The capacitance of a capacitor made by a thin metal foil is `2muF`. If the foil is filled with paperr of thickness 0.15 mm, dielectric constant of paper is 2.5 and width of the paper is 400 mm. what is the length of foil?

Options

A. 0.34 m
B. 1.33 m
C. 13.4 m
D. 33.9 m

Answer

To find the length of the foil in the capacitor, we can use the formula for the capacitance of a parallel plate capacitor filled with a dielectric material. The formula is given by:

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

Where:
- \( C \) is the capacitance,
- \( k \) is the dielectric constant,
- \( \epsilon_0 \) is the permittivity of free space (\( \approx 8.85 \times 10^{-12} \, \text{F/m} \)),
- \( A \) is the area of the plates,
- \( d \) is the distance between the plates.

### Step 1: Identify the known values
- Capacitance \( C = 2 \, \mu F = 2 \times 10^{-6} \, F \)
- Dielectric constant \( k = 2.5 \)
- Thickness of the paper (distance between the plates) \( d = 0.15 \, mm = 0.15 \times 10^{-3} \, m \)
- Width of the paper \( w = 400 \, mm = 400 \times 10^{-3} \, m \)

### Step 2: Calculate the area \( A \)
The area \( A \) of the plates can be expressed as:
\[ A = L \cdot w \]
Where \( L \) is the length of the foil that we want to find.

### Step 3: Substitute \( A \) into the capacitance formula
Substituting \( A \) into the capacitance formula gives:
\[ C = \frac{k \cdot \epsilon_0 \cdot (L \cdot w)}{d} \]

### Step 4: Rearranging the formula to solve for \( L \)
Rearranging the equation to solve for \( L \):
\[ L = \frac{C \cdot d}{k \cdot \epsilon_0 \cdot w} \]

### Step 5: Substitute the known values into the formula
Now, substituting the known values into the formula:
\[ L = \frac{(2 \times 10^{-6}) \cdot (0.15 \times 10^{-3})}{(2.5) \cdot (8.85 \times 10^{-12}) \cdot (400 \times 10^{-3})} \]

### Step 6: Calculate the value of \( L \)
Calculating the numerator:
\[ (2 \times 10^{-6}) \cdot (0.15 \times 10^{-3}) = 3 \times 10^{-7} \]

Calculating the denominator:
\[ (2.5) \cdot (8.85 \times 10^{-12}) \cdot (400 \times 10^{-3}) = 2.5 \cdot 8.85 \cdot 400 \times 10^{-15} = 8850 \times 10^{-15} = 8.85 \times 10^{-12} \]

Now substituting these values into the equation for \( L \):
\[ L = \frac{3 \times 10^{-7}}{8.85 \times 10^{-12}} \approx 33.9 \, m \]

### Final Answer
The length of the foil is approximately \( 33.9 \, m \).