Why is there is no electric field outside a thin, grounded, metallic spherical shell enclosing a point charge at center? 
We have a thin, grounded, metallic spherical shell. While solving for its total energy using the formula
$$U_E = \frac{\epsilon_0}{2}\int|\mathbf{E}|^2\,\mathrm d^3r$$
the book I am following says that the electric field outside the shell is zero. But I think Gauss's law doesn't allow that as any spherical Gaussian surface larger than the shell will enclose a net charge of $q$.
If the field is indeed zero outside the shell, then the charge at the center must somehow be cancelled. I don't understand how can the grounding of the shell cancel the charge. Can anyone explain why this happens?
 A: As the shell is grounded, charge will move either to or from the ground in order to bring the potential of the shell to zero. Let the shell have net charge $x$ on it. Then the potential of shell is given by-
$$V_{total}=V_{shell}+V_q$$
$$V=\frac {kx}r+\frac {kq}r$$
But since $V=0,\; x=-q$. This means that $-q$ charge will flow from the ground to the shell. Hence if we draw a spherical Gaussian surface at any point outside the shell, the net charge contained inside will be  $q+(-q)$ which is zero. Hence net electric field will be zero at all points outside the shell.
A: I am suspicious of the answer given above by Sam.
Suppose the radius of the shell is $a$. Then, the expression of the contribution of the shell, $V_{shell}=\frac{kx}{r}$, to the total potential is only true for $r>a$, i.e., outside the shell. Now, the problem says that the shell is grounded, and that means that $V=0$ inside the shell. So we cannot simply set $V=\frac{kx}{r}+\frac{kq}{r}$ to zero.
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