A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell.Let the delta V b/w the surface of the solid sphere and that of the now given a charge of -3Q the new potential difference b/w the same two surfaces is

a) -2V b) V c) 2V d) 4V

To tackle this problem, we need to analyze the situation involving the solid conducting sphere and the surrounding hollow conducting shell. The key concepts here are electric potential, charge distribution, and the behavior of conductors in electrostatic equilibrium.

Understanding the Initial Setup

Initially, we have a solid conducting sphere with a charge \( Q \). When a conductor is charged, the charge redistributes itself uniformly over its surface. The potential \( V \) at the surface of a charged sphere is given by the formula:

• V = k * Q / r

where \( k \) is Coulomb's constant and \( r \) is the radius of the sphere. Since the hollow shell is uncharged, it does not affect the potential of the inner sphere directly, but it will play a role when we introduce a new charge.

Introducing the New Charge

Now, when we place a charge of \( -3Q \) on the hollow conducting shell, we need to consider how this affects the potential difference between the surfaces of the solid sphere and the shell. The hollow shell will respond to the charge by inducing a charge of \( +3Q \) on its inner surface (to maintain electrostatic equilibrium), while the outer surface of the shell will have a charge of \( -3Q \).

Calculating the New Potential Difference

To find the new potential difference \( \Delta V \) between the surface of the solid sphere and the inner surface of the hollow shell, we can use the following steps:

• The potential at the surface of the solid sphere (with charge \( Q \)) is:
• V_sphere = k * Q / r
• The potential at the inner surface of the hollow shell (which now has an induced charge of \( +3Q \)) is:
• V_shell = k * 3Q / R
• Here, \( R \) is the radius of the hollow shell.

The potential difference \( \Delta V \) is then calculated as:

• ΔV = V_shell - V_sphere
• ΔV = (k * 3Q / R) - (k * Q / r)

Substituting Values

Assuming \( R \) is greater than \( r \) (which it is, as the shell surrounds the sphere), we can simplify this expression. However, since we are interested in the potential difference and the problem gives us specific options, we can analyze the signs:

• Initially, the potential difference was \( V \) (positive) when the sphere had charge \( Q \).
• After introducing the charge \( -3Q \), the potential at the shell becomes significantly more negative compared to the sphere, leading to a new potential difference.

Final Evaluation of Options

Given the options:

• a) -2V
• b) V
• c) 2V
• d) 4V

Since the potential at the shell has increased due to the induced charge, and considering the negative charge on the shell, the potential difference will be negative. The most reasonable conclusion is that the potential difference will be \( -2V \), as the shell's potential becomes more negative relative to the sphere's potential.

Thus, the answer is -2V.