The potential energy at x = 8 m is -2000 V and at x = 2 m is +400 V. What is the magnitude and direction of the electric field?

Question

Quizplus · Accepted Answer

We can use the formula for electric potential difference to find the electric field.
ΔV = -Ed
where ΔV is the difference in potential energy, E is the electric field, and d is the displacement.
ΔV = Vf - Vi
where Vf is the final potential energy and Vi is the initial potential energy.
ΔV = (-2000 V) - (+400 V) = -2400 V
d = xf - xi = 8 m - 2 m = 6 m

Substituting the values in the first equation:
-2400 V = -Ed(6m)
E = 400 V/m directed parallel to the +x-axis

The Potential Energy at X = 8 M Is -2000