Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Electric potential and EPE

  1. Aug 5, 2011 #1
    1. The problem statement, all variables and given/known data

    For this exercise, imagine a large positive charge +Q fixed at some location in otherwise empty space, far from all other charges. A positive test charge of smaller magnitude +q is launched directly towards the fixed charge. Of course, as the test charge gets closer, the repulsive force exerted on it by the fixed charge slows it down. Your job is to explain why the test charge slows down, but in terms of electric potential and EPE, rather than in terms of fields or forces.

    In particular:

    •explain how the electrical potential encountered by the test charge changes as it gets closer to the fixed charge, and why

    •explain how the EPE of the test charge changes as it gets closer to the fixed charge, and why

    On that basis, explain why the test charge +q slows down as it approaches the fixed charge +Q.
    Your response shouldn't mention either fields or forces.

    2. Relevant equations


    3. The attempt at a solution

    Electric potential energy is a type of potential energy. Potential energy is defined as energy that is stored within an object. Since the 2 particles form a closed system the total energy must remain the same in the beginning and the end so : Initial kinetic energy + initial potential energy = final kinetic energy + final potential energy. So there for the total initial energy = total final energy. As the particle moves in the direction towards the other particle its potential energy should decrease. We know this from phy 117. For an example a ball at the top of a hill will have more potential energy than once the ball gets to the bottom of the hill.

    Electric potential is the electric potential energy / charge. The electric potential energy is going to decrease as the particle gets closer, the charge would remain the same. Son since the top half of the fraction is going to decrease while the bottom stays the same this means that the Electric potential will also decrease as the particle gets closer.

    So after I said that this is what my instructor wrote......
    Rather than explaining in terms of equations and math, the explanation should be in terms of potential, charge, distance, and energy.

    Think about the fixed charge +Q first. It creates electric potential in the space surrounding it. That is, at every point in space, there is a certain electric potential due to the charge +Q - 100 V here, 50 V there, 5000 V there. Which of these locations is closest to the charge? Which is farthest from it?

    Thinking about this will tell you what happens to the electric potential encountered by the test charge q as it gets closer to the fixed charge Q: does it rise, or does it fall?

    Once you know how the potential changes as the test charge approaches, you can determine how the EPE of the test charge changes: does it increase, or does it decrease?

    On that basis, using the energy conservation principle, you can explain why the test charge slows down.

    .... so after that was said, I am lost. Can someone please help me
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Aug 5, 2011 #2
    You have the right idea in using energy balance, but the bolded part is backwards. Actually the potential energy increases as the particles approach. Gravity doesn't work this way, because it's always attractive. But because two like charges repel, their potential energy is always positive and increases as they get closer. Knowing this, you can immediately see why the test charge slows down: it gains potential energy as it moves closer to the large positive charge, so it must loose kinetic energy. So that's the answer. Now, about the teacher's comments:

    What they are trying to do here is introduce the concept of potential, or voltage, as distinct from potential energy. So really, when doing this problem they want you to think about the potential field created by the source charge Q first and then use the potential field to find the potential energy of the test charge q.

    A quick-and-dirty explanation of potential: You probably learned that the electric field is the force per charge. If a test charge q is placed at a location where the electric field is E, the force on the test charge is F = qE. So we can measure the electric field at a point by placing a test charge there. But the electric field is a "real" thing in itself; even where there are no test charges, there is still a field. In the exact same way, potential or voltage is potential energy per charge, so we have PE = qV for the test charge. But even if the test charge q is not present, there is still a potential field everywhere around Q. For a single source charge Q, this potential field is,

    V = kQ/r

    (NOT r2!) Where k is the same constant that appears in Coulomb's Law. Because Q is positive, V is positive also. And as r decreases, V increases. So we can picture Q as being at the peak of a "mountain" where height stands for potential. All this exists before we bring in q. But now if we bring it in, its potential energy is,

    PE = qV = kQq/r

    And because q is positive, PE is also positive. And again, it increases as r decreases, so q must slow down.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook