Electric Field & Point Charge Potential Energy

AI Thread Summary
A uniform electric field of 8.5×10^5 N/C is directed along the positive x-axis. To find the change in electric potential energy for an 8.0 µC charge moving 6.0 m, the correct formula is ΔU = q0Ed. Using this equation, the calculation yields ΔU = 8.0×10^-6 C * (-8.5×10^5 N/C) * 6 m, resulting in approximately -40.8 J. The initial attempt incorrectly applied ΔU = qΔV, leading to an erroneous value of 1.13 J. The accurate change in electric potential energy is thus -41 J.
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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 \muC
charge as it moves from the origin to the point (6.0 m , 0).

Homework Equations



1 N/C = 1 V/m
\DeltaU = q\DeltaV
E = -\DeltaV/\Deltas

The Attempt at a Solution



8.5x105 N/C = 8.5x105 V/m

(8.5x105 V/m) / (6m) = 1.41x105V <----> E = -\DeltaV/\Deltas

1.41x105 V * 8.0x10-6C = 1.13J <-----> \DeltaU = q\DeltaV

(the correct answer is -41J; I'm trying to figure out how to get there)
 
Last edited:
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Wrong equation used:

\DeltaU=q0Ed

\DeltaU=8x10-6C* -8.5x105N/C * 6m = -40.8J
 
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