A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates:

Question

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Answer 1

Capacitance $C = ε \frac{A}{d}$ , $A - a r e a o f t h e c a p a c i t o r; d - d i s t a n c e b e t w e e n t h e p l a t e s$
For a charged capacitor $Q = C V$
As the distance between the plates of the capacitor is increased the capacitance $C$ decreases $(∵ C \propto \frac{1}{d})$
As capacitance decreases the potential difference between the plates $V i n c r e a s e s$

Answer 2

Capacitance

C = ε \frac{A}{d}

,

A - a r e a o f t h e c a p a c i t o r; d - d i s t a n c e b e t w e e n t h e p l a t e s

For a charged capacitor

Q = C V

As the distance between the plates of the capacitor is increased the capacitance

C

decreases

(∵ C \propto \frac{1}{d})

As capacitance decreases the potential difference between the plates

V i n c r e a s e s