The capacitance of a parallel plate condenser is `C_(0)` If a dielectric of relative permittivity `epsilon_(r)` and thickness equal to one fourth the plate separation is placed between the plates, then its capacity becomes C. The value of `(C)/(C_(0))` will be -

Options

A. `(5 epsilon_(r))/(4 epsilon_(r)+1)`
B. `(4 epsilon_(r))/(3 epsilon_(r)+1)`
C. `(3 epsilon_(r))/(2 epsilon_(r)+1)`
D. `(2epsilon_(r))/(epsilon_(r)+1)`

Answer

To solve the problem, we need to find the ratio \( \frac{C}{C_0} \) where \( C \) is the capacitance of a parallel plate capacitor with a dielectric inserted, and \( C_0 \) is the original capacitance without the dielectric.

### Step-by-Step Solution:

1. **Identify the Original Capacitance**:
The original capacitance \( C_0 \) of a parallel plate capacitor 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, and \( D \) is the separation between the plates.

**Hint**: Remember that capacitance is dependent on the area of the plates and the distance between them.

2. **Insert the Dielectric**:
A dielectric with relative permittivity \( \epsilon_r \) and thickness \( \frac{D}{4} \) is inserted between the plates. The remaining distance without the dielectric will be \( D - \frac{D}{4} = \frac{3D}{4} \).

3. **Capacitance with Dielectric**:
The capacitor can be viewed as two capacitors in series:
- **Capacitor 1 (with dielectric)**:
\[ C_1 = \frac{\epsilon_r \epsilon_0 A}{\frac{D}{4}} = \frac{4 \epsilon_r \epsilon_0 A}{D} \]
- **Capacitor 2 (without dielectric)**:
\[ C_2 = \frac{\epsilon_0 A}{\frac{3D}{4}} = \frac{4 \epsilon_0 A}{3D} \]

**Hint**: When a dielectric is inserted, the capacitance changes based on the thickness of the dielectric and the remaining air gap.

4. **Calculate the Equivalent Capacitance**:
The total capacitance \( C \) of the two capacitors in series is given by:
\[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the values of \( C_1 \) and \( C_2 \):
\[ \frac{1}{C} = \frac{D}{4 \epsilon_r \epsilon_0 A} + \frac{3D}{4 \epsilon_0 A} \]

**Hint**: When capacitors are in series, the reciprocal of the total capacitance is the sum of the reciprocals of the individual capacitances.

5. **Combine the Terms**:
Finding a common denominator:
\[ \frac{1}{C} = \frac{D}{4 \epsilon_0 A} \left( \frac{1}{\epsilon_r} + 3 \right) \]
Thus,
\[ C = \frac{4 \epsilon_0 A}{D \left( \frac{1}{\epsilon_r} + 3 \right)} \]

6. **Express \( C \) in Terms of \( C_0 \)**:
We know that \( C_0 = \frac{A \epsilon_0}{D} \), so we can express \( C \) as:
\[ C = C_0 \cdot \frac{4 \epsilon_r}{1 + 3 \epsilon_r} \]

7. **Find the Ratio \( \frac{C}{C_0} \)**:
Finally, the ratio \( \frac{C}{C_0} \) becomes:
\[ \frac{C}{C_0} = \frac{4 \epsilon_r}{1 + 3 \epsilon_r} \]

### Final Answer:
\[ \frac{C}{C_0} = \frac{4 \epsilon_r}{1 + 3 \epsilon_r} \]