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## Homework Statement

A uniform electric field of magnitude 8.5×10

^{5}N/C points in the positive x direction.

Find the change in electric potential energy of a 8.0 [tex]\mu[/tex]C

charge as it moves from the origin to the point (6.0 m , 0).

## Homework Equations

1 N/C = 1 V/m

[tex]\Delta[/tex]U = q[tex]\Delta[/tex]V

E = -[tex]\Delta[/tex]V/[tex]\Delta[/tex]s

## The Attempt at a Solution

8.5x10

^{5}N/C = 8.5x10

^{5}V/m

(8.5x10

^{5}V/m) / (6m) = 1.41x10

^{5}V <----> E = -[tex]\Delta[/tex]V/[tex]\Delta[/tex]s

1.41x10

^{5}V * 8.0x10

^{-6}C = 1.13J <-----> [tex]\Delta[/tex]U = q[tex]\Delta[/tex]V

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

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