If a slab of insulating material` 4 xx 10^(–5) `m thick is introduced between the plates of a parallel plate capacitor, the distance between the plates has to be increased by `3.5 xx 10^(–5)` m to restore the capacity to original value. Then the dielectric constant of the material of slab is

Options

A. 6
B. 8
C. 10
D. 12

Answer

To find the dielectric constant (k) of the insulating material slab introduced between the plates of a parallel plate capacitor, we can follow these steps:

### Step 1: Understand the problem
We have a parallel plate capacitor with an insulating slab of thickness \( t = 4 \times 10^{-5} \) m inserted between the plates. To restore the original capacitance, the distance between the plates must be increased by \( x = 3.5 \times 10^{-5} \) m.

### Step 2: Write the relationship for capacitance
The capacitance \( C \) of a parallel plate capacitor is given by:
\[ C = \frac{\varepsilon A}{d} \]
where \( \varepsilon \) is the permittivity of the dielectric material, \( A \) is the area of the plates, and \( d \) is the distance between the plates.

When a dielectric slab is inserted, the effective distance between the plates becomes \( d' = d - t + x \), where \( d \) is the original distance between the plates.

### Step 3: Set up the equation
The original capacitance \( C_1 \) and the new capacitance \( C_2 \) after inserting the slab can be expressed as:
\[ C_1 = \frac{\varepsilon_0 A}{d} \]
\[ C_2 = \frac{k \varepsilon_0 A}{d - t + x} \]
To restore the original capacitance, we set \( C_1 = C_2 \):
\[ \frac{\varepsilon_0 A}{d} = \frac{k \varepsilon_0 A}{d - t + x} \]

### Step 4: Simplify the equation
We can cancel \( \varepsilon_0 A \) from both sides:
\[ \frac{1}{d} = \frac{k}{d - t + x} \]
Cross-multiplying gives:
\[ (d - t + x) = k d \]

### Step 5: Rearranging the equation
Rearranging the equation, we have:
\[ d - t + x = k d \]
\[ d - k d = t - x \]
\[ d(1 - k) = t - x \]

### Step 6: Substitute known values
Substituting \( t = 4 \times 10^{-5} \) m and \( x = 3.5 \times 10^{-5} \) m:
\[ d(1 - k) = 4 \times 10^{-5} - 3.5 \times 10^{-5} \]
\[ d(1 - k) = 0.5 \times 10^{-5} \]

### Step 7: Express \( d \) in terms of \( k \)
From the previous step, we can express \( d \):
\[ d = \frac{0.5 \times 10^{-5}}{1 - k} \]

### Step 8: Substitute \( d \) back into the equation
Now, we can substitute \( d \) back into the equation:
\[ 0.5 \times 10^{-5} = (1 - k) \cdot \frac{0.5 \times 10^{-5}}{1 - k} \]
This simplifies to:
\[ 1 - k = 0.875 \]

### Step 9: Solve for \( k \)
Now we can solve for \( k \):
\[ 1 - 0.875 = \frac{1}{k} \]
\[ 0.125 = \frac{1}{k} \]
\[ k = \frac{1}{0.125} = 8 \]

### Conclusion
The dielectric constant \( k \) of the material of the slab is \( 8 \).