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Q.
A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge, potential and capacitance respectively are
KCETKCET 2015Electrostatic Potential and Capacitance
Solution:
After separation charge is constant.
Capacity, $C=\frac{\varepsilon_{0}A}{d}$
Capacitance decreases with increase in distance and $V=\frac{Q}{C}$
Potential increases with decrease in capacitance (C).