Question

A parallel-plate capacitor with plates of area 720 cm² is charged to a potential difference V and is then disconnected from the voltage source. When the plates are moved 0.3 cm farther apart, the voltage between the plates increases by 100 V.

What is the charge Q on the positive plate of the capacitor?

Answer 1

Given : A = 720 cm² ,( d₂ - d₁) = 0.3 cm , V = 100 V

Constant : ₀ = 8.854 ×10^-12 F/m

Solution :

We know that , capacitance is given by :

For initial position :

For final position :

  

Now arranging as :

= 21.2 ×10^-9 C

Answer : q = 21.2 nC