitor is uniform and has a magnitude of 19.9 A/m2 as the capacitor is being charged. What is dE/dt, the rate at which the electric field strength is changing between the plates

Answer 1

`(dE)/(dt) \approx 2.261 \overline {36} * 10^(12) \ V/ms`

Step-by-step explanation:

The parameters given in the question are;

The magnitude of the displacement current density between the plates of a parallel-plate capacitor, J = 19.9 A/m²

We have;

`i_d = \epsilon _0 * (d \Phi_E)/(dt) =\epsilon _0 * A * (d E)/(dt)`

Therefore;

`\therefore (dE)/(dt) =(i_d)/(\epsilon _0 \cdot A) =(J)/(\epsilon_0) = (19.9)/(8.8 * 10^(-12)) \approx 2.261 \overline {36} * 10^(12)`

Therefore;

`\therefore (dE)/(dt) \approx 2.261 \overline {36} * 10^(12) \ V/ms`