A parallel plate capacitor is charged and then isolated. What is the effect of increasing the plate separation on charge, potential, capacitance, respectively?
Constant, decreases, decreases
Increases, decreases, decreases
Constant, increases, decreases
Constant, decreases, increases
A
Constant, increases, decreases
B
Constant, decreases, decreases
C
Constant, decreases, increases
D
Increases, decreases, decreases
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Solution
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As the capacitor is isolated after charging, charge on it remains constant. Plate separation increases d, decreases C=ϵ0A/d and hence increases potential V=Q/C.
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