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Point charges of charge Q placed at each corner of a cube of side r

  1. May 23, 2012 #1
    Imagine a cube with side lengths 'r', (like a dice) and we put point charges of charge Q at each corner of the cube.



    Issues with 2 subsections of this problem...

    First: What is the total electrostatic potential energy of the arrangement?

    Second: Add a charge -CQ to the center of the cube for what value of C will the total potential energy be negative?


    For the first part I've calculated the potential energy of the system to be

    [tex]
    \frac{Q^2}{4\pi\epsilon}[\frac{12}{r} + \frac{13}{r\sqrt{2}} + \frac{4}{r\sqrt{3}}]
    [/tex]

    Does that seem correct? It was a pretty lengthy algebraic adventure but I feel like it works out, its just a summation of the radii between charges (non-repeating of course).

    For the second part I'm pretty lost:

    I've calculated the total potential at the center to be

    [tex]
    \frac{4Q}{\sqrt{3} \pi \epsilon r}
    [/tex]

    Now, I don't know how to determine the max value of C to see when the whole potential would be thrown into the negative. Do I add the two expressions together and solve for C? I'm pretty lost. I even recalculated the potential from the first part but threw in the charge with the other eight Q's to no avail.

    [tex]
    \frac{Q^2}{4 \pi \epsilon}[\frac{12}{r} + \frac{13}{r \sqrt{2}} + \frac{4}{r \sqrt{3}} - \frac{14C}{r \sqrt{3}}]
    [/tex]

    The roots are giving me trouble to simplify things to a discrete integer. Any pointers would be appreciated.

    Thanks.
     
  2. jcsd
  3. May 23, 2012 #2

    gabbagabbahey

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    Homework Helper
    Gold Member

    A quick BoE calculation gives me almost the same thing, but with [itex]\frac{12}{r\sqrt{2}}[/itex] instead of [itex]\frac{13}{r\sqrt{2}}[/itex]. I think you may have double counted one of the radii. The simplest method for this calculation is probably to calculate the electrostatic potential at one corner due to the other 7 charges, then appeal to symmetry to conclude that the potential at the other corners will be identical, and then plug that into the general formula for the energy of a collection of point charges [itex]q_i[/itex] at locations [itex]\mathbf{r}_i[/itex]:

    [tex]W = \frac{1}{2}\sum_{i}q_iV(\mathbf{r}_i)[/tex]

    It looks like you've only taken account of the work done against the electrostatic force between the center charge and each of the other 8 charges when assembling the distribution. What about all the interactions between just the 8 corner charges?

    Again, I'd suggest using the same formula for [itex]W[/itex] given above. This time, the electrostatic potential for each corner will have an extra term due to the center charge, and your sum will have an extra term for the center charge multiplied by the potential at the center due to the 8 corner charges.
     
  4. May 23, 2012 #3
    I was pushing numbers around to follow up with a different subsection which asks to find the energy if we were given charge magnitudes and radii at each edge; working backwards I figured out that my radii counting was in fact off by a very very small amount. So you were completely right in that regard. Thank you. As for the last part (part II), I'm gonna have to mess around with it a bit more later on as I'm typing this whilst sitting in another class. Thanks again, I'll check back with any other issues.
     
  5. May 23, 2012 #4
    I'm pretty sure there's an error in my processing...

    I'm getting as the total potential

    [tex]
    \frac{Q^2}{4\pi \epsilon}[ \frac{12}{r} + \frac{12}{r \sqrt{2}} + \frac{4}{r \sqrt{3}} + \frac{8C}{r \sqrt{3}}]
    [/tex]

    To determine the minimum value of C for which the entire potential is negative it looks like I'm going to end up getting some continuous rather than discrete value. Is that right? I feel like it should be a discrete answer, so I think I did something wrong. Can you point out where I'm going wrong?

    Thanks again.
     
  6. May 24, 2012 #5

    gabbagabbahey

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    Homework Helper
    Gold Member

    For the last term, I get [tex]\frac{6C}{r \sqrt{3}}[/tex] instead of [tex]\frac{8C}{r \sqrt{3}}[/tex], so you may have double counted somewhere again.

    As for find what value(s) of C make the enrgy negative, you basically have a line in the W-C plane, so anywhere before the C-intercept, the work will be negative.

    As a side note, I really dislike the wording of the 2nd part of the question. Potential energy is really only defined up to a constant, so you can change what value of C makes it zero, simply by adding a constant to the energy (the dynamics that result will be identical!). I'd much prefer if the question was asking about the work needed to assemble the charge distribution from infinity, because that is a unique quantity, and probably what the questioner expects you to use as "the" potential energy....just something you may wish to point out to your professor (or not, depending on their temperment)
     
  7. May 25, 2012 #6
    Great! Thanks for the help. As for the wording...it's funny you mention that because when I first read the question I was confused for a little bit, "What is this asking?". Stranger still we're using the Griffiths book and he mentions what a horrible "misnomer" Potential is. Guess I'm getting a first hand account.

    Thanks again.
     
  8. Apr 7, 2013 #7
    ED

    please upload the complete derivation..thanks
     
  9. Oct 12, 2015 #8
    20151012_212227 (960x1280).jpg 20151012_212236 (960x1280).jpg
    [itex]U_e=\frac{q^2}{4\pi \epsilon_oa}\left ( 13+\frac{11}{\sqrt{2}} +\frac{4}{\sqrt{3}}\right )[/itex]
     
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