A parallel plate condenser is connected with the terminals of a battery. The distance between the plates is `6 mm`. If a glass plate (dielectric constant `K = 9)` of `4.5 mm` is introduced between them, then the capacity will become

Options

A. `2` times
B. the same
C. `3` times
D. `4` times

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the initial conditions
We have a parallel plate capacitor with:
- Distance between plates, \( D = 6 \, \text{mm} \)
- Dielectric constant of the glass plate, \( K = 9 \)
- Thickness of the glass plate, \( t = 4.5 \, \text{mm} \)

### Step 2: Calculate the effective distance between the plates after inserting the dielectric
Since the glass plate is inserted between the plates of the capacitor, the effective distance \( D' \) between the plates can be calculated as follows:
- The distance without the glass plate is \( D - t = 6 \, \text{mm} - 4.5 \, \text{mm} = 1.5 \, \text{mm} \)
- The glass plate effectively reduces the distance between the plates. The effective distance contributed by the glass plate is given by:
\[ D_g = \frac{t}{K} = \frac{4.5 \, \text{mm}}{9} = 0.5 \, \text{mm} \]
- Therefore, the new effective distance \( D' \) is:
\[ D' = (D - t) + D_g = 1.5 \, \text{mm} + 0.5 \, \text{mm} = 2.0 \, \text{mm} \]

### Step 3: Calculate the initial and final capacitance
The capacitance \( C \) of a parallel plate capacitor is given by:
\[ C = \frac{\epsilon_0 A}{D} \]
Where \( \epsilon_0 \) is the permittivity of free space and \( A \) is the area of the plates.

- The initial capacitance \( C_1 \) (before inserting the dielectric) is:
\[ C_1 = \frac{\epsilon_0 A}{D} = \frac{\epsilon_0 A}{6 \, \text{mm}} \]

- The final capacitance \( C_2 \) (after inserting the dielectric) is:
\[ C_2 = \frac{\epsilon_0 A}{D'} = \frac{\epsilon_0 A}{2.0 \, \text{mm}} \]

### Step 4: Relate the capacitances
Since capacitance is inversely proportional to the distance between the plates, we can write:
\[ \frac{C_2}{C_1} = \frac{D_1}{D_2} \]
Where \( D_1 = 6 \, \text{mm} \) and \( D_2 = 2.0 \, \text{mm} \):
\[ \frac{C_2}{C_1} = \frac{6 \, \text{mm}}{2.0 \, \text{mm}} = 3 \]
Thus, we find:
\[ C_2 = 3 \times C_1 \]

### Conclusion
The capacitance after the glass plate is introduced becomes three times the initial capacitance.

### Final Answer
The capacity will become \( 3C_1 \).

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