Asked by SHUBHANGI

The work done by an external force to move a -7.75 µC charge from point a to point b is 25.0 * 10-4 J. If the charge was started from rest and had 4.73*10-4 J of kinetic energy when it reached point b, what is the magnitude of the potential difference between a and b?

15 years ago

Answered by drwls

The charge times the potential difference is the increase in potential energy. That plus the acquired kinetic energy equals the work done.

(Vb - Va)*Q + 4.73*10^-4 J
= (25.0)*10^-4 J

15 years ago

Answered by Tessa

I don't understand how you get the answer. What is the Vb, Va and Q?

15 years ago

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