Physics (please help) electricity

  • Thread starter blackout85
  • Start date
  • #1
28
1
The work in joules required to carry a 6.0 C charge from a 5.0 V equipotential surface to a 6.0V equipotential surface and back again to the 5.0V surface is:

A) 0
B) 1.2 X 10^-5
C) 3.0 X 10^-5
D) 6.0 X 10^-5
E) 6.0X10^-6

Can someone please explain how to start off doing this problem. I thought that it involved the equation U= U2-U1= W. But, I am unsure how you connect the two.

Thank you
 

Answers and Replies

  • #2
261
2
You can think of this in terms of the net change in potential energy and the total work done, as you did.

Or you can think of it this way -- how much work did it take to move the charge one way? Then how much work did it take to move the charge back?

It helps your intuition if you think of the situation as rolling a ball up a hill. How much work does gravity do when you roll a ball up a hill? How much work does gravity do when the ball rolls back down?
 
  • #3
28
1
The answer then would have to be zero. Am I right to think that
 

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