In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C. T

Q: Why does the charge divide equally when $S_2$ is closed?

Because both capacitors have the same capacitance $C$. When connected in parallel, they must have the same potential difference $V$. Since $Q = CV$, equal capacitance and equal voltage mean equal charge.

Q: What happens to the charge on $C_1$ when $S_2$ is opened?

Once $S_2$ is opened, $C_1$ is completely isolated from the rest of the circuit. Therefore, its charge cannot change and remains at $+CV_0$.

Q: Why is the charge on the upper plate of $C_2$ negative at the end?

Look at the symbol for the $V_0$ battery. The longer line (positive terminal) is at the bottom, and the shorter, thicker line (negative terminal) is at the top. This means the upper plate of $C_2$ is connected to the negative terminal, giving it a charge of $-CV_0$.

Q: Does the initial charge on $C_2$ matter when $S_3$ is closed?

No. When a capacitor is connected directly across an ideal battery, the battery will supply or absorb whatever charge is necessary to ensure the capacitor's voltage matches the battery's voltage. The final charge is determined entirely by $Q = CV$.

Q: Is energy conserved during the charge sharing when $S_2$ is closed?

No, energy is not conserved. Whenever charge is redistributed between capacitors, some energy is always lost as heat in the connecting wires, even if their resistance is negligible.

Question

In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance $C$. The switch $S_1$ is pressed first to fully charge the capacitor $C_1$ and then released. The switch $S_2$ is then pressed to charge the capacitor $C_2$. After some time, $S_2$ is released and then $S_3$ is pressed. After some time (2013 Adv.)

Options:

(A) the charge on the upper plate of $C_1$ is $2 CV_0$

(B) the charge on the upper plate of $C_1$ is $CV_0$

(C) the charge on the upper plate of $C_2$ is 0

(D) the charge on the upper plate of $C_2$ is $-CV_0$

Sigma Percentile · Accepted Answer

The problem presents a fascinating sequential switching circuit involving two capacitors and two batteries. It tests our understanding of charge distribution, parallel connections, and the critical importance of battery polarity. Let's break down the sequence of events step-by-step.

### Analyzing the Setup
We are given two identical capacitors, $C_1$ and $C_2$, each with capacitance $C$. The circuit also features two batteries: a $2V_0$ battery on the left and a $V_0$ battery on the right.

Before we touch any switches, we must carefully observe the polarities of the batteries. The $2V_0$ battery has its longer line (positive terminal) at the top. However, the $V_0$ battery has its longer line at the bottom, meaning its positive terminal is downwards and its negative terminal is upwards. This subtle detail is the key to solving the final part of the problem!

### The First Switch: Charging $C_1$
The sequence begins by pressing switch $S_1$. This action connects capacitor $C_1$ directly across the $2V_0$ battery.

Since the upper plate of $C_1$ is connected to the positive terminal of the battery, it will acquire a positive charge. The fundamental equation for a capacitor is $Q = CV$. Here, the voltage is $2V_0$, so the charge acquired by $C_1$ is:
$$Q_1 = C(2V_0) = 2CV_0$$
Thus, the upper plate of $C_1$ holds a charge of $+2CV_0$, and the lower plate holds $-2CV_0$.

### Charge Sharing: Closing $S_2$
Next, $S_1$ is released, which isolates $C_1$ from the $2V_0$ battery. Then, $S_2$ is pressed. This connects the charged capacitor $C_1$ in parallel with the uncharged capacitor $C_2$.

When capacitors are connected in parallel, they share their total charge until they reach a common potential difference. Because $C_1$ and $C_2$ have the exact same capacitance $C$, they will share the total charge equally.

The total initial charge is $2CV_0$. Dividing this equally between the two capacitors gives:
$$Q_1' = Q_2' = \frac{2CV_0}{2} = CV_0$$
Now, both $C_1$ and $C_2$ have a charge of $+CV_0$ on their upper plates.

### The Final Connection: Closing $S_3$
Finally, $S_2$ is released, isolating $C_1$. Because $C_1$ is now disconnected from the rest of the circuit, its charge cannot change. The charge on the upper plate of $C_1$ remains permanently at $+CV_0$.

Then, $S_3$ is pressed, connecting $C_2$ directly across the $V_0$ battery. As we noted earlier, the $V_0$ battery has its negative terminal at the top. This means the upper plate of $C_2$ is now forced to be at the potential of the negative terminal.

The battery will supply or absorb charge until the voltage across $C_2$ matches $V_0$. Since the upper plate is connected to the negative terminal, its final charge will be:
$$Q_2'' = C(-V_0) = -CV_0$$

### Conclusion
We have traced the entire sequence of events. The final charge on the upper plate of $C_1$ is $+CV_0$, and the final charge on the upper plate of $C_2$ is $-CV_0$. Comparing this with our options, we find that statements (b) and (d) are the correct ones.

Select Answer:

Visualized Solution

\text{Circuit Analysis}

\text{Closing } S_1

Q_1 = C_1 V

\text{Closing } S_2

\text{Charge Redistribution}

\text{Closing } S_3

Q_2'' = C(-V_0)

\text{Final Charges}

The Sigma Insight: Combination of Capacitors

Analyzing the Setup We are given two identical capacitors, $C_{1}$ and $C_{2}$ , each with capacitance $C$

The First Switch

Charge Sharing

The Final Connection

Conclusion We have traced the entire sequence of events

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