A set of charged plates 0.00262 m apart has an electric field of 155 N/C between them. What is the potential difference between the plates? V

Answer: The potential difference between the plates = 0.4061V

Explanation:

Given that the

Electric field strength E = 155 N/C

Distance d = 0.00262 m

From the definition of electric field strength, is the ratio of potential difference V to the distance between the plates. That is

E = V/d

Substitute E and d into the above formula

155 = V/0.00262

Cross multiply

V = 155 * 0.00262

V = 0.4061 V

The potential difference between the plates is 0.4061 V