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 a capacitor, we can use the relationship between charge, capacitance, and potential difference. Here’s a step-by-step solution:

### Step 1: Understand the given values
- Capacitance (C) = 2 µF (microfarads) = 2 x 10^-6 F
- Charge on one plate (Q1) = +10 µC (microcoulombs) = 10 x 10^-6 C
- Charge on the other plate (Q2) = +40 µC (microcoulombs) = 40 x 10^-6 C

### Step 2: Calculate the net charge difference
The potential difference (V) between the plates of a capacitor is related to the difference in charge on the plates. We can calculate the difference in charge (ΔQ):
\[ \Delta Q = Q2 - Q1 = 40 \, \mu C - 10 \, \mu C = 30 \, \mu C = 30 \times 10^{-6} C \]

### Step 3: Use the formula for potential difference
The potential difference (V) across a capacitor is given by the formula:
\[ V = \frac{\Delta Q}{C} \]
Substituting the values we have:
\[ V = \frac{30 \times 10^{-6} C}{2 \times 10^{-6} F} \]

### Step 4: Perform the calculation
\[ V = \frac{30}{2} = 15 \, V \]

### Final Answer
The potential difference between the plates is **15 Volts**.

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