To tackle this problem, we need to understand how electric potential and charge distribution work in conductors, particularly when dealing with spherical geometries. Let's break it down step by step.
Initial Setup
We start with a solid conductor that has a charge \( Q \) and is surrounded by an uncharged conducting hollow spherical shell. The potential difference \( V \) between the surface of the solid sphere and the outer surface of the hollow shell can be expressed in terms of the electric field and the geometry of the spheres.
Understanding Electric Potential
The electric potential \( V \) at a distance \( r \) from a point charge \( Q \) is given by the formula:
where \( k \) is Coulomb's constant. For a solid sphere, the potential at its surface (radius \( R \)) is:
- \( V_{\text{solid}} = \frac{kQ}{R} \)
For the hollow shell, since it is initially uncharged, the potential at its inner surface (radius \( R_{\text{shell}} \)) is equal to that of the solid sphere:
- \( V_{\text{shell}} = \frac{kQ}{R_{\text{shell}}} \)
Introducing the Charge on the Shell
Now, when the hollow shell is given a charge of \( -3Q \), we need to analyze how this affects the potential difference. The charge on the shell will redistribute itself due to the influence of the solid sphere's charge. The inner surface of the shell will acquire a charge of \( -Q \) to neutralize the field inside the conductor, while the outer surface will have a charge of \( -3Q + Q = -2Q \).
Calculating the New Potential Difference
Now, we can calculate the new potentials:
- The potential at the inner surface of the shell (due to the solid sphere's charge \( Q \)) remains:
- \( V_{\text{inner}} = \frac{kQ}{R_{\text{shell}}} \)
- The potential at the outer surface of the shell (due to the charge \( -2Q \)) is given by:
- \( V_{\text{outer}} = \frac{k(-2Q)}{R_{\text{outer}}} \)
Final Expression for the Potential Difference
The new potential difference \( V' \) between the surface of the solid sphere and the outer surface of the hollow shell can now be expressed as:
- \( V' = V_{\text{inner}} - V_{\text{outer}} \)
- \( V' = \frac{kQ}{R_{\text{shell}}} - \frac{k(-2Q)}{R_{\text{outer}}} \)
Substituting Values
Assuming \( R_{\text{shell}} \) and \( R_{\text{outer}} \) are the respective radii of the inner and outer surfaces of the shell, we can simplify this expression. The potential difference will now depend on the specific values of \( R_{\text{shell}} \) and \( R_{\text{outer}} \), but generally, we can conclude that:
- The potential difference will increase due to the additional negative charge on the shell.
In summary, the new potential difference between the surfaces of the solid sphere and the outer shell, after the shell is given a charge of \( -3Q \), will be greater than the initial potential difference \( V \). The exact value will depend on the radii of the spheres involved.
