Question

The area of each plate of a parallel–plate capacitor is 100 cm2 and the intensity of electric field between the plates is 100 NC-1. Find the charge on each plate.

Answer 1

Answer:

Explanation:

We know that capacitance C= Aε0 /d

Also C=Q/V , so Q=CV

V=E*d

=> Q=Aε0 /d x Exd

Q=Aε0 E=100x10^-4 x8.85x10^-12 x 100= 88500x10^-8 = 8.85x10^-4

Answer 2

Answer:

ans = Eo = 8.854 * (10^-12)

Explanation:We know that capacitance C= Aε0 /d

Also C=Q/V , so Q=CV

V=E*d

=> Q=Aε0 /d x Exd

Q=Aε0 E = Eo = 8.854 * (10^-12)