The capacitance of a parallel plate capacitor is `2 mu F` and the charge on its positive plate is `2 muC`. If the charge on its plates is doubled, the capacitance of the capacitor

Options

A. remains `2 mu F`
B. becomes `1 mu F`
C. becomes `4 mu F`
D. data insufficient

Answer

To solve the problem, we need to understand the relationship between capacitance, charge, and voltage in a parallel plate capacitor.

### Step-by-Step Solution:

1. **Understanding Capacitance**:
The capacitance (C) of a parallel plate capacitor is defined as:
\[ C = \frac{Q}{V} \]
where \( Q \) is the charge on one plate and \( V \) is the voltage across the plates.

2. **Given Values**:
- Initial capacitance, \( C = 2 \, \mu F \)
- Initial charge on the positive plate, \( Q = 2 \, \mu C \)

3. **Doubling the Charge**:
If the charge on the plates is doubled, the new charge \( Q' \) becomes:
\[ Q' = 2Q = 2 \times 2 \, \mu C = 4 \, \mu C \]

4. **Calculating the New Voltage**:
Using the formula \( V = \frac{Q}{C} \), we can find the initial voltage \( V \) when the charge is \( 2 \, \mu C \):
\[ V = \frac{Q}{C} = \frac{2 \, \mu C}{2 \, \mu F} = 1 \, V \]
Now, with the new charge \( Q' = 4 \, \mu C \), the new voltage \( V' \) becomes:
\[ V' = \frac{Q'}{C} = \frac{4 \, \mu C}{2 \, \mu F} = 2 \, V \]

5. **Understanding Capacitance Independence**:
The capacitance of a capacitor is determined by its physical characteristics (area of the plates, distance between the plates, and the permittivity of the dielectric material between the plates). It does not depend on the charge or voltage. Therefore, even though the charge has doubled, the capacitance remains:
\[ C' = C = 2 \, \mu F \]

### Conclusion:
The capacitance of the capacitor remains unchanged at \( 2 \, \mu F \) even after the charge is doubled.