The capacity of a parallel plate capacitor is 10 μF when the distance between its plates is 9 cm. What will be its capacity if the distance between the plates is reduced by 6 cm.

Answer

To solve the problem of finding the new capacitance of a parallel plate capacitor when the distance between the plates is reduced, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the formula for capacitance**: The capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{\varepsilon_0 A}{D} \]
where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( D \) is the distance between the plates.

2. **Identify the initial conditions**: We know that the initial capacitance \( C_1 \) is 10 μF and the distance between the plates \( D_1 \) is 9 cm.

3. **Calculate the product \( \varepsilon_0 A \)**: From the initial conditions, we can express:
\[ C_1 = \frac{\varepsilon_0 A}{D_1} \]
Substituting the known values:
\[ 10 \, \mu F = \frac{\varepsilon_0 A}{9 \, \text{cm}} \]
Rearranging gives:
\[ \varepsilon_0 A = 10 \, \mu F \times 9 \, \text{cm} = 90 \, \mu F \cdot \text{cm} \]

4. **Determine the new distance**: The distance between the plates is reduced by 6 cm. Therefore, the new distance \( D_2 \) is:
\[ D_2 = D_1 - 6 \, \text{cm} = 9 \, \text{cm} - 6 \, \text{cm} = 3 \, \text{cm} \]

5. **Calculate the new capacitance**: Now we can find the new capacitance \( C_2 \) using the same formula:
\[ C_2 = \frac{\varepsilon_0 A}{D_2} \]
Substituting \( \varepsilon_0 A = 90 \, \mu F \cdot \text{cm} \) and \( D_2 = 3 \, \text{cm} \):
\[ C_2 = \frac{90 \, \mu F \cdot \text{cm}}{3 \, \text{cm}} = 30 \, \mu F \]

6. **Final answer**: The new capacitance when the distance is reduced to 3 cm is:
\[ C_2 = 30 \, \mu F \]