Inside a parallel plate capacitor of width 'd' a di-electric plate with dielectric constant k & width 3d/4 is inserted the new capacitance is C'. Before insertion of dielectric the capacitance was `C_0`. Find relation between `C_0 & C' `

Answer

To find the relationship between the initial capacitance \( C_0 \) and the new capacitance \( C' \) after inserting a dielectric plate in a parallel plate capacitor, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Capacitance**:
The initial capacitance \( C_0 \) of a parallel plate capacitor with air (or vacuum) between the plates is given by the formula:
\[ C_0 = \frac{A \epsilon_0}{d} \]
where:
- \( A \) is the area of the plates,
- \( \epsilon_0 \) is the permittivity of free space,
- \( d \) is the separation between the plates.

2. **Insert the Dielectric Plate**:
A dielectric plate with dielectric constant \( k \) and thickness \( t = \frac{3d}{4} \) is inserted between the plates. The remaining air gap will have a thickness of:
\[ d - t = d - \frac{3d}{4} = \frac{d}{4} \]

3. **Calculate the Capacitance with Dielectric**:
The capacitor now consists of two sections in series:
- The section with the dielectric: thickness \( t = \frac{3d}{4} \) and capacitance \( C_1 \):
\[ C_1 = \frac{A \epsilon_0 k}{t} = \frac{A \epsilon_0 k}{\frac{3d}{4}} = \frac{4A \epsilon_0 k}{3d} \]
- The section with air: thickness \( \frac{d}{4} \) and capacitance \( C_2 \):
\[ C_2 = \frac{A \epsilon_0}{\frac{d}{4}} = \frac{4A \epsilon_0}{d} \]

4. **Combine the Capacitances in Series**:
The total capacitance \( C' \) of the capacitor with the dielectric is given by the formula for capacitors in series:
\[ \frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the values of \( C_1 \) and \( C_2 \):
\[ \frac{1}{C'} = \frac{3d}{4A \epsilon_0 k} + \frac{d}{4A \epsilon_0} \]

5. **Find a Common Denominator**:
The common denominator for the right-hand side is \( 4A \epsilon_0 k \):
\[ \frac{1}{C'} = \frac{3d}{4A \epsilon_0 k} + \frac{d \cdot k}{4A \epsilon_0 k} = \frac{3d + dk}{4A \epsilon_0 k} \]
Therefore:
\[ \frac{1}{C'} = \frac{d(3 + k)}{4A \epsilon_0 k} \]

6. **Invert to Find \( C' \)**:
Taking the reciprocal gives:
\[ C' = \frac{4A \epsilon_0 k}{d(3 + k)} \]

7. **Relate \( C' \) to \( C_0 \)**:
Now, substituting \( C_0 = \frac{A \epsilon_0}{d} \) into the equation for \( C' \):
\[ C' = \frac{4k C_0}{3 + k} \]

### Final Relation:
Thus, the relationship between the initial capacitance \( C_0 \) and the new capacitance \( C' \) after inserting the dielectric is:
\[ C' = \frac{4k C_0}{3 + k} \]