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 it was filled with air. The dielectric constant of air, k = 1

Capacitance, C, is given by the formula,

`"C" = ("k"in_0"A")/"d"`

= `(in_0"A")/"d"` .......(i)

Where,

A = Area of each plate

`in_0` = Permittivity of free space

If the distance between the plates is reduced to half, then the new distance, d^’ = `"d"/2`

Dielectric constant of the substance filled in between the plates, k' = 6

Hence, the capacitance of the capacitor becomes

`"C'" = ("k'"in_0"A")/"d"`

= `(6in_0"A")/("d"/2)` ........(ii)

Taking ratios of equations (i) and (ii), we obtain

C' = 2 × 6 C

= 12 C

= 12 × 8

= 96 pF

Therefore, the capacitance between the plates is 96 pF.