The electric field between two parallel plates has a magnitude of 1250 n/c. the positive plate is 0.05 m away from the negative plate. the electric potential difference between the plates, rounded to the tenths place, is v.

62.5 V

Explanation:

The relationship between electric field, potential difference and distance between the plates is:

where

E is the electric field magnitude

V is the potential difference

d is the distance between the plates

Here we have

E = 1250 N/C

d = 0.05 m

Solving the formula for V, we find

what are the options.

explanation:

i believe the answer is a hope it and is correct