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,

${\displaystyle \text{C}=\frac{\text{k}{\in }_0\text{A}}{\text{d}}}$

= ${\displaystyle \frac{{\in }_0\text{A}}{\text{d}}}$ .......(i)

Where,

A = Area of each plate

${\displaystyle {\in }_0}$ = Permittivity of free space

If the distance between the plates is reduced to half, then the new distance, d^’ = ${\displaystyle \frac{\text{d}}{2}}$

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

Hence, the capacitance of the capacitor becomes

${\displaystyle \text{C'}=\frac{\text{k'}{\in }_0\text{A}}{\text{d}}}$

= ${\displaystyle \frac{6{\in }_0\text{A}}{\frac{\text{d}}{2}}}$ ........(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.