A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on the capacitance is :

capacitance remains constant

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capacitance decreases

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capacitance increases

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cannot determine

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Solution

The correct option is B capacitance decreases
The capacitance of a parallel plate capacitor is $C = \frac{A ϵ_{0}}{d}$ where $A =$ area of plate and $d =$ separation between plates.
Thus, when d is increased the capacitance will decrease.

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A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on the capacitance is :

The correct option is B capacitance decreasesThe capacitance of a parallel plate capacitor is C=Aϵ0d where A= area of plate and d= separation between plates.Thus, when d is increased the capacitance will decrease.

The correct option is B capacitance decreases
The capacitance of a parallel plate capacitor is $C = \frac{A ϵ_{0}}{d}$ where $A =$ area of plate and $d =$ separation between plates.
Thus, when d is increased the capacitance will decrease.