Question

A parallel plate capacitor consisting of square plates is charged. One edge of the plates is 0.4 (m). When the current is 5 (A), the rate of change of the electric field between the plates is approximately how many V / (m.s). Eo = 8.85x10-12 (SI)

Answer 1

Answer:

Let a parallel plate capacitor of area A and distance between the plates d be charged by a current whose instantaneous value is i and let the instantaneous charge on the capacitor of capacitance C be q and the instantaneous voltage across the plates be v.

We know, q = C v. Now, let the electric field normal to the capacitor plates be E and we also know, E = v/d or v = E d, so, we get q = C E d (C and d being constants here). Differentiating both sides with repsect to time (t), we get:

We know, dq/dt = i and C = A/d, so, putting these two into the above equation, we get:

Substituting the given values in the problem into the above boxed equation, we get the value of the rate of change of the electric field between the plates of the capacitor as follows:

(correct upto 3 significant figures)

Cheers!