In the figure shown below, the charge on the left plate of the 10 microfarad capacitor is -30 microc

Q: Why is the net charge of the isolated system zero?

Because the plates forming that H-shaped section are not connected to any external source of charge (like a battery terminal). Assuming the capacitors were initially uncharged, the total charge trapped in that isolated conducting section must remain zero due to the law of conservation of charge.

Q: Why did we completely ignore the $2\mu\text{F}$ capacitor?

The question provides enough local information (the charge on the $10\mu\text{F}$ capacitor) to solve for the charge distribution within the isolated section. The $2\mu\text{F}$ capacitor would only matter if we were trying to find the total equivalent capacitance or if we were given the total voltage across the entire circuit instead of a specific plate's charge.

Q: Why does charge distribute proportionally to capacitance in a parallel circuit?

Capacitors in parallel are connected across the same two points, so they must have the same potential difference ($V$). Since the formula for charge is $Q = CV$, and $V$ is the same for both, the charge $Q$ is directly proportional to the capacitance $C$.

Q: What if the isolated system had an initial charge before the circuit was connected?

If the isolated system had an initial net charge (say, $Q_0$), then the sum of the charges on those three plates would equal $Q_0$ instead of zero. However, in standard JEE problems, unless specified otherwise, we assume capacitors are initially uncharged.

Q: Why is the final answer positive when the distributed charge was negative?

The charge distribution calculation gave us the charge on the *left* plate of the $6\mu\text{F}$ capacitor, which is part of the isolated system balancing the $+30\mu\text{C}$. So the left plate got $-18\mu\text{C}$. The question specifically asked for the charge on the *right* plate, which must be equal and opposite to the left plate, hence $+18\mu\text{C}$.

Question

In the figure shown below, the charge on the left plate of the $10\mu	ext{F}$ capacitor is $-30\mu	ext{C}$. The charge on the right plate of the $6\mu	ext{F}$ capacitor is

Options:

(A) $+ 12\mu	ext{C}$

(B) $+ 18\mu	ext{C}$

(C) $- 12\mu	ext{C}$

(D) $- 18\mu	ext{C}$

Sigma Percentile · Accepted Answer

### The Secret of the Isolated Plates: Mastering Charge Distribution

Welcome to an intriguing puzzle from the world of electrostatics! At first glance, a circuit with multiple capacitors might look like a tangled web, but by applying a few fundamental principles, we can unravel it beautifully. Let's dive into the logic behind finding the exact charge on a specific plate.

### The Principle of Facing Plates

Our journey begins with a crucial piece of information: the left plate of the $10\mu	ext{F}$ capacitor carries a charge of $-30\mu	ext{C}$.

Before we even look at the rest of the circuit, we must recall the most fundamental property of a capacitor: **the facing plates of a capacitor always carry equal and opposite charges**. This happens because the electric field inside the conducting material of the plates must be zero. Therefore, if the left plate has $-30\mu	ext{C}$, the right plate of the $10\mu	ext{F}$ capacitor must hold exactly $+30\mu	ext{C}$.

### The Isolated System

Now, let's trace the wire connected to that right plate. It branches out and connects to the left plates of both the $6\mu	ext{F}$ and $4\mu	ext{F}$ capacitors.

Notice something special about this H-shaped section of wire and plates? It is completely disconnected from any external battery or ground. It forms an **isolated system**. Assuming the capacitors were initially uncharged before any external connections were made, the law of conservation of charge dictates that the net charge of this isolated system must remain zero.

Mathematically, this means:
$$q_{10R} + q_{6L} + q_{4L} = 0$$

Substituting the value we found:
$$+30\mu	ext{C} + q_{6L} + q_{4L} = 0$$

$$q_{6L} + q_{4L} = -30\mu	ext{C}$$

This tells us that a total charge of $-30\mu	ext{C}$ is distributed between the left plates of the parallel combination.

### Parallel Distribution

How does this $-30\mu	ext{C}$ split between the $6\mu	ext{F}$ and $4\mu	ext{F}$ capacitors? Because they are connected in parallel, they share the same potential difference ($V$).

Since $Q = CV$, and $V$ is constant, the charge $Q$ is directly proportional to the capacitance $C$. The larger capacitor will grab a larger share of the charge. We can use the charge division rule:

$$q_{6L} = \left( \frac{C_6}{C_6 + C_4} ight) 	imes q_{	ext{total}}$$

$$q_{6L} = \left( \frac{6}{6 + 4} ight) 	imes (-30\mu	ext{C})$$

$$q_{6L} = \frac{6}{10} 	imes (-30\mu	ext{C}) = -18\mu	ext{C}$$

### The Final Trap

We've successfully calculated that the left plate of the $6\mu	ext{F}$ capacitor holds $-18\mu	ext{C}$. But wait! Read the question carefully. It asks for the charge on the **right plate** of the $6\mu	ext{F}$ capacitor.

This is a classic trap. We must apply our first principle again. If the left plate has $-18\mu	ext{C}$, the right plate must have the equal and opposite charge:

$$q_{6R} = -(-18\mu	ext{C}) = +18\mu	ext{C}$$

And there we have it! By systematically applying charge conservation and the rules of parallel circuits, we bypassed the irrelevant $2\mu	ext{F}$ capacitor and arrived at the correct answer.

Animated Solution for Physics - Electrostatics Potential and Capacitance: In the figure shown below, the charge on the left plate of the $10 μ F$ capacitor is $- 30 μ C$ . The charge on the right plate of the $6 μ F$ capacitor is

Select Answer:

Visualized Solution

Visual Anchor

Property of Facing Plates

Identifying the Isolated System

Charge Conservation

Charge Distribution in Parallel

Calculating $q_{6 L}$

Final Answer

The Way Forward

The Sigma Insight: Combination of Capacitors and Energy Stored in a Capacitor

The Secret of the Isolated Plates

The Principle of Facing Plates

The Isolated System

Parallel Distribution

The Final Trap