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 1: Understand the given values.**
- The capacitance \( C \) of the capacitor is given as \( 2 \, \mu F \) (microfarads).
- The charge \( Q \) on the positive plate is given as \( 2 \, \mu C \) (microcoulombs).

**Step 2: Recall the formula for capacitance.**
The capacitance \( C \) of a parallel plate capacitor is defined by the formula:
\[ C = \frac{Q}{V} \]
where \( Q \) is the charge on one plate and \( V \) is the voltage across the plates.

**Step 3: Analyze the effect of doubling the charge.**
According to the problem, if the charge on the plates is doubled, the new charge \( Q' \) will be:
\[ Q' = 2 \times Q = 2 \times 2 \, \mu C = 4 \, \mu C \]

**Step 4: Determine the effect on capacitance.**
It is important to note that the capacitance \( C \) of a capacitor is determined by its physical characteristics (area of the plates, distance between the plates, and the dielectric material) and does not depend on the charge stored. Therefore, even if the charge is doubled, the capacitance remains the same.

Thus, the capacitance remains:
\[ C' = C = 2 \, \mu F \]

**Final Answer:**
The capacitance of the capacitor remains \( 2 \, \mu F \).

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