Between the plates of a parallel plate condenser, a plate of thickness `t_(1)` and dielectric constant `k_(1)` is placed. In the rest of the space, there is another plate of thickness `t_(2)` and dielectric constant `k_(2)`. The potential difference across the condenser will be

Options

A. `(Q)/(Aepsilon_(0))(t_(1)/k_(1)+t_(2)/(k_(2)))`
B. `(epsilon_(0)Q)/(A)(t_(1)/k_(1)+t_(2)/(k_(2)))`
C. `(Q)/(Aepsilon_(0))"(t_(1)/k_(1)+t_(2)/(k_(2)))`
D. `(epsilon_(0)Q)/(A)(k_(1)t_(1)+k_(2)t_(2))`

Answer

To find the potential difference across a parallel plate capacitor with two dielectric materials, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Configuration**:
We have a parallel plate capacitor with two different dielectric materials. The first dielectric has thickness \( t_1 \) and dielectric constant \( k_1 \), and the second dielectric has thickness \( t_2 \) and dielectric constant \( k_2 \).

2. **Capacitance Calculation**:
The capacitance of a capacitor filled with a dielectric material is given by:
\[ C = \frac{k \epsilon_0 A}{d} \]
where \( k \) is the dielectric constant, \( \epsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation between the plates.

3. **Calculate Individual Capacitances**:
- For the first dielectric (thickness \( t_1 \)):
\[ C_1 = \frac{k_1 \epsilon_0 A}{t_1} \]
- For the second dielectric (thickness \( t_2 \)):
\[ C_2 = \frac{k_2 \epsilon_0 A}{t_2} \]

4. **Combine Capacitances in Series**:
Since the two dielectrics are in series, the equivalent capacitance \( C_{eq} \) can be calculated using the formula for capacitors in series:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the values of \( C_1 \) and \( C_2 \):
\[ \frac{1}{C_{eq}} = \frac{t_1}{k_1 \epsilon_0 A} + \frac{t_2}{k_2 \epsilon_0 A} \]

5. **Simplify the Equation**:
Factoring out \( \epsilon_0 A \):
\[ \frac{1}{C_{eq}} = \frac{t_1 k_2 + t_2 k_1}{k_1 k_2 \epsilon_0 A} \]
Therefore, the equivalent capacitance \( C_{eq} \) is:
\[ C_{eq} = \frac{k_1 k_2 \epsilon_0 A}{t_1 k_2 + t_2 k_1} \]

6. **Calculate the Potential Difference**:
The potential difference \( V \) across the capacitor is given by:
\[ V = \frac{Q}{C_{eq}} \]
Substituting \( C_{eq} \):
\[ V = \frac{Q (t_1 k_2 + t_2 k_1)}{k_1 k_2 \epsilon_0 A} \]

### Final Expression:
Thus, the potential difference across the capacitor is:
\[ V = \frac{Q (t_1 k_2 + t_2 k_1)}{k_1 k_2 \epsilon_0 A} \]