Question

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.

Answer 1

Answer : The electric potential difference between the plates is 63 volts.

Explanation :

Given that,

Electric field between two parallel plates, E = 1250 N/C

Separation between plates, d = 0.05 m

The relation between the electric field and electric potential is given by :

`E=(V)/(d)`

So,
`V=E* d`

`V=1250\ N/C* 0.05\ m`

`V=62.5\ V`

or

V = 63 Volts

Hence, this is the required solution.

Answer 2

In electronics, electric field is simply defined as the electric force per unit charge, or Newton/Coulomb. When you multiply the electric field with the distance between the charged parallel plates, it will yield the voltage. Thus,

E = V*d
1250 N/C = V*0.05 m
V = 25,000 volts