A parallel plate capacitor is connected to a battery as shown in figure. Consider two situations :
.
A : Key `K` is kept closed and plates of capacitors are moved apart using insulting handle.
B : Key `K` is opened and plates of capacitors are moved apart using insulting handle. Choose the correct options (s).

Options

A. In (i), Q remains same but C charges.
B. In (ii) V remains same but C charges.
C. In (i) V remains same and hence Q changes.
D. In (ii) both Q and V changes.

Answer

C

Explanation

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.