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Question:

If there are two conducting spherical shells and the inner shell is grounded, what will be >the charge density in the inner shell if there is a charge Q placed on the outer shell?

Yes, this is a HW problem, but I am not asking you guys to solve it for me... just show me the way :D

If there is a charge on the outer shell with radius $a$, the charge density will be $\frac{Q}{4\pi a^2}$.

That should induce, a charge density on the inner side of outer shell -- $\frac{Q}{4\pi (a-x)^2}$ where $x$ is thickness of the outer shell.

Now, if the inner shell was never grounded, $\frac{Q}{4\pi b^2}$ ($b$ = radius of the inner shell) charge density would have been induced on the inner shell, now that it is grounded there would be no charge on it, right?

There is no need for a charge to be induced for the electric field to be 0 inside the inner shell.

This seems to be my conclusion, but I fear it is too simple for the question. There might be something I am missing.

Edit: the hint of the problem says when a sphere is grounded, potential is infinity.

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    $\begingroup$ An infinite potential sounds wrong intuitively. A grounded terminal is by definition at zero potential. $\endgroup$
    – Noldorin
    Jan 24, 2011 at 1:36
  • $\begingroup$ My understanding is that the grounding of the inner sphere allows it to draw any amount of charge it needs to balance the charge on the outer shell. $\endgroup$
    – stl_piznaul
    Jan 24, 2011 at 2:07
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    $\begingroup$ My bad, when grounded, the inner sphere has same potential as potential at infinity. Which is zero potential. $\endgroup$
    – Ender
    Jan 24, 2011 at 3:38
  • $\begingroup$ I think you want to minimize the total energy in the electric field. This is proportional to the square of the E field strength integrated over all space. By spherical symmetry the electric field at radius r will be proportional to the total enclosed charge inside this radius. So the E field outside of the outer shell is proportional to (Qin+Q)/r**2, integrate this from Router to infinity. A similar (definite) integral gives you the concentration of field energy in the region in between the two shells. Adjust Qinner, until the sum of these two integrals is minimized. $\endgroup$
    – Omega Centauri
    Jan 24, 2011 at 3:39

3 Answers 3

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Now, if the inner shell was never grounded, $\frac{Q}{4\pi b^2}$ ($b$ = radius of the inner shell) charge density would have been induced on the inner shell

If the inner shell were not grounded, it wouldn't be connected to anything. And if it wasn't connected to anything, how would it be able to have any charge density other than zero? Where would the extra charge come from?

now that it is grounded there would be no charge on it, right?

No, as pointed out in the comments, the potential is zero, but that doesn't mean the charge is zero.

Can you figure out the potential of the outer sphere? Once you know the potentials of both spheres, what else can you calculate using that information?

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As the problem is spherically symmetric, you know that the potential must go as $1/r$ in free space. Let $Q$ be the charge at the outer sphere and $Q'$ the charge at the inner sphere; then the potential must be expressed as $\frac{A\,(Q+Q')}{r}$ for the space outside the outer shell ($r>a$) and as $\frac{A\,Q'}{r}+K$ at the space between the inner shell and the outer shell ($a>r>b$).

You can get the value of $Q'$ and $K$ by matching both expressions for the potential at $r=a$ (the potential doesn't jump at surface charge distributions) and by forcing a zero potential at $r=b$ (you have two equations with two unknowns).

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  • $\begingroup$ If You had read all the thread, You had seen that Q = Q' ! And charge density on those speres is simply Q/surface. $\endgroup$
    – Georg
    Jan 29, 2011 at 11:11
  • $\begingroup$ I have read the whole thread. :-) $Q$ cannot be equal to $Q'$, because in this case the potential at the inner sphere cannot be zero! (The electric field doesn't change its direction between the spheres.) If you mean $Q = -Q'$, then the potential is zero at the outer sphere, but the inner sphere will be charged and will have a nonzero potential. $\endgroup$
    – mmc
    Jan 29, 2011 at 15:55
  • $\begingroup$ Before You make wrong assumptions about charges, first look for potentials in the problem setup. I wrote something on the typical errors You (made. $\endgroup$
    – Georg
    Jan 29, 2011 at 18:50
  • $\begingroup$ Hmm. Do you accept Gauss's Law (en.wikipedia.org/wiki/Gauss's_law)? Do you see why $Q = -Q'$ implies that the potential of the outer sphere is zero? $\endgroup$
    – mmc
    Jan 30, 2011 at 11:03
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Firstly assume that you know the charge on the inner shell $q_{in}$. Then calculate the electric field between the shells $E_{in}$ and the field outside shells $E_{out}$ by using Gauss law. Know by knowing that the potential at the $\infty$ is 0 and that when conductor is grounded has the same potential, you can write potential difference $U$ in two ways:

$$U = \int_R^\infty E_{out}\,dr$$ $$U = -\int_r^RE_{in} \, dr$$

Where $r$ is inner but R is outer shell radius.

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