Correct Answer - C
When key K is kept closed, condenser C is charged to potenital V. When plates of capacitors are moved apart, its capacitance, `C = (K epsi_(0)A)/(d)` decreases.
As potential of condenser remain same, charge Q = CV decreases. So option (c) is correct.
Once key K is closed , condenser gets charged , Q = CV Now if key K is opend ,battery is disconnected no more charging can occur i.e., Q remains same .
As plates of capacitor moved apart , its capacity `C = (K epsi_(0)A)/(d)` decreases.
Therefore , its potenital `V = (q)/(C)` increase.