Question

The capacitance of a parallel plate capacitor is 12 pF. If area of both the plates is doubled and the distance b/w them is reduced to half, the capacity of the capacitor will be: (A) 48 pf (C) 196 pF (B) 124 pF (D) 187 F​

Answer 1

Answer:

The capacity of the capacitor will be 48 pF.

Explanation:

Given:

• c = 12 pF
• d' = d/2
• A' = 2A

Formula to be used:

• c = ε₀A/d

Solution:

=> c = ε₀A/d

=> ε₀A/d = 12 pF ---(i)

New capacity will be,

c' = ε₀A'/d' = ε₀2A/(d/2)

c' = 4ε₀A/d = 4×12

c' = 48 pF

∴ Therefore, The capacity of the capacitor will be 48 pF.