Two spherical conductors of radius R and 2R having potential 4V and 2V are connected with a wire. Find the loss in energy if they are connected by a wire.

To find the loss in energy when two spherical conductors are connected by a wire, we first need to understand how the potentials and charges on these conductors will change once they are connected. Let's break this down step by step.

Initial Conditions

We have two spherical conductors:

• The first conductor has a radius of R and an initial potential of 4V.
• The second conductor has a radius of 2R and an initial potential of 2V.

Calculating Initial Charges

The potential \( V \) of a spherical conductor is related to its charge \( Q \) and radius \( R \) by the formula:

\( V = \frac{Q}{4\pi \epsilon_0 R} \)

From this, we can express the charge on each conductor:

• For the first conductor (radius R, potential 4V):

\( Q_1 = 4V \cdot 4\pi \epsilon_0 R = 16\pi \epsilon_0 VR \)

• For the second conductor (radius 2R, potential 2V):

\( Q_2 = 2V \cdot 4\pi \epsilon_0 (2R) = 16\pi \epsilon_0 VR \)

Final Conditions After Connection

When the two conductors are connected by a wire, they will reach a common potential \( V_f \). The total charge \( Q_{total} \) will be conserved:

\( Q_{total} = Q_1 + Q_2 = 16\pi \epsilon_0 VR + 16\pi \epsilon_0 VR = 32\pi \epsilon_0 VR \)

The total capacitance \( C \) of the system can be calculated as follows:

• The capacitance of the first conductor is \( C_1 = 4\pi \epsilon_0 R \).
• The capacitance of the second conductor is \( C_2 = 4\pi \epsilon_0 (2R) = 8\pi \epsilon_0 R \).

The total capacitance when they are connected in parallel is:

\( C_{total} = C_1 + C_2 = 4\pi \epsilon_0 R + 8\pi \epsilon_0 R = 12\pi \epsilon_0 R \)

Finding the Final Potential

The final potential \( V_f \) can be calculated using the total charge and total capacitance:

\( V_f = \frac{Q_{total}}{C_{total}} = \frac{32\pi \epsilon_0 VR}{12\pi \epsilon_0 R} = \frac{32V}{12} = \frac{8V}{3} \)

Calculating Energy Before and After Connection

The energy stored in a capacitor is given by the formula:

\( U = \frac{1}{2} C V^2 \)

Calculating the initial energy for each conductor:

• For the first conductor:

\( U_1 = \frac{1}{2} (4\pi \epsilon_0 R)(4V)^2 = \frac{1}{2} (4\pi \epsilon_0 R)(16V^2) = 32\pi \epsilon_0 RV^2 \)

• For the second conductor:

\( U_2 = \frac{1}{2} (8\pi \epsilon_0 R)(2V)^2 = \frac{1}{2} (8\pi \epsilon_0 R)(4V^2) = 16\pi \epsilon_0 RV^2 \)

The total initial energy \( U_{initial} \) is:

\( U_{initial} = U_1 + U_2 = 32\pi \epsilon_0 RV^2 + 16\pi \epsilon_0 RV^2 = 48\pi \epsilon_0 RV^2 \)

Now, calculating the final energy after they are connected:

\( U_{final} = \frac{1}{2} C_{total} V_f^2 = \frac{1}{2} (12\pi \epsilon_0 R) \left(\frac{8V}{3}\right)^2 = \frac{1}{2} (12\pi \epsilon_0 R) \frac{64V^2}{9} = \frac{384\pi \epsilon_0 RV^2}{18} = \frac{64\pi \epsilon_0 RV^2}{3} \)

Energy Loss Calculation

The loss in energy \( \Delta U \) when the conductors are connected is:

\( \Delta U = U_{initial} - U_{final} = 48\pi \epsilon_0 RV^2 - \frac{64\pi \epsilon_0 RV^2}{3} \)

To simplify this, we can express \( 48\pi \epsilon_0 RV^2 \) as \( \frac{144\pi \epsilon_0 RV^2}{3} \):

\( \Delta U = \frac{144\pi \epsilon_0 RV^2}{3} - \frac{64\pi \epsilon_0 RV^2}{3} = \frac{80\pi \epsilon_0 RV^2}{3} \)

Thus, the loss in energy when the two conductors are connected by a wire is:

\( \Delta U = \frac{80\pi \epsilon_0 RV^2}{3} \)

This result shows how energy is redistributed when conductors with different potentials are connected, leading to a loss of energy in the form of heat or electromagnetic radiation. This is a fundamental concept in electrostatics and helps us understand energy conservation in electrical systems.