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The potentials of the two plates of capa...

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

A

2 F

B

4 F

C

0.5 F

D

0.25 F

Text Solution

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The correct Answer is:
To find the capacitance of the capacitor, we can use the formula: \[ C = \frac{Q}{V} \] where: - \( C \) is the capacitance, - \( Q \) is the charge on one of the plates, - \( V \) is the potential difference between the two plates. ### Step 1: Identify the potential difference The potential of one plate is \( +10V \) and the potential of the other plate is \( -10V \). The potential difference \( V \) between the two plates can be calculated as: \[ V = V_1 - V_2 = 10V - (-10V) = 10V + 10V = 20V \] ### Step 2: Identify the charge on the plates We are given that the charge \( Q \) on one of the plates is \( 40C \). ### Step 3: Substitute values into the capacitance formula Now we can substitute the values of \( Q \) and \( V \) into the capacitance formula: \[ C = \frac{Q}{V} = \frac{40C}{20V} \] ### Step 4: Calculate the capacitance Now performing the calculation: \[ C = \frac{40}{20} = 2F \] Thus, the capacitance of the capacitor is \( 2 \, \text{Farads} \). ### Final Answer The capacitance of the capacitor is \( 2 \, \text{F} \). ---
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