a conducting sphere of radius R carrying charge Q lies inside inside an uncharged conducting shell of radius 2R. if they are joined by a wire then amount of heat produced will be? Please explain in detail and step wise about the heat

To determine the amount of heat produced when a conducting sphere of radius R carrying charge Q is connected to an uncharged conducting shell of radius 2R by a wire, we need to analyze the situation step by step. This involves understanding the distribution of charge, the potential difference, and the energy transformation into heat.

Understanding the System

We have two conductors: a solid conducting sphere and a surrounding conducting shell. Initially, the sphere has a charge Q, while the shell is uncharged. When they are connected by a wire, charge will redistribute between the two conductors until they reach the same electric potential.

Step 1: Initial Charge Distribution

• The conducting sphere has a charge Q.
• The conducting shell has no charge initially, so its total charge is 0.

Step 2: Final Charge Distribution

When the wire connects the two conductors, charge will flow until both the sphere and the shell reach the same electric potential. The potential \( V \) of a charged sphere is given by the formula:

\( V = \frac{kQ}{R} \)

where \( k \) is Coulomb's constant. For the shell, the potential at its inner surface (which is at radius 2R) due to the charge on the sphere is:

\( V_{shell} = \frac{kQ}{2R} \)

Step 3: Equating Potentials

At equilibrium, the potentials must be equal:

\( \frac{kQ}{R} = \frac{kQ_{shell}}{2R} \)

Here, \( Q_{shell} \) is the charge that will reside on the shell after connection. Rearranging gives:

\( Q_{shell} = 2Q \)

Thus, the total charge in the system is conserved:

\( Q + 0 = Q + Q_{shell} \)

From this, we can deduce that the charge on the sphere after redistribution will be:

\( Q_{final} = Q - Q_{shell} = Q - 2Q = -Q \)

Step 4: Energy Considerations

The energy stored in the electric field of the charged conductors can be calculated using the formula for electric potential energy:

\( U = \frac{1}{2} Q V \)

Initially, the energy in the sphere is:

\( U_{initial} = \frac{1}{2} Q \left(\frac{kQ}{R}\right) = \frac{kQ^2}{2R} \)

After redistribution, the energy in the sphere becomes:

\( U_{final} = \frac{1}{2} (-Q) \left(\frac{k(-Q)}{R}\right) = \frac{kQ^2}{2R} \)

For the shell, the energy is:

\( U_{shell} = \frac{1}{2} (2Q) \left(\frac{k(2Q)}{2R}\right) = \frac{2kQ^2}{2R} = \frac{kQ^2}{R} \)

Step 5: Heat Produced

The heat produced during the process is the difference in energy before and after connecting the conductors:

\( Q_{heat} = U_{initial} - (U_{final} + U_{shell}) \)

Substituting the values:

\( Q_{heat} = \frac{kQ^2}{2R} - \left(\frac{kQ^2}{2R} + \frac{kQ^2}{R}\right) \)

Calculating this gives:

\( Q_{heat} = \frac{kQ^2}{2R} - \frac{3kQ^2}{2R} = -\frac{2kQ^2}{2R} = -\frac{kQ^2}{R} \)

Since heat cannot be negative, we interpret this as the energy being released as heat during the redistribution of charge.

Final Thoughts

The amount of heat produced when the conducting sphere and shell are connected is \( \frac{kQ^2}{R} \). This process illustrates how charge redistribution in conductors can lead to energy transformations, specifically into thermal energy, which is a fundamental concept in electrostatics and thermodynamics.