Question

(IV)
The capacitance of a parallel plate capacitor with plate separation of 1 mm is 2 pF. If the
plate separation is increased to 2 mm, the capacitance becomes
() 4pF
3pF
1pF
(iv) 0.5pF​

Answer 1

Given :

The capacitance of a parallel capacitor is 2pF when distance b/w both plates is 1mm.

To Find :

Capacitance of parallel plate capacitor when separation is increased to 2mm.

Solution :

❒ Capacitance of a parallel plate capacitor is given by

• C = ε｡ × A / d

where,

>> C denotes capacitance

>> ε｡ denotes permittivity of air

>> A denotes are of plate

>> d denotes distance b/w plates

In this question, Area of plate remains constant so we can say that capacitance of parallel plate capacitor is directly proportional to the distance b/w both plates$.$

Mathematically, C ∝ d

Let initial separation b/w both plates be d and final separation be d'.

➠ C / C' = d / d'

➠ 2 / C' = 1 / 2

➠ C' = 2 × 2

➠ C' = 4 pF

∴ New capacitance of parallel plate capacitor will be 4 pF.

Answer 2

Given :

The capacitance of a parallel capacitor is 2pF when distance b/w both plates is 1mm.

To Find :

Capacitance of parallel plate capacitor when separation is increased to 2mm.

Solution :

Capacitance of a parallel plate capacitor is given by C = ε｡ × A / d

where,

• C denotes capacitance

• ε｡ denotes permittivity of air

• A denotes are of plate

• d denotes distance b/w plates

In this question,

• Area of plate remains constant so we can say that capacitance of parallel plate capacitor is directly proportional to the distance b/w both plates..

➤ C ∝ d

Let initial separation b/w both plates be d and final separation be d'.

➔ C / C' = d / d'

➔ 2 / C' = 1 / 2

➔ C' = 2 × 2

➔ C' = 4 pF

∴ New capacitance of parallel plate capacitor will be 4 pF.