The potentials of the two plates of capacitor are + 10V and -10 V. The charge on one of the plates is 40 C. The capacitance of the capacitor is
[AFMC 2005]

2 F

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4 F

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0.5 F

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0.25 F

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Solution

The correct option is A 2 F
Potential difference between the plates
= 10 - (-10)
= 20 V

Q = 40 C
Q = CV
$\Rightarrow C = \frac{Q}{V} = \frac{40}{20} = 2 F$
C = 2F

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The potentials of the two plates of capacitor are + 10V and -10 V. The charge on one of the plates is 40 C. The capacitance of the capacitor is [AFMC 2005]

The correct option is A 2 FPotential difference between the plates = 10 - (-10) = 20 V Q = 40 C Q = CV ⇒C=QV=4020=2F C = 2F

The potentials of the two plates of capacitor are + 10V and -10 V. The charge on one of the plates is 40 C. The capacitance of the capacitor is
[AFMC 2005]

The correct option is A 2 F
Potential difference between the plates
= 10 - (-10)
= 20 V

Q = 40 C
Q = CV
$\Rightarrow C = \frac{Q}{V} = \frac{40}{20} = 2 F$
C = 2F