The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively `C_(0)` and `W_(0)`. If the air is replaced by glass (dielectric constant `= 5`) between the plates, the capacity of the plates and the energy stored in it will respectively be

Options

A. `5C_(0),5W_(0)`
B. `5C_(0),(W_(0))/(5)`
C. `(C_(0))/(5),5W_(0)`
D. `(C_(0))/(5),(W_(0))/(5)`

Answer

To solve the problem, we need to determine the new capacitance and energy stored in a parallel plate capacitor when the dielectric medium is changed from air to glass.

### Step-by-Step Solution:

1. **Understand the Initial Conditions:**
- The initial capacitance with air as the dielectric is given as \( C_0 \).
- The initial energy stored in the capacitor is given as \( W_0 \).

2. **Capacitance with Dielectric:**
- The capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{\epsilon_0 A}{d} \]
where \( \epsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the distance between the plates.
- When a dielectric material (in this case, glass with a dielectric constant \( K = 5 \)) is introduced, the capacitance becomes:
\[ C = K \cdot C_0 \]
- Substituting the value of \( K \):
\[ C = 5 \cdot C_0 \]

3. **Energy Stored in the Capacitor:**
- The energy stored in a capacitor is given by:
\[ W = \frac{1}{2} \frac{Q^2}{C} \]
where \( Q \) is the charge on the capacitor.
- The charge \( Q \) remains the same when the dielectric is changed, so we can express the new energy \( W \) when the dielectric is glass:
\[ W = \frac{1}{2} \frac{Q^2}{5C_0} \]
- Since \( W_0 = \frac{1}{2} \frac{Q^2}{C_0} \), we can relate the new energy to the initial energy:
\[ W = \frac{W_0}{5} \]

4. **Final Results:**
- The new capacitance with glass as the dielectric is:
\[ C = 5C_0 \]
- The new energy stored in the capacitor is:
\[ W = \frac{W_0}{5} \]

### Conclusion:
The capacity and the energy stored in the parallel plate capacitor with glass between its plates are respectively \( 5C_0 \) and \( \frac{W_0}{5} \).