Spherical charged shells with grounding

Question

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?

Answer 1

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 $V = \dfrac {1}{4 \pi \epsilon_o} \dfrac {Q_i}{a}$

Answer 2

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.