A 10 uF capacitor is fully charged to a potential difference of 50 V. After removing the source volt

Q: Why does the potential difference drop from 50 V to 20 V?

When the charged capacitor is connected to an uncharged one, charge flows from the higher potential to the lower potential until both reach an equilibrium (common potential). Since the charge is now spread over a larger total capacitance, the voltage must decrease according to $V = Q / C_{eq}$.

Q: Why are the capacitors considered to be in parallel?

When the positive plates of both capacitors are connected together, and their negative plates are connected together, they form a parallel combination. This ensures they both have the exact same potential difference across them.

Q: Is energy conserved in this process?

No, energy is not conserved. When charge flows through the connecting wires, some of the initial electrostatic potential energy is dissipated as heat and electromagnetic radiation. Only charge is strictly conserved.

Q: Can we use the formula $V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$ directly?

Yes, absolutely! This is the standard formula for common potential. Here, $V_2 = 0$ since the second capacitor is initially uncharged. Plugging in the values will give you the same result for $C_2$.

Q: What if the second capacitor was also initially charged?

If the second capacitor was charged, you would need to know its initial voltage and polarity. You would calculate its initial charge $Q_2 = C_2 V_2$, and the total charge would be $Q_{total} = Q_1 \pm Q_2$ depending on whether plates of the same or opposite polarity were connected together.

Question

A $10 \mu	ext{F}$ capacitor is fully charged to a potential difference of 50 V. After removing the source voltage it is connected to an uncharged capacitor in parallel. Now, the potential difference across them becomes 20 V. The capacitance of the second capacitor is

Options:

(A) $15 \mu	ext{F}$

(B) $30 \mu	ext{F}$

(C) $20 \mu	ext{F}$

(D) $10 \mu	ext{F}$

Sigma Percentile · Accepted Answer

### The Beauty of Charge Conservation

Imagine you have a bucket full of water, and you suddenly connect it to an empty bucket using a pipe at the bottom. What happens? The water flows from the full bucket to the empty one until the water levels in both buckets are exactly the same.

This is precisely what happens in electrostatics when a charged capacitor is connected in parallel with an uncharged one. The "water" is the electric charge, and the "water level" is the electric potential (voltage). Let's dive into the mathematics of this beautiful physical phenomenon.

### Analyzing the Setup

We start with a single capacitor, let's call it $C_1$, which has a capacitance of $10 \mu	ext{F}$. It is fully charged using a $50	ext{ V}$ battery.

Before we do anything else, we must determine how much "water" (charge) is in our bucket. The fundamental equation relating charge, capacitance, and voltage is:

$$Q = C_1 V_1$$

Substituting our known values:

$$Q = 10 \mu	ext{F} 	imes 50	ext{ V} = 500 \mu	ext{C}$$

So, our initial system holds a total charge of $500 \mu	ext{C}$. This is a crucial piece of information because, in an isolated system, charge can neither be created nor destroyed.

### The Parallel Connection

Next, the battery is removed, isolating the charged capacitor. It is then connected in parallel to a second, completely uncharged capacitor, $C_2$.

Because they are connected in parallel, charge will flow from $C_1$ to $C_2$ until the potential difference across both capacitors is identical. The problem states that this new equilibrium voltage, known as the **common potential**, is $V = 20	ext{ V}$.

Since we know the new voltage across $C_1$, we can easily calculate how much charge it retained after the sharing process:

$$Q_1 = C_1 V$$

$$Q_1 = 10 \mu	ext{F} 	imes 20	ext{ V} = 200 \mu	ext{C}$$

### The Master Equation: Conservation of Charge

Here is where the magic happens. We started with $500 \mu	ext{C}$ of charge. After connecting the second capacitor, $C_1$ only holds $200 \mu	ext{C}$. Where did the remaining charge go?

By the principle of conservation of charge, the lost charge must have migrated to the second capacitor, $C_2$. Therefore, the charge on $C_2$ is:

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

$$Q_2 = 500 \mu	ext{C} - 200 \mu	ext{C} = 300 \mu	ext{C}$$

### Final Calculation

We now have all the pieces of the puzzle for the second capacitor. We know it holds a charge $Q_2 = 300 \mu	ext{C}$, and we know the potential difference across it is $V = 20	ext{ V}$.

Using the capacitance formula rearranged for $C$:

$$C_2 = \frac{Q_2}{V}$$

$$C_2 = \frac{300 \mu	ext{C}}{20	ext{ V}} = 15 \mu	ext{F}$$

And there we have it! The capacitance of the second capacitor is **$15 \mu	ext{F}$**.

### An Alternative Approach: The Common Potential Formula

If you prefer a more direct algebraic route, you can use the standard formula for the common potential of two capacitors connected in parallel:

$$V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$$

Since the second capacitor was initially uncharged, $V_2 = 0$. Plugging in our values:

$$20 = \frac{10 	imes 50 + C_2 	imes 0}{10 + C_2}$$

$$20(10 + C_2) = 500$$

$$200 + 20 C_2 = 500$$

$$20 C_2 = 300$$

$$C_2 = 15 \mu	ext{F}$$

Both methods rely on the exact same physical principle—conservation of charge—and both elegantly lead us to the correct answer.

Select Answer:

Visualized Solution

$C_{1}$ Initial State

Initial Charge

Parallel Connection

Charge on $C_{1}$

Calculating $Q_{1}$

Conservation of Charge

Charge on $C_{2}$

Capacitance $C_{2}$

Alternative Method

The Sigma Insight: Combination of Capacitors

The Beauty of Charge Conservation

Analyzing the Setup

The Parallel Connection

The Master Equation

Final Calculation

An Alternative Approach

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