A capacitor has two circular plates whose radius are 8 cm and distance between them is 1mm.When mica (dielectric constant =6) is placed between the plates, the capacitance of this capacitor and the energy stored when it is given potetial of 150 volt respectively are

Options

A. `1.06xx10^(5)F,1.2xx10^(-9)J`
B. `1.068x10^(-9)F,1.2xx10^(-5)J`
C. `1.2xx10^(-9)F,1.068xx10^(-5)J`
D. `1.6xx10^(-9)F,1.208xx10^(-5)J`

Answer

To solve the problem step by step, we will calculate the capacitance of the capacitor with the dielectric and then find the energy stored when a potential of 150 volts is applied.

### Step 1: Convert the given dimensions to SI units
- The radius of the plates is given as 8 cm. We convert this to meters:
\[ r = 8 \text{ cm} = 0.08 \text{ m} \]
- The distance between the plates is given as 1 mm. We convert this to meters:
\[ d = 1 \text{ mm} = 1 \times 10^{-3} \text{ m} \]

### Step 2: Calculate the area of the circular plates
The area \( A \) of a circular plate is given by the formula:
\[ A = \pi r^2 \]
Substituting the value of \( r \):
\[ A = \pi (0.08)^2 = \pi \times 0.0064 \approx 0.0201 \text{ m}^2 \]

### Step 3: Use the formula for capacitance with a dielectric
The capacitance \( C \) of a capacitor with a dielectric is given by:
\[ C = k \frac{\epsilon_0 A}{d} \]
Where:
- \( k = 6 \) (dielectric constant)
- \( \epsilon_0 = 8.85 \times 10^{-12} \text{ F/m} \) (permittivity of free space)

Substituting the values:
\[ C = 6 \times \frac{(8.85 \times 10^{-12}) \times (0.0201)}{1 \times 10^{-3}} \]
Calculating this:
\[ C = 6 \times \frac{(8.85 \times 10^{-12}) \times (0.0201)}{1 \times 10^{-3}} \approx 1.068 \times 10^{-9} \text{ F} \text{ (or Farads)} \]

### Step 4: Calculate the energy stored in the capacitor
The energy \( U \) stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} C V^2 \]
Where \( V = 150 \text{ V} \). Substituting the values:
\[ U = \frac{1}{2} \times (1.068 \times 10^{-9}) \times (150)^2 \]
Calculating \( V^2 \):
\[ 150^2 = 22500 \]
Now substituting back into the energy formula:
\[ U = \frac{1}{2} \times (1.068 \times 10^{-9}) \times 22500 \approx 1.2 \times 10^{-5} \text{ J} \text{ (or Joules)} \]

### Final Answers
- Capacitance \( C \approx 1.068 \times 10^{-9} \text{ F} \)
- Energy \( U \approx 1.2 \times 10^{-5} \text{ J} \)