In the figure shown, after the switch 'S' is turned from position 'A' to position 'B', the energy di

Q: Why is energy lost when we connect two capacitors?

When charges flow from one capacitor to another, they form a transient current. This current passes through the connecting wires, which always have some microscopic resistance, dissipating energy as heat. Even in ideal wires with zero resistance, energy is radiated away as electromagnetic waves!

Q: Why do we use charge conservation instead of energy conservation?

Energy is not conserved in this process because of the heat/radiation loss mentioned above. However, since the circuit is isolated from the battery, the total number of electrons (and thus the total charge) has nowhere else to go, making charge strictly conserved.

Q: How do we know the capacitors are in parallel and not in series?

When the switch is at B, the top plates of both capacitors are connected together by the top wire, and their bottom plates are connected together by the bottom wire. This means they share the exact same potential difference, which is the defining condition for a parallel combination.

Q: Can I use the formula $\frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2$ for energy loss?

Absolutely! That is a direct derived formula for this exact scenario. Here, $C_1 = C$, $C_2 = 3C$, $V_1 = \varepsilon$, and $V_2 = 0$. Plugging these in gives $\frac{1}{2} \frac{C \cdot 3C}{4C} (\varepsilon - 0)^2 = \frac{3}{8} C\varepsilon^2$, which perfectly matches our derived result.

Q: Why did we substitute $\varepsilon$ with $Q/C$ at the end?

Always keep an eye on the options! The options provided in the question are expressed in terms of the total charge $Q$ and capacitance $C$. Since our intermediate answer was in terms of $\varepsilon$, we used the relation $Q = C\varepsilon$ to translate our answer into the required format.

Question

In the figure shown, after the switch 'S' is turned from position 'A' to position 'B', the energy dissipated in the circuit in terms of capacitance 'C' and total charge 'Q' is

Options:

(A) $\frac{3}{4} \cdot \frac{Q^2}{C}$

(B) $\frac{5}{8} \cdot \frac{Q^2}{C}$

(C) $\frac{1}{8} \cdot \frac{Q^2}{C}$

(D) $\frac{3}{8} \cdot \frac{Q^2}{C}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you have a bucket of water representing our charged capacitor, and you suddenly connect it to an empty bucket. The water will rush from the full bucket to the empty one until the water levels in both are exactly the same. This is precisely what happens in our circuit!

Initially, when the switch $S$ is at position $A$, the capacitor $C$ is directly connected across the battery. It acts like a bucket being filled to the brim. The capacitor charges up until the potential difference across its plates exactly matches the battery's EMF, $\varepsilon$. At this steady state, the charge stored on the capacitor is given by the fundamental relation:

$Q = C\varepsilon$

Because the capacitor is fully charged, it stores electrostatic potential energy. The formula for the energy stored in a capacitor is $\frac{1}{2} C V^2$. In our case, the voltage $V$ is simply $\varepsilon$. So, the initial energy of our system is:

$U_1 = \frac{1}{2} C \varepsilon^2$

### The Master Equation

Next, the switch is thrown from position $A$ to position $B$. This is the critical moment. The battery is completely disconnected from the circuit. Our charged capacitor $C$ is now connected in a closed loop with the uncharged capacitor $3C$. They are now in parallel with each other, meaning they must share the same potential difference.

Since the battery is no longer in the picture, the **total charge in the system must remain conserved**. The initial charge $Q$ is trapped. This charge will flow from capacitor $C$ to capacitor $3C$ until they reach a common potential, $V$. Because they are in parallel, their equivalent capacitance is simply the sum of their individual capacitances:

$C_{eq} = C + 3C = 4C$

We can easily find the new common potential by dividing the total conserved charge by the total equivalent capacitance:

$V = \frac{Q_{total}}{C_{eq}} = \frac{C\varepsilon}{4C} = \frac{\varepsilon}{4}$

### Final Calculation

With the common potential known, we can now calculate the final energy of the entire system. We use the same energy formula, but this time with the equivalent capacitance and the new common potential:

$U_2 = \frac{1}{2} C_{eq} V^2 = \frac{1}{2} (4C) \left(\frac{\varepsilon}{4}\right)^2$

Squaring the term inside the parenthesis gives $\frac{\varepsilon^2}{16}$. Multiplying this by $4C$ and dividing by $2$ simplifies neatly to:

$U_2 = \frac{1}{8} C\varepsilon^2$

Now for the main question: how much energy was dissipated? When charges redistribute through connecting wires, some energy is always lost as heat or electromagnetic radiation. This dissipated energy, $\Delta U$, is simply the difference between the initial energy $U_1$ and the final energy $U_2$:

$\Delta U = U_1 - U_2 = \frac{1}{2} C\varepsilon^2 - \frac{1}{8} C\varepsilon^2 = \frac{3}{8} C\varepsilon^2$

But wait, look at the options! They are given in terms of the initial charge $Q$, not $\varepsilon$. Since $Q = C\varepsilon$, we can write $\varepsilon = \frac{Q}{C}$. Substituting this into our result gives:

$\Delta U = \frac{3}{8} C \left(\frac{Q}{C}\right)^2 = \frac{3}{8} \frac{Q^2}{C}$

This elegant result shows exactly how much energy is inevitably lost during the charge sharing process.

Animated Solution for Physics - Electrostatics: In the figure shown, after the switch 'S' is turned from position 'A' to position 'B', the energy dissipated in the circuit in terms of capacitance 'C' and total charge 'Q' is

Select Answer:

Visualized Solution

Visual Anchor

Charging the Capacitor

Initial Energy

Switching to Position B

Charge Conservation

Common Potential

Final Energy Setup

Final Energy Calculation

Energy Dissipated

Final Answer

The Way Forward

The Sigma Insight: Combination of Capacitors

Analyzing the Setup

The Master Equation

Final Calculation

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