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))((k_(1))/(t_(1))+(k_(2))/(t_(2)))`
D. `(epsilon_(0)Q)/A(k_(1)t_(1)+k_(2)t_(2))`

Explanation

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

### Step 1: Understand the Configuration
We have a parallel plate capacitor with two dielectric slabs:
- The first slab has thickness \( t_1 \) and dielectric constant \( k_1 \).
- The second slab has thickness \( t_2 \) and dielectric constant \( k_2 \).

### Step 2: Model the Capacitor
The two dielectric slabs can be treated as two capacitors in series:
- Capacitor 1 (with dielectric \( k_1 \) and thickness \( t_1 \))
- Capacitor 2 (with dielectric \( k_2 \) and thickness \( t_2 \))

### Step 3: Calculate Individual Capacitances
The capacitance of a capacitor filled with a dielectric 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 distance between the plates.

For the two capacitors:
- For Capacitor 1:
\[ C_1 = \frac{k_1 \epsilon_0 A}{t_1} \]
- For Capacitor 2:
\[ C_2 = \frac{k_2 \epsilon_0 A}{t_2} \]

### Step 4: Find the Equivalent Capacitance
Since the two capacitors are in series, the equivalent capacitance \( C_{eq} \) is given by:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the expressions for \( C_1 \) and \( C_2 \):
\[ \frac{1}{C_{eq}} = \frac{t_1}{k_1 \epsilon_0 A} + \frac{t_2}{k_2 \epsilon_0 A} \]
This simplifies to:
\[ \frac{1}{C_{eq}} = \frac{1}{\epsilon_0 A} \left( \frac{t_1}{k_1} + \frac{t_2}{k_2} \right) \]
Thus, the equivalent capacitance is:
\[ C_{eq} = \frac{\epsilon_0 A}{\frac{t_1}{k_1} + \frac{t_2}{k_2}} \]

### Step 5: Calculate the Potential Difference
The potential difference \( V \) across the capacitor can be expressed using the formula:
\[ V = \frac{Q}{C_{eq}} \]
Substituting for \( C_{eq} \):
\[ V = Q \cdot \frac{\frac{t_1}{k_1} + \frac{t_2}{k_2}}{\epsilon_0 A} \]

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