The circular plates A and B of a parallel plate air capacitor have a diameter of 0.1 m and are `2 xx 10^(-3) m` apart. The plates C and D of a similar capacitor have a diameter of `0.12 m` and are `3 xx 10^(-3)m` apart. Plate A is earthed. Plates B and D are connected together. Plate C is connected to the positive pole of a 120V battery whose negative is earthed. The energy stored in the system is

Options

A. `0.1224 mu J`
B. `0.2224 mu J`
C. `0.3224 mu J`
D. `0.4224 mu J`

Answer

To solve the problem, we need to find the energy stored in the system of two parallel plate capacitors. Let's break down the steps involved in the calculation.

### Step 1: Calculate the Capacitance of Capacitor A and B

The capacitance \( C_1 \) of a parallel plate capacitor is given by the formula:

\[ C = \frac{\varepsilon_0 \cdot A}{d} \]

where:
- \( \varepsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{F/m} \)
- \( A \) is the area of one of the plates
- \( d \) is the separation between the plates

**Given for Capacitor A and B:**
- Diameter \( D_1 = 0.1 \, \text{m} \) → Radius \( R_1 = \frac{D_1}{2} = 0.05 \, \text{m} \)
- Distance \( d_1 = 2 \times 10^{-3} \, \text{m} \)

**Area Calculation:**
\[ A_1 = \pi R_1^2 = \pi (0.05)^2 = \pi \times 0.0025 \approx 0.00785 \, \text{m}^2 \]

**Capacitance Calculation:**
\[ C_1 = \frac{(8.85 \times 10^{-12}) \cdot (0.00785)}{2 \times 10^{-3}} \approx 3.47 \times 10^{-10} \, \text{F} \, \text{(or 347 pF)} \]

### Step 2: Calculate the Capacitance of Capacitor C and D

**Given for Capacitor C and D:**
- Diameter \( D_2 = 0.12 \, \text{m} \) → Radius \( R_2 = \frac{D_2}{2} = 0.06 \, \text{m} \)
- Distance \( d_2 = 3 \times 10^{-3} \, \text{m} \)

**Area Calculation:**
\[ A_2 = \pi R_2^2 = \pi (0.06)^2 = \pi \times 0.0036 \approx 0.0113 \, \text{m}^2 \]

**Capacitance Calculation:**
\[ C_2 = \frac{(8.85 \times 10^{-12}) \cdot (0.0113)}{3 \times 10^{-3}} \approx 3.33 \times 10^{-10} \, \text{F} \, \text{(or 333 pF)} \]

### Step 3: Find the Equivalent Capacitance of the System

Since Capacitors A and B are in series with Capacitors C and D, the equivalent capacitance \( C_{eq} \) can be calculated as:

\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \]

Calculating \( C_{eq} \):
\[ \frac{1}{C_{eq}} = \frac{1}{3.47 \times 10^{-10}} + \frac{1}{3.33 \times 10^{-10}} \approx 5.23 \times 10^{9} \, \text{F}^{-1} \]
\[ C_{eq} \approx 1.91 \times 10^{-10} \, \text{F} \, \text{(or 191 pF)} \]

### Step 4: Calculate the Energy Stored in the System

The energy \( U \) stored in a capacitor is given by:

\[ U = \frac{1}{2} C V^2 \]

**Given Voltage \( V = 120 \, \text{V} \)**:
\[ U = \frac{1}{2} \cdot (1.91 \times 10^{-10}) \cdot (120)^2 \]
\[ U \approx \frac{1}{2} \cdot (1.91 \times 10^{-10}) \cdot 14400 \approx 1.37 \times 10^{-6} \, \text{J} \, \text{(or 1.37 µJ)} \]

### Final Answer
The energy stored in the system is approximately \( 1.37 \, \mu J \).

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