Question

In a Capacitor of 0.000031 F. a charge of 0.000022 C. exists between the plates.If the separation is 0.7 mm, What is the electric field?

Answer 1

1014 V/m

Step-by-step explanation:

C = 0.000031 F. Q = 0.000022 C, d = 0.7 mm,

V = Q/C = 0.000022/0.000031 = 0.7097 V

E = V/d = 0.7097/0.0007 = 1014 V/m