Work done by external forces on an equipotential surface.

[nwyatt]

The question I am having trouble with shows an image of 2 different conductors, one is at +300 volts, while the other is -600 volts with a few lines going in between which circle around. The points on the equipotential surface are on lines labeled as a (which lies on a -200V line) and the other point is b (lies on -400V line). The question is asking how much work is done by an external agent if the electron is moved from A to B? I have had so much trouble finding an equation to fit the information given. I also have no idea on how to approach this problem.

 

[berkeman]

The question I am having trouble with shows an image of 2 different conductors, one is at +300 volts, while the other is -600 volts with a few lines going in between which circle around. The points on the equipotential surface are on lines labeled as a (which lies on a -200V line) and the other point is b (lies on -400V line). The question is asking how much work is done by an external agent if the electron is moved from A to B? I have had so much trouble finding an equation to fit the information given. I also have no idea on how to approach this problem.

Welcome to the PF. The lines going from one conductor to the other are Electric Field lines. The Equipotential surfaces are at right angles to the Electric field lines.

When they ask for the work done in moving a charge between equipotential surfaces, you should use an equation that expresses the work done in terms of the force applied and the distance, or you can use the simpler relation that the work done is equal to the change in total energy of the particle, change in TE = change in (KE + PE). There is a reason the "equipotential" surfaces have the word "potential" in them. What is the relation between the voltage level and the potential energy of a charged particle?

 

[berkeman]

Another hint -- look up an "electron volt" (eV). It's a unit of energy...

 

[nwyatt]

Your answer has helped me a great deal! So if my understanding is correct, the change is -200, and i would multiply this by -1.602x10^-19. Since it is an external force the sign changes to a +3.2x10^-17. I am not sure if my understanding is right, but that's the first thing that came to mind when you mentioned the difference in potential energies and when I looked up what an eV was.

 

[berkeman]

Your answer has helped me a great deal! So if my understanding is correct, the change is -200, and i would multiply this by -1.602x10^-19. Since it is an external force the sign changes to a +3.2x10^-17. I am not sure if my understanding is right, but that's the first thing that came to mind when you mentioned the difference in potential energies and when I looked up what an eV was.

Good. I didn't check your work, mainly because you didn't include units with your post. It's always good to include units with your equations and answers, because that helps you to check the consistency of your work, and helps us to understand what you are posting. But otherwise, it sounds like you've taken the right approach to the question.