This is one of the "fast answer" exercises I've been given to train (should be answerable in around 6-7 minutes). I can only think of a very long-round way to solve this. The question is as following:

Two point charges 2Q and Q are located at the opposite corners of a square of length l (2Q at the top right corner). The point A is in the lower left corner and the point B is located halfway the right side of the square. Determine the work $W_{AB}$ required to move a particle with charge q from A to B.

There are several methods I can think of, though non come to an answer. My best guess is to calculate the magnitude of the force acting on the particle in A, and in B. And then substract these... However to simply calculate the effective force in position B proves itself to be lengthy task.

So there must be a simpler mechanic?

  • $\begingroup$ Why don't you calculate the Potential Energy at each point? Then $W = -\Delta P.E$ $\endgroup$ – mikhailcazi Aug 10 '13 at 12:25
  • $\begingroup$ Yes, I'm pretty sure that would be the way to do it - why do you feel it is so long? $\endgroup$ – mikhailcazi Aug 10 '13 at 12:28
  • $\begingroup$ Please right down the "method s you have thought of". meta.physics.stackexchange.com/questions/714/… $\endgroup$ – Abhimanyu Pallavi Sudhir Aug 10 '13 at 12:45
  • $\begingroup$ @mikhailcazi Well can one say that the potential energy of a particle inside an electric field that is caused by two charges to be the sum of the potential energy of each of the particle's electric field? Do I not have to to first combine the electric fields into a single field and then calculate the potentials? Combining the electric fields to a single vector function is what took me a lot of time. $\endgroup$ – paul23 Aug 10 '13 at 13:33
  • $\begingroup$ Yes, you can. Haven't you learnt about superposition? If you take the charge 2Q and the point in consideration, they will have a separate energy. It isn't a vector, so remember not to take direction. Only magnitude. You can also take the equivalent field at that point and do it. Why do you feel it is difficult? Use Pythagoras! $\endgroup$ – mikhailcazi Aug 10 '13 at 13:56

calculating the potential energy $$ W=\frac{q_1.q_2}{4\pi \epsilon_o r} $$ would give the workdone required to move the charge


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