Question

Two identical capacitors have same capacitance C. One of them is charged to the potential V and other to the potential 2V. The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is

• A. \(\frac{CV^2}{4}\)
• B. 2CV²
• C. \(\frac{1}{2}CV^2\)
• D. \(\frac{3}{4}CV^2\)

Answer 1

The Correct Option is A

When the capacitors are connected, the potential of the combined system is given by:

\[ V_c = \frac{q_{net}}{C_{net}} = \frac{CV + 2CV}{2C} = \frac{3V}{2} \]

The loss of energy in the system can be calculated as:

\[ \text{Loss of energy} = \frac{1}{2}CV^2 + \frac{1}{2}C(2V)^2 - \frac{1}{2}C \left(\frac{3V}{2}\right)^2 \]

Simplifying this expression:

\[ \text{Loss of energy} = \frac{1}{2}CV^2 + \frac{1}{2}C(4V^2) - \frac{1}{2}C \left(\frac{9V^2}{4}\right) \]

\[ \text{Loss of energy} = \frac{CV^2}{2} + 2CV^2 - \frac{9CV^2}{8} \]

\[ \text{Loss of energy} = \frac{4CV^2}{8} + \frac{16CV^2}{8} - \frac{9CV^2}{8} \]

\[ \text{Loss of energy} = \frac{CV^2}{4} \]

Answer 2

We are given two identical capacitors, each with capacitance C. One is charged to a potential V, and the other to a potential 2V. They are then connected in parallel, with plates of the same polarity joined together. We need to find the decrease in the electrostatic potential energy of this combined system.

Concept Used:

The solution involves the principles of charge conservation and energy storage in capacitors.

1. Charge on a capacitor: The charge \(Q\) stored on a capacitor is given by \( Q = CV \), where \(C\) is the capacitance and \(V\) is the potential difference.
2. Energy stored in a capacitor: The electrostatic potential energy \(U\) stored in a capacitor is given by \( U = \frac{1}{2}CV^2 \).
3. Parallel Combination of Capacitors: When capacitors are connected in parallel, the total charge is the sum of the individual charges (\(Q_{total} = Q_1 + Q_2\)), and the equivalent capacitance is the sum of the individual capacitances (\(C_{eq} = C_1 + C_2\)). The system reaches a common final potential \(V_f = \frac{Q_{total}}{C_{eq}}\).
4. Energy Loss: When capacitors at different potentials are connected, charge flows from the higher potential capacitor to the lower one until they reach a common potential. This redistribution of charge involves current flow through the connecting wires, which have some resistance, resulting in a loss of energy in the form of heat. The decrease in energy is calculated as \( \Delta U = U_{initial} - U_{final} \).

Step-by-Step Solution:

Step 1: Calculate the initial charge and energy for each capacitor before connection.

For the first capacitor:

\[ C_1 = C, \quad V_1 = V \] \[ Q_1 = C_1V_1 = CV \] \[ U_1 = \frac{1}{2}C_1V_1^2 = \frac{1}{2}CV^2 \]

For the second capacitor:

\[ C_2 = C, \quad V_2 = 2V \] \[ Q_2 = C_2V_2 = C(2V) = 2CV \] \[ U_2 = \frac{1}{2}C_2V_2^2 = \frac{1}{2}C(2V)^2 = \frac{1}{2}C(4V^2) = 2CV^2 \]

Step 2: Calculate the total initial energy (\(U_{initial}\)) of the system.

The total initial energy is the sum of the energies stored in the two separate capacitors.

\[ U_{initial} = U_1 + U_2 = \frac{1}{2}CV^2 + 2CV^2 = \frac{5}{2}CV^2 \]

Step 3: Determine the properties of the combined system after connection.

When the capacitors are connected in parallel (positive to positive, negative to negative), the total charge is conserved and is the sum of the individual charges. The equivalent capacitance is the sum of the individual capacitances.

Total charge:

\[ Q_{total} = Q_1 + Q_2 = CV + 2CV = 3CV \]

Equivalent capacitance:

\[ C_{eq} = C_1 + C_2 = C + C = 2C \]

Step 4: Calculate the final common potential (\(V_f\)) across the parallel combination.

The final potential is the total charge divided by the equivalent capacitance.

\[ V_f = \frac{Q_{total}}{C_{eq}} = \frac{3CV}{2C} = \frac{3}{2}V \]

Step 5: Calculate the total final energy (\(U_{final}\)) of the combined system.

The final energy is calculated using the equivalent capacitance and the final common potential.

\[ U_{final} = \frac{1}{2}C_{eq}V_f^2 = \frac{1}{2}(2C)\left(\frac{3}{2}V\right)^2 \] \[ U_{final} = C\left(\frac{9}{4}V^2\right) = \frac{9}{4}CV^2 \]

Final Computation & Result:

The decrease in energy of the system is the difference between the initial energy and the final energy.

\[ \Delta U = U_{initial} - U_{final} = \frac{5}{2}CV^2 - \frac{9}{4}CV^2 \]

To subtract, we find a common denominator:

\[ \Delta U = \frac{10}{4}CV^2 - \frac{9}{4}CV^2 = \frac{1}{4}CV^2 \]

The decrease in energy of the combined system is \( \frac{1}{4}CV^2 \).