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Where electrical potential is zeo

  1. Dec 7, 2009 #1
    1. The problem statement, all variables and given/known data

    A +3.00 mirco C charge is separated 4.00cm from a +4.50 mirco C charge.
    Is the electrical potential zero anywhere? If so, find the location.
    Explain how the problem changes if the 4.50 mirco C charge is negative.

    2. Relevant equations

    V = E/q V = w/q v = kq/r

    3. The attempt at a solution

    My teacher just taught us this today and gave us a horrible explaination of what electrical potential is so my classmates and I have no idea what to do.
  2. jcsd
  3. Dec 7, 2009 #2
    Well forgetting the math and explanation given in class, imagine that the + particles give off a wind. The force of the wind depends on the size of the particle (charge) and the distance away.

    Also say, negative charges create a suction force, in other words create a wind directed at them instead of away.

    Which of the above situations +/+ or +/- is more likely to have a location where no wind is felt? Imagine that winds of the same strength from opposite directions cancel each other out completely.
  4. Dec 7, 2009 #3
    Wouldn't your "wind" correspond to electric field rather than electric potential? Assuming that wind will flow away from areas of high pressure/concentration, I would think that potential would correspond to pressure rather than wind flow.
    Likewise, a point with no electric field does not necessary have zero potential (just zero gradient or zero change near that point, i.e. it is at a local maximum or minimum).

    Also, the original question does not define a reference point. Typically in systems described by point charges, the reference point is V=0 at an infinite distance away.
    Assuming that is the case, your given equation shows that V = k q /r.
    This is the electric potential of a single point charge with nothing else in the universe around it.
    What does your equation mean physically?
    It means that if you dropped a test charge of magnitude q_1 in the vicinity of your first charge, it would have q_1*V potential energy shared between it and your original charge. For the sake of argument, let's say that your original charge is positive, and so is your test charge. You will have positive electric potential V in the vicinity of your original charge, and your test charge q_1 will have positive potential energy q_1*V.
    The test charge will experience an electric force given by Coulomb's law and will accelerate away from your original charge. If the reference point is V=0 at an infinite distance away, then as the test charge gets further and further away, it's potential energy will get smaller and smaller. At an (effectively) infinite distance away, all of this potential energy will be converted to kinetic energy and your charge will have a kinetic energy equal to q_1*V.
    This is what the potential of a point charge means. It represents how much energy per unit of test charge a given system can and will impart to a test charge if the charge is moved by the system from it's current position to the zero reference point. If you would like a more practical picture, then instead of waiting until the particle gets an (effectively) infinite distance away, stop and measure it's kinetic energy at some point earlier in it's trajectory. The kinetic energy will be equal to the negative change in the electric potential between the two points multiplied by the value of the test charge.

    Intuitively, how would you expect the potential to change if you added another point charge (potential source) to your system?
    Hint: Superposition
    Last edited: Dec 7, 2009
  5. Dec 7, 2009 #4
    Mmmiguel, you are absolutely correct. This is an argument based on fields. I'm hoping it might actually make matters easier to understand. If electric fields can be added and there is a point where the net is zero, the notion of a test charge is irrelevant.

    The first post was just intended to get intuition working w/o math which is the enemy of understanding for far too many these days IMHO.
  6. Dec 7, 2009 #5
    I understand, but if the field is zero, that says nothing about the potential being zero.
    For example, if you have two positive charges of 1 Coulomb at x=1 and x=-1,
    then the electric field will be zero at the halfway point in between them (the winds will cancel). However the potential will not be zero (assuming the conventional reference point). The potential contributions from each charge will be equal and positive.
    You cannot in general deduce the zeros of potential from the zeros of electric field, just like you cannot find the zeros of a polynomial by knowing the zeros of its derivative.
  7. Dec 7, 2009 #6
    Pm in your box.
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