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Electric Field & Point Charge Potential Energy

  • Thread starter Xels
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  • #1
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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 [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



8.5x105 N/C = 8.5x105 V/m

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

1.41x105 V * 8.0x10-6C = 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)
 
Last edited:

Answers and Replies

  • #2
19
0
Wrong equation used:

[tex]\Delta[/tex]U=q0Ed

[tex]\Delta[/tex]U=8x10-6C* -8.5x105N/C * 6m = -40.8J
 
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