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Consider two concentric spherical shells of radii $a$ and $2a$ respectively. Let the inner shell have potential $V_0$ and the outer shell be grounded. What is the potential $V(r)$ as a function of the distance to the center of the shells, and what are the charges on the shells?

This is how I would approach the problem. Let $Q_i$ and $Q_o$ be the charge on the inner and outer shell respectively. Put $$V(r) = -\int_{2a}^{r} \vec{E}(r') \cdot d\hat{r}'.$$

Since $E(r) = 0$ for $r < a$,

$E(r) = Q_i/(4\pi\epsilon_0r^2)$ for $a < r <2a$

$E(r) = (Q_i+Q_o)/(4\pi\epsilon_0r^2)$, this gives us

$V(r) = Q_i/(8\pi\epsilon_0a)$ for $r < a$, $V(r) = Q_i(2a/r-1)/(8\pi\epsilon_0a)$ for $a < r < 2a$

$V(r) = (Q_i+Q_o)(2a/r-1)/(8\pi\epsilon_0a)$ for $2a < r.$

Now requiring that $V(a)=V_0$, we get $Q_i = 8\pi\epsilon_0aV_0$. But what about $Q_o?$ From the form of the potential we immediately get $V(2a) = 0$, regardless of the value of $Q_o$.

In the answer to the problem it states that $Q_o = -Q_i$. It is also directly states that $V(r) = 0$ for $r > 2a$, but frankly I don't see why that follows from the assumption that the outer shell is grounded.

Does it all rest on the implicit assumption that $V(\infty) = 0$, which makes my definition of $V(r)$ wrong? Certainly we are not required to make this assumption?

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With the outer shell grounded once you pot a charge of $+Q_i$ on the outer side of the inner shell then a charge of $-Q_i$ will be induced on the inner side of the outer shell.
Think of it as no electric field inside a conductor so every electric field line which starts on a charge on the outside surface of the inner sphere must finish on an opposite charge on the inside surface of the outer sphere.

So there is only an electric field between the inner and the outer shell and so this is the only region where the electric potential changes.

If earth is taken to be the zero of potential which is often the case then the outer shell must also be at zero potential if it is connected to earth.

If you think about Gauss's law and consider a spherical Gaussian surface centred at the centre of the spherical shells then if $a \le r \le 2a$ the charge enclosed by the surface is $+Q_i$.
Once you have $r>2a$ the enclosed charge is zero, $+Q_i- Q_i =0$ and so the electric field outside the outer sphere is zero.

$r>2a$ then $E=0$ and $V=0$

$a \le r \le 2a$ then $E = \dfrac {1}{4 \pi \epsilon_o} \dfrac {Q_i}{r^2}$ and $V = \dfrac {1}{4 \pi \epsilon_o} \dfrac {Q_i}{r}$

$r< a$ then $E=0$ and $V = \dfrac {1}{4 \pi \epsilon_o} \dfrac {Q_i}{a}$

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  • $\begingroup$ Thanks! But what if we change the wording of the problem a bit. Rather than saying the outer shell is grounded, we say its potential is zero. Also, we need not assume the shells are metallic sheets. Let's just say we have two charge distributions localized to two spherical shells? $\endgroup$ Nov 10 '16 at 23:07
  • $\begingroup$ If the outer shell is not grounded then there would be a charge of $-Q_i$ on the inside surface of the outer shell and a charge of $+Q_i$ on the outside surface of the outer shell. $\endgroup$
    – Farcher
    Nov 10 '16 at 23:20
  • $\begingroup$ So the outer shell would have not net charge? Then we would not have $V(2a) = 0$? $\endgroup$ Nov 11 '16 at 8:17
  • $\begingroup$ Relative to ground (the zero of potential) the outer sphere would have a non zero potential because work would have to be done in moving charge from the ground to the outer sphere. $\endgroup$
    – Farcher
    Nov 11 '16 at 9:03
  • $\begingroup$ Sorry for my late reply. I don't quite understand what you mean. Surely it would be possible to have a spherical shell with charge $Q_i$ surrounded by a nongrounded spherical shell with charge $-Q_i$. Then the potential on and outside of the outer shell would be zero. Correct? What is then the difference between saying the shell is grounded and saying its potential is zero? $\endgroup$ Nov 11 '16 at 18:41
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Your definition of the field for$r>2a$ is correct. But for $Q_i=-Q_0$ the field is zero. The charge induced on the inner surface of the grounded outer sphere must, by Gauss law, be $-Q_i=Q_0$ (there is no field in the metal). In addition, because of the outer sphere being at ground potential, there can be no field outside the sphere (no induced outer surface charges) because there is no potential difference to the environment.

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  • $\begingroup$ Thanks! But what if we change the wording of the problem a bit. Rather than saying the outer shell is grounded, we say its potential is zero. Also, we need not assume the shells are metallic sheets. Let's just say we have two charge distributions localized to two spherical shells? $\endgroup$ Nov 10 '16 at 23:07
  • $\begingroup$ @Etienne Bezout - If you assume the outer shell potential to be zero then the question is if you also assume, as usual, that the outer potential is zero for $r->infinity$. If this is the case, then there is no outer field as in the grounded situation. If you assume two (different) immobile charges localized evenly on concentric spheres, you have, in general, an outside field, and you can use Gauss law to determine it. $\endgroup$
    – freecharly
    Nov 10 '16 at 23:52
  • $\begingroup$ Alright, so then I think it can be concluded that my definition of the potential is wrong, because it is implicitly assumed that we have fixed $V(\infty) = 0$, and the values $V(a) = V_0$, $V(2a) = 0$ rest on this assumption. $\endgroup$ Nov 11 '16 at 8:14

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