Calculating Work on Equipotential Surfaces

[threewingedfury]

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

I was thinkin the work is 0, but then again that seems too easy

 

[Mentz114]

Do you have a reason for thinking it's 0 ?

 

[AznBoi]

Is it 0 because of $$W=F*d=qEd$$?? The work done would be equal to 6C(1V)=6J when you bring it from 5V to 6V (6-5=1) and the work done would be equal to 6C(-1V)=-6J when it is going down the voltage from 6V to 5V (5-6=-1). Therefore when you add 6J and -6J you get 0? Is my solution correct?

 

[Mentz114]

I can't argue with that.

 

[threewingedfury]

thats what I was thinking, I just didn't know because mult choice questions are so tricky!

thanks!

 

[krateesh]

Zero..Work is of equal manitude but of opposite signs..they cancel to make net work 00000