voltage and work

[dextor2003]

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When -10 C of charge are moved 1cm from point A to point B, 90 J of work is done.

The voltage between point A and B is:
0.9V
9V
90V
900V

i did 90 = 10 x .01 x V, and solved for V to get 900....which is wrong. the correct answer's 9V. EK just says that V= J/C, so 90/10 = 9...i understand that work is conservative, so the path of the object doesnt matter, but dont the charges still have to travel that .01m to do that work ? i dont understand why the distance isnt being taken into account in this case....or if it is, can someone help clarify how ?

thanks

 

[Rabolisk]

The charges do have to travel, and this movement is achieved by the voltage gradient or the electrostatic force exerted due to the electric field. Realize that voltage is just the change in potential energy per coulomb of charge. All conservative work converts kinetic energy into potential energy and vice-versa. If I say that two points have a certain voltage difference, that means that the change in energy, or the work done to move a charge there is voltage difference times the charge via W=-U=-Vq regardless of the physical separation of the two points.

To further explain, consider two parallel equipotential lines with a deltaV of x V. No matter which path I take, the same work is done. I can move perpendicularly, which would be the shortest path, or I can take a longer slanted route, but the work required will be the same.

Don't confuse this concept with W=Fd. If two points with x voltage difference were separated further from each other but still have the same deltaV, you would have less force to make up for the increase in distance. This is verifiable by using W=Fd, F=kq1q2/r^2 and V=Ed. Remember when distance is necessary for calculation and when it is not.

 

[Rabolisk]

What I meant to write was F=Eq, my apologies.