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Homework Help: Work done by moving a point-charge

  1. Nov 19, 2016 #1
    Hello everyone!

    1. The problem statement, all variables and given/known data

    Calculate the work needed to move a point-charge with the charge q from infinity to the center of the chargedistribution given by:
    [itex]\rho(\vec{r})=\rho_0 e^{-\alpha r}[/itex]

    2. Relevant equations
    3. The attempt at a solution
    I did the integral by using integration by parts:
    [itex]\frac{1}{4\pi\epsilon_0}\int_{V}\frac{\rho(\vec{r})}{r}dV=\frac{1}{4\pi\epsilon_0}\int_0^{\infty}\int_0^{2\pi}\int_0^{\pi}\frac{\rho_0 e^{-\alpha r}}{r}r^2=\frac{\rho_0}{\epsilon_0 \alpha^2}[/itex]
    The work should then be, for moving the charge from infinity to 0, [itex]U=\frac{\rho_0 q}{\epsilon_0 \alpha^2} [/itex]
    Is this correct? Please let me know if I should elaborate some steps.
  2. jcsd
  3. Nov 19, 2016 #2

    Simon Bridge

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    You cannot use V for volume and for voltage in the same equation.
    In the equation for potential at position ##\vec r##, V is not going to be the entire volume enclosing the charge.
    The potential energy wanted is the work needed to move the charge from infinity to the origin.
    How would you figure the potential energy to put the same charge a distance R from the origin?
  4. Nov 19, 2016 #3
    I think I would do a path integral over [itex]\vec{E}[/itex].
    [itex]-\int_0^Rq\vec{E}\cdot d\vec{S}[/itex].
    To do that I'm trying to calculate the field, I got stuck at that though.
    [itex]\vec{E}(\vec{r})=\frac{1}{4\pi\epsilon_0}\int_{volume}d\vec{r}'\rho_0 e^{-\alpha r'}\frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^3}[/itex]
    How do I integrate the fraction in the integrand? I have to integrate over the whole space in spherical coordinates?
  5. Nov 19, 2016 #4

    Simon Bridge

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    ... so first off I notice that you would not use the potential equation like you did before. Why not? What changed in your reasoning?

    How about this: how much work is done moving the charge from radius r+dr an infinitesimally short distance to radius r?

    But yes - by definition, the change in potential energy between point P and point Q is given by the path integral ##\Delta U = \int_P^Q \vec F \cdot d\vec s##
    For a conservative force, this equation becomes ##\Delta U = U(Q)-U(P)## - you've seen these equations before right?
    ... also by definition, the potential energy at a point is this integral, but P is at infinity.

    Notes: for a spherically symmetric charge distribution:
    The charge below radius r is ##Q(r)=4\pi\int_0^r \rho(r')r'^2\;dr'## (check);
    The potential at radius r is ##\phi(r)=Q(r)/r## where ##U=q\phi## ... equipotential surfaces are spherical shells.
  6. Nov 20, 2016 #5
    I thought if the potential equation wasn't right in this case, I'd calculate the work in the same way as for a gravitational field.

    [itex]dW=q\vec{E}(\vec{r})dr[/itex], could I then integrate r from 0 to [itex]\infty[/itex] and use the fact that [itex]\phi(r)=-\nabla \vec{E}(r)[/itex] in spherical coordinates?

    This field is conservative because it is time independent? I think we derived these equations in classical mechanics for the gravitational field.

    I tried evaluating the charge below r, which gave the potential at r: [itex]\phi(r)=\frac{4\pi\rho_0(e^{-\alpha r}(-\alpha^2 r^2 - 2\alpha r -2 ) +2)}{\alpha^3 r}[/itex]
    If I then try to get the potential difference between 0 and [itex]\infty[/itex] I get 0: [itex]lim_{P->\infty}U(P)=0[/itex] and [itex]lim_{Q->0}U(Q)=0[/itex] then [itex]\Delta U = U(Q)-U(P)=U(0)-U(\infty)=0[/itex]
    I think I'm missing some important point.
  7. Nov 20, 2016 #6

    Simon Bridge

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    So basically, in response to questions, you have come up with at least three different approaches, and you are as confused as ever.

    One of your sticking points is not understanding the relations you use - the other may be a confusion about how you apparently have to integrate over the radius twice (once to get a Q(r) and again to get U).

    I think you've shown you are OK with the actual maths part - it's just figuring what it all means. So here's the start of a walkthrough:

    The potential energy at a point is the work needed to get the charge there from infinity ...
    The electric field is conservative##^1##, so the path does not matter.... so pick a path that makes the maths simple:
    Since the force is always radially outwards, come in directly along the radius.
    ##\vec F(\vec r)\cdot d\vec r = -F(r)\; dr## (force points opposite to the path taken##^2##.)
    So $$U = -\int_\infty^0 f(r)\;dr = \int_0^\infty F(r)\;dr$$ ... follow me so far##^3##?

    Since ##d\vec F=q\; d\vec E## ... $$F(r)=\frac{q}{4\pi\epsilon_0}\frac{Q(r)}{r^2}$$... where Q(r) is the total charge inside a ball of radius r.

    That would be: $$Q(r)=4\pi\int_0^r u^2\rho(u)\; du$$ ... "u" plays the role of a dummy variable radius coordinate##^4##.

    I have a feeling that should help you.

    [1] a conservative field is one where the total work done moving around any closed path is zero.
    [2] I may have this wrong ... but it's easy to troubleshoot.
    [3] dU = F(r)dr ... the work done to get q from r+dr to r is the same as that done by the force getting q from r to r+dr.
    [4] I'd normally use r' in this role but I used a u-substitution u=r' because I want to make the distinction clear - in the equation for Q, "r" is a specific radius, and u is a variable on the radial axis. In the equation for U, you are summing over a range of the specific values of radius the charge passes through to get to the origin.
    The ##4\pi## out the front is because the charge between r and r+dr is ##dQ=4\pi r^2\rho(r)\; dr##
  8. Nov 22, 2016 #7
    Thanks for the detailed answer.
    I tried following your answer, I think I understand it now.
    My calculation looks like this:

    First I calculated the total charge below the pointcharge q at R.
    [itex]Q(R)=4\pi \int_0^R u^2\rho(u)du=4\pi\rho_0\int_0^Ru^2e^{-\alpha u}du[/itex]
    I did the integral in maple and got this:
    [itex]Q(R)=\frac{4\pi\rho_0}{\alpha^3}[e^{-\alpha R}(-R^2\alpha-2R\alpha-2)+2][/itex]

    The force from the charge below acting on the pointcharge q is then given by:
    [itex]F(r)=\frac{q}{4\pi\epsilon_0}\frac{Q(r)}{r^2}=\frac{q\rho_0}{\alpha^3 \epsilon_0 }\frac{1}{r^2}[e^{-\alpha r}(-r^2\alpha-2r\alpha-2)+2][/itex]

    Moving it dr requires the work: [itex]dU=F(r)dr[/itex]
    Integrating over the path from [itex]\infty[/itex] to 0 then gives the total work required:
    [itex]U=\frac{\rho_0 q}{\alpha^3 \epsilon_0}\int_0^\infty (-e^{-\alpha r} \alpha^2 - 2\alpha e^{-\alpha r} r^{-1} - 2e^{-\alpha r}r^{-2}+2r^{-2})=\frac{\rho_0 q}{\alpha^3 \epsilon_0} \alpha=\frac{\rho_0 q}{\alpha^2 \epsilon_0}[/itex] (I did this integral in maple aswell)

    Why am I still getting the result from post #1? Is my reasoning equivalent to yours?
  9. Nov 22, 2016 #8


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    Why not? This is potential, not field.
  10. Nov 26, 2016 #9
    My tutor showed us the correct result in case anyone wanted to know: [itex]W=\frac{4\pi \rho_o q}{\epsilon_0 \alpha^2}[/itex] (I forgot the [itex]4 \pi[/itex] from the charge [itex]Q(r)[/itex] below q)
    Thanks for making the physics behind the calculation more clear.
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