A capacitor has a capacitance of 27.0 microfarads. If we triple the area of the plates of the capacitor and cut the distance between the plates to 1/3 of its original value, what is the new capacitance of the capacitor?

243 microfarads

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9.0 microfarads

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3.0 microfarads

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81.0 microfarads

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27.0 microfarads

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Solution

The correct option is A 243 microfarads
Capacitance of parallel plate capacitor $C = \frac{A ϵ_{o}}{d} = 27.0$ microfarads
New area and the distance between the plates $A^{'} = 3 A$ and $d^{'} = d / 3$

Thus new capacitance of the capacitor $C^{'} = \frac{(3 A) ϵ_{o}}{d / 3} = 9 \frac{A ϵ_{o}}{d} = 9 \times 27 = 243$ microfarads

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A capacitor has a capacitance of 27.0 microfarads. If we triple the area of the plates of the capacitor and cut the distance between the plates to 1/3 of its original value, what is the new capacitance of the capacitor?

The correct option is A 243 microfaradsCapacitance of parallel plate capacitor C=Aϵod=27.0 microfaradsNew area and the distance between the plates A′=3A and d′=d/3 Thus new capacitance of the capacitor C′=(3A)ϵod/3=9Aϵod=9×27=243 microfarads