A) Increases, decreases, decreases
B) Constant, increases, decreases
C) Constant, decreases, decreases
D) Constant, decreases, increases
Correct Answer: B
Solution :
: As the capacitor is isolated after charging, charge Q on it remains constant. Plate separation d increases, capacitance decreases as \[C=\frac{{{\varepsilon }_{0}}A}{d}\] and hence, potential increases as \[V=\frac{Q}{C}\].
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