Question

The electric field between two parallel plates has a magnitude of 875 N/C. The positive plate is 0.002 m away from the negative plate. What is the electric potential difference between the plates? 2.3 × 10-6 V 1.8 × 100 V 8.8 × 102 V 4.4 × 104 V

Answer 1

Answer : Electric potential,
`V=1.8* 10^0\ V`

Explanation :

It is given that,

Magnitude of electric field,
`E=875\ N/C`

Distance between two plates,
`d=0.002\ m`

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

`E=(V)/(d)`

`V=E* d`

`V= 875\ N/C* 0.002\ m`

`V=1.75\ V`

or

`V=1.8* 10^0\ V`

So, the magnitude of electric potential is given by option (2).

Hence, this is the required solution.

Answer 2

Given:
E = 875 N/C electric field
d = 00.002 m, distance between parallel plates

Note that
1 V = 1 J/C

The electric potential difference is

`V=E*d \\\\ =(875 \, (N)/(C))*(0.002 \, m) \\\\ = 1.75 \, (J)/(C) \\\\ =1.75 \, V`

Answer: 1.8 V (nearest tenth)