A capacitor is connected to a battery of voltage V. Now a di-electric slab of di-electric constant k is completely inserted between the plates, then the final charge on the capacitor will be : (If initial charge is `q_(0)`)

Options

A. `(epsi_(0)A)/(d) V`
B. `(kepsi_(0)A)/(d) V`
C. `(epsi_(0)A)/(kd) A`
D. zero

Explanation

To solve the problem step by step, we will analyze the situation of a capacitor connected to a battery and how the insertion of a dielectric slab affects the charge on the capacitor.

### Step-by-Step Solution:

1. **Understanding Initial Conditions**:
- The capacitor is connected to a battery of voltage \( V \).
- The initial charge on the capacitor is given as \( q_0 \).

2. **Initial Charge Calculation**:
- The capacitance \( C_0 \) of the capacitor without the dielectric can be expressed as:
\[ C_0 = \frac{\varepsilon_0 A}{d} \]
where \( A \) is the area of the plates and \( d \) is the separation between them.
- The initial charge \( q_0 \) on the capacitor is given by:
\[ q_0 = C_0 \cdot V = \frac{\varepsilon_0 A}{d} \cdot V \]

3. **Effect of Inserting the Dielectric**:
- When a dielectric slab of dielectric constant \( k \) is inserted between the plates, the capacitance increases. The new capacitance \( C' \) becomes:
\[ C' = k \cdot C_0 = k \cdot \frac{\varepsilon_0 A}{d} \]

4. **Final Charge Calculation**:
- Since the capacitor remains connected to the battery, the voltage across the capacitor remains constant at \( V \). Therefore, the final charge \( q \) on the capacitor after the dielectric is inserted is:
\[ q = C' \cdot V = \left(k \cdot \frac{\varepsilon_0 A}{d}\right) \cdot V \]
- This simplifies to:
\[ q = k \cdot \left(\frac{\varepsilon_0 A}{d} \cdot V\right) = k \cdot q_0 \]

5. **Conclusion**:
- The final charge on the capacitor after the dielectric slab is fully inserted is:
\[ q = k \cdot q_0 \]

### Final Answer:
The final charge on the capacitor after inserting the dielectric slab is \( k \cdot q_0 \).

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