# Electric Field & Point Charge Potential Energy

## Homework Statement

A uniform electric field of magnitude 8.5×105 N/C points in the positive x direction.

Find the change in electric potential energy of a 8.0 $$\mu$$C
charge as it moves from the origin to the point (6.0 m , 0).

## Homework Equations

1 N/C = 1 V/m
$$\Delta$$U = q$$\Delta$$V
E = -$$\Delta$$V/$$\Delta$$s

## The Attempt at a Solution

8.5x105 N/C = 8.5x105 V/m

(8.5x105 V/m) / (6m) = 1.41x105V <----> E = -$$\Delta$$V/$$\Delta$$s

1.41x105 V * 8.0x10-6C = 1.13J <-----> $$\Delta$$U = q$$\Delta$$V

(the correct answer is -41J; I'm trying to figure out how to get there)

Last edited:

$$\Delta$$U=q0Ed
$$\Delta$$U=8x10-6C* -8.5x105N/C * 6m = -40.8J