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Work needed to charge a sphere

  1. Jul 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Calculate the work that must be done to charge a spherical shell of radius [tex]R[/tex] to a total charge [tex]Q[/tex].
    No diagram was provided.
    (Another user posted the same question at https://www.physicsforums.com/showthread.php?t=109489, but there was no solution and the thread is ancient.)

    2. Relevant equations
    I'm not sure what equations are needed, so I'm just guessing here.
    [tex]U = k_e\frac{q_1q_2}{r_{12}}[/tex]

    [tex]W = -\Delta U[/tex]

    [tex]V = k_e\frac{q}{r}[/tex] for a sphere

    3. The attempt at a solution
    I know the answer is [tex]\frac{k_eQ^2}{2R}[/tex] from the back of the book, but I don't know how to get it. I've reread the chapter and all the examples, but I can't find anything talking about energy with just one object. Everything deals with pairs or movement in a field. I thought about making up an identical sphere with charge [tex]-Q[/tex] next to the existing one and finding the change in potential energy. This gets me the right answer, but I doubt that this is a correct process.

    Any help would be appreciated.
     
  2. jcsd
  3. Jul 11, 2008 #2
    You correctly said [tex]W = - \Delta U[/tex]. Also, keep in mind [tex]\Delta U = -q \int^{A}_{B} E dr[/tex]. Now you can find out what E equals and integrate
     
  4. Jul 11, 2008 #3

    alphysicist

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    Homework Helper

    Hi thirteen,

    There are several ways to approach this problem. Does your textbook have an expression for the energy density of an electric field? If so, you can integrate that expression over all space to find the energy.

    Another approach is to think of the stored energy in a set of discrete point charges, and follow that idea to write out the correct expression for a continuous charge distribution.
     
  5. Jul 11, 2008 #4
    Thanks, but I thought that equation was for the movement of a test charge q in a different field. I don't see how it can be used.

    Thanks, I'll look into the latter, since I don't have anything on energy density.
     
  6. Jul 12, 2008 #5
    It can be used because that is essentially what you are doing: moving a charge from infinity to R against an electric field supplied by the charges you already moved. Another hint:
    [tex] E = \frac{kq}{r^2}[/tex] so [tex]\Delta U = -q \int^{A}_{B} \frac{kq}{r^2} dr[/tex]. Integrate that from infinity to R.
     
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