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Change in electrostatic energy on two spheres

  1. May 16, 2017 #1
    1. The problem statement, all variables and given/known data
    We have a spehere of radius ##r_1## and on of ##r_2## far away from each other. The first sphere has a charge ##Q##. What is the change in electro static energy when they are connected together?

    2. Relevant equations
    Potential of a charged sphere: ## V = \frac{Q}{4\pi\epsilon_0 r}##
    Elctrostatic energy: ##W = q \Delta V ##

    3. The attempt at a solution
    Initially, the potential of the system is: ## V = \frac{Q}{4\pi\epsilon_0 r_1}##
    When the spheres are connected together, they become equipotential and will have charges:
    ## Q_1 = Q \frac{r_1}{r_1+r_2} ## and ## Q_2 = Q \frac{r_2}{r_1+r_2} ##. Hence the change in voltage is: ## \Delta V = \frac{Q}{4\pi\epsilon_0 r_1} - \frac{Q_1}{4\pi\epsilon_0 r_1} = \frac{Q}{4\pi\epsilon_0} (\frac{r_2}{r_1(r_1+r_2)}) ##. So the change in potential energy is: ## W=\frac{Q^2}{4\pi\epsilon_0} (\frac{r_2}{r_1(r_1+r_2)})##.

    However, this is wrong. The answer is supposed to be: ## W=\frac{Q^2}{8\pi\epsilon_0} (\frac{r_2}{r_1(r_1+r_2)})##. Where do I loose the factor of 1/2?
     
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  3. May 16, 2017 #2

    TSny

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    I don't follow how you can write the change in potential energy as ##W## as ##Q \Delta V##. This would imply that all of the charge ##Q## undergoes the same change in potential ##\Delta V##. Also, your expression for ##\Delta V## does not appear to account for the change in potential of the second sphere.

    As you noted, the potential ##V## at the surface of an isolated sphere of charge ##Q## and radius ##r## is ##V = \frac{Q}{4 \pi \epsilon_0 r}## (taking ##V = 0## at inifinity). But the potential energy of the sphere is not ##W = QV##.

    The potential energy of an isolated sphere can be derived using integration of ##dW = V \, dq##, where ##q## varies between ##0## and ##Q## and ##V## is the potential when the sphere has a charge ##q##.
     
    Last edited: May 16, 2017
  4. May 23, 2017 #3
    How do I account for the potential of the second sphere in my equation of the potential change?
    Also, I see you point with the integral, thank you very much. But if I am to perform it, what potential should I calculate with? The potential of the first or second sphere?
     
  5. May 23, 2017 #4

    TSny

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    You have found the final charge of each sphere correctly.

    The spheres are so far apart that you may consider the total energy of the system be the sum of the energy of each sphere alone. So, you just need to know how to calculate the energy of an isolated sphere in terms of its charge and radius. You can do this by integration as mentioned. Or, if you are familiar with the capacitance of an isolated spherical conductor, then you can use the formula for the energy of a capacitor.
     
  6. May 26, 2017 #5
    Oh I see, so we don't have to integrate at all! Am I rigth in that the potential energy of a charged sphere is simply its potential multiplied by its charge? I tried to apply:
    ## \Delta V = \frac{Q^2}{4\pi\epsilon_0 r_1} - \frac{Q_1^2}{4\pi\epsilon_0 r_1}- \frac{Q_2^2}{4\pi\epsilon_0 r_2} ## but ended up with the exact same answer. :(
     
  7. May 26, 2017 #6

    haruspex

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    No. Imagine letting a small charge dq from the sphere go to infinity. The work done is Vdq. But now the charge on the sphere is a bit less.
    As all the charge leaks away, what is the average potential?
     
  8. May 27, 2017 #7
    Are the spheres conductors ?

    I also have a similar problem, so I am asking.
     
  9. May 27, 2017 #8

    haruspex

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    If they are not then, trivially, nothing changes, so it is safe to assume they are.
     
  10. May 28, 2017 #9
    If they are insulators then charge won't flow from shere ##r_1## to sphere ##r_2##, then the surface won't become equipotential right ?
     
  11. May 28, 2017 #10

    haruspex

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    Right.
     
  12. May 28, 2017 #11
    Then why they are trivally same ?
     
  13. May 28, 2017 #12

    haruspex

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    I didn't say they were. I wrote that when you connect them nothing changes.
     
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