Question

A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10⁻¹² F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

Answer 1

Capacitance between the parallel plates of the capacitor,

C = 8 pF

Initially, the distance between the parallel plates was d, and the space between them was filled with air.

Capacitance C is given by the formula,

`C = (kε_0A)/d`

= `(ε_0A)/d`   

= 8 pF

= 8 × 10⁻¹²     ...(i)

If the distance is reduced to half,

d' = `d/2`

The new dielectric constant of the substance is filled in between the plates, 

k = 6

Let the new capacitance be C'.

`C' = (kε_0A)/(d')`

= `(6ε_0A)/(d/2)`  

= `6 xx 2 xx (ε_0A)/d`      ...[Using equation (i)]

= 12 × 8 × 10⁻¹²  

= 96 × 10⁻¹²

= 96 pF

Therefore, the new capacitance between the plates is 96 pF.