The potentials of the two plates of capacitor are `+10V` and `-10 V`. The charge on one of the plate is `40 C`. The capacitance of the capacitor is

Options

A. `2F`
B. `4F`
C. `0.5 F`
D. `0.25 F`

Explanation

To find the capacitance of the capacitor given the potentials of its plates and the charge on one of the plates, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Potential of one plate, \( V_1 = +10 \, \text{V} \)
- Potential of the other plate, \( V_2 = -10 \, \text{V} \)
- Charge on one plate, \( Q = 40 \, \text{C} \)

2. **Calculate the Potential Difference (\( \Delta V \)):**
\[ \Delta V = V_1 - V_2 = 10 \, \text{V} - (-10 \, \text{V}) = 10 \, \text{V} + 10 \, \text{V} = 20 \, \text{V} \]

3. **Use the Capacitance Formula:**
The relationship between charge (\( Q \)), capacitance (\( C \)), and potential difference (\( \Delta V \)) is given by:
\[ Q = C \Delta V \]
Rearranging this formula to find capacitance:
\[ C = \frac{Q}{\Delta V} \]

4. **Substitute the Values:**
\[ C = \frac{40 \, \text{C}}{20 \, \text{V}} = 2 \, \text{F} \]

5. **Final Result:**
The capacitance of the capacitor is \( 2 \, \text{F} \).