A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential

Q: Why do we subtract the charges instead of adding them?

Because the problem explicitly states that the positive terminal of one capacitor is connected to the negative terminal of the other. When you connect a plate with $+4CV$ charge to a plate with $-CV$ charge, they partially neutralize, leaving a net charge of $+3CV$.

Q: Why is the equivalent capacitance $C_1 + C_2$ even though they are connected with opposite polarities?

Capacitance is a purely geometric and material property of the system. When you connect the terminals of two capacitors together, they are in parallel regardless of their initial charge states. In parallel, equivalent capacitance always adds up: $C_{eq} = C_1 + C_2$.

Q: Is energy conserved in this process?

No. Whenever capacitors are connected and charge redistributes, some energy is always lost as heat in the connecting wires due to resistance, or radiated away as electromagnetic waves. You can verify this by calculating the initial total energy and comparing it to the final energy.

Q: What would happen if they were connected with the same polarity?

If positive was connected to positive, the net charge would be the sum of the initial charges ($Q_{net} = Q_1 + Q_2 = 5CV$). This would lead to a higher common potential and a different final energy.

Q: Can we use the formula for energy loss directly to solve this?

Yes, you can use the energy loss formula $\Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 + V_2)^2$ (note the plus sign because of opposite polarity) to find the energy lost, and then subtract it from the initial total energy. However, finding the final energy directly using the common potential is usually faster and less prone to algebraic errors.

Question

A parallel plate capacitor of capacitance $C$ is connected to a battery and is charged to a potential difference $V$. Another capacitor of capacitance $2C$ is similarly charged to a potential difference $2V$. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

Options:

(A) zero

(B) $\frac{3}{2}CV^2$

(C) $\frac{25}{6}CV^2$

(D) $\frac{9}{2}CV^2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Let's visualize the physical setup of our problem. We are given two parallel plate capacitors that have been charged independently.

The first capacitor has a capacitance of $C$ and is charged to a potential difference of $V$. The second capacitor has a capacitance of $2C$ and is charged to a potential difference of $2V$.

Before we connect them, let's determine the initial charge stored on each capacitor. Using the fundamental relation $Q = CV$, we can find:

For the first capacitor:
$$Q_1 = C_1 V_1 = CV$$

For the second capacitor:
$$Q_2 = C_2 V_2 = (2C)(2V) = 4CV$$

### The Master Equation for Charge Sharing

Now, here is the crucial part of the problem. The capacitors are connected in parallel, but with a specific condition: **the positive terminal of one is connected to the negative terminal of the other**.

Because of this opposite polarity connection, the charges on the connected plates will partially neutralize each other. The net charge that will be shared across the equivalent system is the difference between their initial charges.

$$Q_{net} = Q_2 - Q_1$$

Substituting our calculated values:

$$Q_{net} = 4CV - CV = 3CV$$

This $3CV$ is the total charge that will redistribute itself between the two capacitors once the switch is closed.

### Calculating the Common Potential

Since the capacitors are connected in parallel, their equivalent capacitance is simply the sum of their individual capacitances.

$$C_{eq} = C_1 + C_2 = C + 2C = 3C$$

When the charges redistribute, the system will reach an electrostatic equilibrium where both capacitors share a **common potential**, let's call it $V'$. This common potential is the total net charge divided by the equivalent capacitance.

$$V' = \frac{Q_{net}}{C_{eq}}$$

$$V' = \frac{3CV}{3C} = V$$

Fascinatingly, the final common potential across the combination is exactly $V$.

### Final Energy Calculation

Finally, we need to find the total final energy of the configuration. The energy $U_f$ stored in a capacitor system is given by the formula:

$$U_f = \frac{1}{2} C_{eq} (V')^2$$

Substituting our equivalent capacitance and common potential into the equation:

$$U_f = \frac{1}{2} (3C) (V)^2$$

$$U_f = \frac{3}{2} CV^2$$

This matches option (b). The beauty of this problem lies in carefully handling the sign conventions during the charge redistribution phase!

Select Answer:

Visualized Solution

$C_{1} = C, V_{1} = V C_{2} = 2 C, V_{2} = 2 V$

$Q_{1} = C V Q_{2} = 4 C V$

$Q_{n e t} = Q_{2} - Q_{1}$

$Q_{n e t} = 4 C V - C V$

$Q_{n e t} = 3 C V$

$C_{e q} = C_{1} + C_{2} = 3 C$

$V^{'} = \frac{Q _{n e t}}{C _{e q}} = V$

$U_{f} = \frac{3}{2} C V^{2}$

$Q_{n e t} = Q_{1} + Q_{2}$

The Sigma Insight: Combination of Capacitors

Analyzing the Setup

The Master Equation for Charge Sharing

Calculating the Common Potential

Final Energy Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​=C,V1​=VC2​=2C,V2​=2V

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q1​=CVQ2​=4CV

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Qnet​=Q2​−Q1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Qnet​=4CV−CV

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Qnet​=3CV

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ceq​=C1​+C2​=3C

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V′=Ceq​Qnet​​=V

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Uf​=23​CV2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Qnet​=Q1​+Q2​