One plate of a capacitor of capacitance `2muF` has total charge `+10muC` and its other plate has total charge `+40mUC`. The potential difference between the plates is (in Volt).

Options

A. 7.5
B. 15
C. 25
D. 50

Explanation

To find the potential difference between the plates of the capacitor, we can follow these steps:

### Step 1: Understand the given values
We have a capacitor with:
- Capacitance \( C = 2 \, \mu F = 2 \times 10^{-6} \, F \)
- Charge on the first plate \( Q_1 = +10 \, \mu C = 10 \times 10^{-6} \, C \)
- Charge on the second plate \( Q_2 = +40 \, \mu C = 40 \times 10^{-6} \, C \)

### Step 2: Calculate the net charge on the capacitor
The net charge \( Q \) on the capacitor can be calculated as:
\[ Q = Q_2 - Q_1 = 40 \, \mu C - 10 \, \mu C = 30 \, \mu C = 30 \times 10^{-6} \, C \]

### Step 3: Relate charge, capacitance, and potential difference
The relationship between charge \( Q \), capacitance \( C \), and potential difference \( V \) is given by the formula:
\[ Q = C \cdot V \]
From this, we can express the potential difference \( V \) as:
\[ V = \frac{Q}{C} \]

### Step 4: Substitute the values into the formula
Now, substituting the values of \( Q \) and \( C \) into the equation:
\[ V = \frac{30 \times 10^{-6} \, C}{2 \times 10^{-6} \, F} \]

### Step 5: Calculate the potential difference
Calculating the above expression gives:
\[ V = \frac{30}{2} = 15 \, V \]

### Final Answer
The potential difference between the plates of the capacitor is \( 15 \, V \).

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