Question

Two parallel conducting plates are separated by 4.00 cm. The electric field strength between the two plates is 5.70×10⁴ V/m.
What is the magnitude of the potential difference between the plates?

Answer 1

Answer:

2280 V

Explanation:

distance between the plates, d = 4 cm = 0.04 m

Electric field strength, E = 5.7 x 10^4 V/m

The formula for potential difference is given by

V = E x d

V = 5.7 x 10^4 x 0.04

V = 2280 V

thus, the potential difference between the plates is 2280 V.