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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?

Respuesta :

Kazeemsodikisola

Answer:

Explanation:

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