Animated Solution for Physics - Electrostatics: A spherical metal shell $A$ of radius $R_A$ and a solid metal sphere $B$ of radius $R_B(<R_A)$ are kept far apart and each is given charge $+Q$. Now they are connected by a thin metal wire. Then

Imagine two metallic spheres, one is a hollow shell $A$ with a larger radius $R_A$, and the other is a solid sphere $B$ with a smaller radius $R_B$. Initially, they are far apart and both carry a charge of $+Q$.

When we connect them with a thin conducting wire, we are essentially allowing charges to flow freely between them. This flow of charge will continue until electrostatic equilibrium is reached.

Let's evaluate the first statement regarding the electric field inside shell $A$.

According to Gauss's Law, the net charge enclosed by any Gaussian surface entirely within the conducting material of the shell is zero. Since all the charge resides on the outer surface of the conductor, the electric field inside the hollow region of the shell is strictly zero.

When the two spheres are connected by the wire, they form a single equipotential system. This means charge will redistribute until the potential on the surface of sphere $A$ equals the potential on the surface of sphere $B$.

We know the formula for the potential of a charged sphere is $V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R}$. Substituting this into our equipotential condition gives:

Since we are given that $R_A>R_B$, for the equality to hold true, the charge on sphere $A$ must be greater than the charge on sphere $B$.

Now, let's explore the relationship between their surface charge densities, ${\sigma }_A$ and ${\sigma }_B$. The potential of a sphere can also be expressed in terms of its surface charge density. Since $Q=\sigma \cdot 4\pi R^2$, the potential becomes:

Using the equipotential condition $V_A=V_B$ again, we can write:

Rearranging this equation yields the ratio of their surface charge densities:

Finally, let's determine the electric field on the surface of each sphere. The electric field just outside the surface of a conductor is given by:

This shows that the electric field is directly proportional to the surface charge density ($E\propto \sigma$).

From our previous derivation, we know that ${\sigma }_A{R}_A={\sigma }_B{R}_B$. Since $R_A>R_B$, it mathematically follows that ${\sigma }_A<{\sigma }_B$.

Because the surface charge density of sphere $A$ is less than that of sphere $B$, the electric field on the surface of $A$ must also be less than the electric field on the surface of $B$.

In conclusion, all four statements are physically and mathematically sound!