# How much charge will flow to the earth?

Consider a thin spherical conducting shell of radius $$R$$, which carries a total charge $$Q$$ on its surface. Two point charges $$Q$$ and $$2Q$$ are at A and B respectively as shown in the figure ($$C$$ is the center of the shell).

If now the shell is earthed. How much total charge will flow into the earth ?

My attempt: Initially, the inner surface of the sphere will have a total charge $$-Q$$ distributed non-uniformly. Hence the outer surface will have a total charge $$2Q$$ (including the induced charges, whose sum amounts to zero) distributed non-uniformly.

After earthing, let the charge on the outer surface be $$Q_1$$. If $$\vec{E}=0$$ inside the shell, we get $$V_{centre}= V_{shell}=0$$. Putting the values with $$k=\frac{1}{4π\epsilon_0}$$, one gets $$\frac{kQ}{R/2}+\frac{k(2Q)}{2R}+\frac{k(-Q)}{R}+\frac{kQ_1}{R}=0.$$ (Since every point on the shell is equidistant from the centre , the charge distribution doesn't matter while calculating $$V_{centre}$$).

The above equation results in $$Q_1=-2Q$$, which further implies that $$4Q$$ charge flowed from the shell to the earth.

But the answer given is $$3Q$$. This is contradicting the fact that $$\vec{E}=0$$ inside the shell, which I can't understand since in my opinion existence of an electric field inside the shell will result in the movement of the charge inside, causing instability in the overall configuration. Please clear any misconceptions you find in this solution.

• The R in your equation has not been defined. Aug 3 '20 at 15:04

• Thanks a lot! I understood that the charges $2Q$ outside and the charge $Q_1$ on the outer surface contribute zero electric field inside the sphere, collectively. So a net electric field exists inside the sphere .Now the electric field lines inside, cannot penetrate the inner wall of the conductor , which seems to conclude that the potential due to the charges $Q$ and $-Q$ , at the inner wall must be equal to that on the outer surface , which further implies that their combined potential at the inner surface must be 0 , for $V_{shell}$ to be 0 ( which I can't understand ). Aug 3 '20 at 7:52
• Thanks , All this time I was more focused on the $"\Sigma\frac{kq}{r}"$ definition of the electric potential . Now one last thing I want to make sure is that ,wouldn't the electric field inside will result in the movement of the charge $Q$ ? If it does , wouldn't it affect the stability of the system? Aug 3 '20 at 9:11