A 2 microfarad capacitor C1 is first charged to a potential difference of 10 V using a battery. Then

Q: Why are the capacitors considered to be in parallel when connected this way?

When switch $S_2$ is closed, the top plates of both capacitors are connected by a single wire, and the bottom plates are connected by another wire. This forces them to share the exact same potential difference, which is the defining characteristic of a parallel circuit.

Q: Why doesn't the voltage stay at 10V?

The 10V was maintained by the battery. Once the battery is disconnected, the system only has a fixed amount of charge ($20\mu\text{C}$). When this charge spreads out over a larger total capacitance ($10\mu\text{F}$ instead of $2\mu\text{F}$), the voltage must drop according to $V = \frac{Q}{C}$.

Q: Is the total energy conserved during this process?

No! This is a classic trap. While charge is strictly conserved, energy is lost. When charge flows through the connecting wires from $C_1$ to $C_2$, it encounters resistance, dissipating some of the stored electrical energy as heat.

Q: What happens to the charge on $C_1$?

It decreases. It started with $20\mu\text{C}$. Since $16\mu\text{C}$ moved to $C_2$, $C_1$ is left with $4\mu\text{C}$. You can verify this using $q_1 = C_1 V = 2\mu\text{F} \times 2\text{V} = 4\mu\text{C}$.

Q: Can I just use the formula $Q_1 + Q_2 = Q_{\text{total}}$ directly?

Yes, absolutely. You can write $C_1 V + C_2 V = Q_{\text{total}}$, which factors to $(C_1 + C_2)V = Q_{\text{total}}$. This is exactly the logic we used to derive the common potential formula!

Question

A $2\mu	ext{F}$ capacitor $C_1$ is first charged to a potential difference of 10 V using a battery. Then, the battery is removed and the capacitor is connected to an uncharged capacitor $C_2$ of $8\mu	ext{F}$. The charge in $C_2$ on equilibrium condition is ............ $\mu	ext{C}$.
(Round off to the nearest integer)

Sigma Percentile · Accepted Answer

### The Beauty of Charge Conservation

Imagine two water tanks connected by a pipe with a closed valve. One tank is filled with water to a certain height, representing high pressure, while the other is completely empty. What happens when you open the valve? Water rushes from the full tank to the empty one until the water levels in both tanks equalize.

This intuitive physical phenomenon is exactly what happens in electrical circuits involving capacitors. In this problem, we explore the elegant principle of charge conservation and the concept of common potential when a charged capacitor is connected to an uncharged one.

### Phase 1: Filling the First Tank

Our journey begins with a single capacitor, $C_1 = 2\mu	ext{F}$, connected to a $10	ext{V}$ battery. When the switch $S_1$ is closed, the battery acts like a powerful pump, pushing electrons onto the plates of the capacitor until the potential difference across the plates perfectly matches the battery's voltage.

The amount of charge stored is determined by the fundamental capacitor equation:

$$Q_0 = C_1 V_0$$

Substituting our known values:

$$Q_0 = 2\mu	ext{F} 	imes 10	ext{V} = 20\mu	ext{C}$$

At this moment, our first 'tank' holds $20\mu	ext{C}$ of electrical 'water' under a 'pressure' of $10	ext{V}$.

### Phase 2: The Critical Transition

Now, we perform a critical maneuver. We open switch $S_1$, completely disconnecting the battery. This isolates the $20\mu	ext{C}$ of charge. It has nowhere to escape.

Next, we close switch $S_2$, connecting the charged capacitor $C_1$ directly across an uncharged capacitor, $C_2 = 8\mu	ext{F}$. Because their positive plates are wired together and their negative plates are wired together, they are now in a **parallel combination**.

### Phase 3: Reaching Equilibrium

Just like opening the valve between the water tanks, charge immediately begins to flow from $C_1$ to $C_2$. This flow continues until the electrical pressure—the potential difference—is identical across both capacitors. We call this the **Common Potential ($V$)**.

Because the isolated system cannot create or destroy charge, the total initial charge must equal the total final charge. This is the Law of Conservation of Charge.

$$Q_{	ext{initial}} = Q_{	ext{final}}$$

$$Q_0 = q_1 + q_2$$

Since $q_1 = C_1 V$ and $q_2 = C_2 V$, we can write:

$$Q_0 = (C_1 + C_2) V$$

Rearranging to solve for the common potential $V$:

$$V = \frac{Q_0}{C_1 + C_2}$$

Let's plug in our numbers. The total trapped charge is $20\mu	ext{C}$, and the equivalent capacitance of the parallel combination is $(2\mu	ext{F} + 8\mu	ext{F}) = 10\mu	ext{F}$.

$$V = \frac{20\mu	ext{C}}{10\mu	ext{F}} = 2	ext{V}$$

The electrical pressure has equalized at $2	ext{V}$.

### Phase 4: The Final State

The question specifically asks for the final charge residing on the second capacitor, $C_2$. Now that we know the equilibrium voltage across it, this is a straightforward calculation.

$$q_2 = C_2 V$$

$$q_2 = 8\mu	ext{F} 	imes 2	ext{V} = 16\mu	ext{C}$$

And there we have it! Out of the original $20\mu	ext{C}$, $16\mu	ext{C}$ migrated to the larger capacitor $C_2$, leaving exactly $4\mu	ext{C}$ behind on $C_1$. The system is in perfect, stable equilibrium.

Enter Numerical Value:

Visualized Solution

Circuit Analysis

Charging $C_{1}$

Initial Charge Calculation

Initial Charge Calculation

Charge Redistribution Setup

Common Potential

Calculating Common Potential

Calculating Common Potential

Final Charge on $C_{2}$

Final Charge on $C_{2}$

Final Answer

Energy Loss (Food for Thought)

The Sigma Insight: Combination of Capacitors

The Beauty of Charge Conservation

Phase 1

Phase 2

Phase 3

Phase 4

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Enter Numerical Value:

Visualized Solution

Circuit Analysis

Charging .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

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