A uniform electric field, with a magnitude of 650 N/C, is directed parallel to the positive x-axis. If the potential at x = 3.0 m is 1 700 V, what is the potential at x = 1.0 m?

Given that,

Electric Field is

E = 650 N/C

The potential at x1 = 3 is

V1 = 1700V

What is the potential at x2 = 1

V2 = ?

Electric potential is given as

V = Ed

Then,

E = V/d

Therefore, the electric field is gradient of the potential and the position.

So,

E = - ∆V / ∆x

E = - (V2 - V1) / (x2 - x1)

E = - (V2-1700) / (1-3)

650 = - (V2-1700) / - 2

650 * - 2 = - V2 + 1700

-1300 = - V2 + 1700

V2 = 1700+1300

V2 = 3000 V