The plates in a parallel plates capacitor are seperated by a distance d with air as the medium between the plates. In order to increase the capacity by 66% a dielectric slab of thickness`(3d)/5` is introduced between the plates. What is the dielectric constant of the dielectric slab ?

Options

A. 5
B. 3
C. 1.66
D. 1.5

Explanation

To solve the problem, we need to find the dielectric constant \( k \) of the dielectric slab that increases the capacitance of a parallel plate capacitor by 66% when a dielectric slab of thickness \( \frac{3d}{5} \) is introduced between the plates.

### Step-by-Step Solution:

1. **Initial Capacitance Calculation:**
The initial capacitance \( C \) of a parallel plate capacitor with air as the medium is given by:
\[ C = \frac{\varepsilon_0 A}{d} \]
where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation between the plates.

2. **Capacitance After Introducing Dielectric:**
When a dielectric slab of thickness \( t = \frac{3d}{5} \) is introduced, the remaining gap between the plates is:
\[ d - t = d - \frac{3d}{5} = \frac{2d}{5} \]
The capacitance \( C' \) with the dielectric slab can be calculated using the formula:
\[ C' = \frac{\varepsilon_0 A}{d - t} + \frac{\varepsilon_0 A t}{k} \]
Substituting the values:
\[ C' = \frac{\varepsilon_0 A}{\frac{2d}{5}} + \frac{\varepsilon_0 A \cdot \frac{3d}{5}}{k} \]

3. **Simplifying the Capacitance Expression:**
This simplifies to:
\[ C' = \frac{5\varepsilon_0 A}{2d} + \frac{3\varepsilon_0 A}{5k} \]

4. **Setting Up the Relationship Between Capacitances:**
We know that the capacitance increases by 66%, so:
\[ C' = C + 0.66C = 1.66C \]
Therefore:
\[ C' = 1.66 \cdot \frac{\varepsilon_0 A}{d} \]

5. **Equating the Two Expressions for Capacitance:**
Now we equate the two expressions for \( C' \):
\[ \frac{5\varepsilon_0 A}{2d} + \frac{3\varepsilon_0 A}{5k} = 1.66 \cdot \frac{\varepsilon_0 A}{d} \]

6. **Dividing Through by \( \varepsilon_0 A \):**
Cancel \( \varepsilon_0 A \) from both sides:
\[ \frac{5}{2d} + \frac{3}{5k} = \frac{1.66}{d} \]

7. **Multiplying Through by \( 10kd \) to Eliminate Denominators:**
\[ 25k + 6d = 16.6k \]

8. **Rearranging the Equation:**
Rearranging gives:
\[ 25k - 16.6k = -6d \]
\[ 8.4k = 6d \]
\[ k = \frac{6d}{8.4} = \frac{60}{84} = \frac{5}{7} \approx 1.5 \]

9. **Final Result:**
The dielectric constant \( k \) of the dielectric slab is approximately \( 1.5 \).