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 capacitors. We have two parallel plate capacitors, and we will calculate the capacitance for each and then find the total energy stored.

### Step 1: Calculate the Capacitance of Capacitor A-B (C1)

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

\[ C = \frac{\epsilon_0 A}{d} \]

Where:
- \( \epsilon_0 \) 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

The diameter of plates A and B is \( 0.1 \, \text{m} \), so the radius \( r_1 \) is:

\[ r_1 = \frac{0.1}{2} = 0.05 \, \text{m} \]

The area \( A_1 \) is:

\[ A_1 = \pi r_1^2 = \pi (0.05)^2 = \pi (0.0025) \approx 0.00785 \, \text{m}^2 \]

The separation \( d_1 \) is \( 2 \times 10^{-3} \, \text{m} \).

Now substituting the values into the capacitance formula:

\[ C_1 = \frac{\epsilon_0 A_1}{d_1} = \frac{8.85 \times 10^{-12} \times 0.00785}{2 \times 10^{-3}} \approx 3.47 \times 10^{-10} \, \text{F} \]

### Step 2: Calculate the Capacitance of Capacitor C-D (C2)

For plates C and D, the diameter is \( 0.12 \, \text{m} \), so the radius \( r_2 \) is:

\[ r_2 = \frac{0.12}{2} = 0.06 \, \text{m} \]

The area \( A_2 \) is:

\[ A_2 = \pi r_2^2 = \pi (0.06)^2 = \pi (0.0036) \approx 0.0113 \, \text{m}^2 \]

The separation \( d_2 \) is \( 3 \times 10^{-3} \, \text{m} \).

Now substituting the values into the capacitance formula:

\[ C_2 = \frac{\epsilon_0 A_2}{d_2} = \frac{8.85 \times 10^{-12} \times 0.0113}{3 \times 10^{-3}} \approx 3.34 \times 10^{-10} \, \text{F} \]

### Step 3: Calculate the Effective Capacitance (C)

Since capacitors C1 and C2 are connected in series, the effective capacitance \( C \) is given by:

\[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \]

Substituting the values we calculated:

\[ \frac{1}{C} = \frac{1}{3.47 \times 10^{-10}} + \frac{1}{3.34 \times 10^{-10}} \]

Calculating this gives:

\[ \frac{1}{C} \approx 2.88 \times 10^{9} \, \text{F}^{-1} \]

Thus,

\[ C \approx \frac{1}{2.88 \times 10^{9}} \approx 3.47 \times 10^{-10} \, \text{F} \]

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

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

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

Where \( V \) is the voltage across the capacitor. Here, \( V = 120 \, \text{V} \).

Substituting the values:

\[ E = \frac{1}{2} \times 3.47 \times 10^{-10} \times (120)^2 \]

Calculating this gives:

\[ E \approx \frac{1}{2} \times 3.47 \times 10^{-10} \times 14400 \approx 2.5 \times 10^{-6} \, \text{J} = 0.25 \, \text{microjoules} \]

### Final Answer

The energy stored in the system is approximately \( 0.25 \, \text{microjoules} \).

---