Equipotential Surfaces: Calculating Work Required to Move a Charge

[daimoku]

[SOLVED] Equipotential surfaces

Homework Statement

http://personalpages.tds.net/~locowise/test/equipot1.jpg
Fig. 1 -- Some equipotential surfaces

In the figure above, you see a set of equipotentials representing an electric field in the region and some labeled points (A..G).

What is the work required by you to move a 3.4 x 10^-6 C charge from E to F?

We know k=8.99*10^9 N*m^2/C^2

Homework Equations

r = kq/v
Work = force * displacement
V2-V1=-Welec/q

The Attempt at a Solution

Calculated Re to be 196.81m and Rf to be 305.66m. <---not sure if this is necessary
Vf-Ve=-80V

-80V=-Welec/(3.4*10^-6C)
Welec comes out to be 2.72*10-4 J

Could someone point out where I made a mistake or even if I used the correct method? Many thanks for your help!

 

[Shooting Star]

Calculated Re to be 196.81m and Rf to be 305.66m. <---not sure if this is necessary

What is this? Looks like you are trying to find the radii of the two equipotential surfaces. Do they look spherical?

Vf-Ve=-80V

-80V=-Welec/(3.4*10^-6C)
Welec comes out to be 2.72*10-4 J

Could someone point out where I made a mistake or even if I used the correct method? Many thanks for your help!

The answer looks correct, except that it should have negative sign, since the work is being done by the charge.

 

[daimoku]

Thanks alot! You were correct-o-mundo on the negative sign!

 

[kihr]

The question is the work done BY ME, and NOT by the electric field. The work done by the electric field is positive as the force and displacement are in the same direction (since the charge moves from a point of higher to lower potential). Since the work is done by moving the charge without acceleration, the force given by me on the charge has to be equal and opposite to the force of the electric field on the charge. Hence the work done by me is negative.