KCET 2015 Physics Solved Paper

Given, parallel plate is charged and then isolated. Therefore,Q = constant.
Now, capacitance is given as C =

ε0A

d

⇒C∝

1

d

So, if the separation between plates is increased then, C decreases.
Also, we know Q=CV⇒V=

Q

C

⇒V∝

1

C

So, if C decreases then potential increases.
Thus, on increasing the plate separation charge remains constant, potential increases and capacitance decreases.