# Why are the charges pushed to the earth?

Question:

A cross-section of two conducting concentric spheres is shown in the figure. The larger sphere of radius $R$ has a charge $Q$ on it. The smaller sphere of radius $r$ has no charge on it initially. Now, the smaller sphere is earthed. How much charge leaves the inner sphere?

Intuition:

There are initially no charges on the inner sphere. Also the electric field due to the larger sphere is zero at all points inside it because of the fact that it is a conductor. I don't see any forces that could push charges from the sphere to the earth.

Solution:

The potential at all points on the inner sphere initially is the same and is equal to $V=\frac{kQ}{R}$ (taking the potential at infinity to be zero). The potential of the earth is zero. As long as there is a difference in potential between the earth and the sphere there will also be an electric field that pushes positive charges to the region of lower potential. After the sphere is earthed if a total charge $P$ has flown out of the inner sphere when it is at equilibrium, $$\frac{kQ}{R}-\frac{kP}{r} =0 \implies P=\frac{rQ}{R}$$ $\rule{17cm}{0.4pt}$

The math tells me that there is an electric field that pushes charges from the sphere to the earth. What is creating the electric field?