# A parallel plate capacitor with air between the plates has a capacitance of

Question:

A parallel plate capacitor with air between the plates has a capacitance of $8 \mathrm{pF}\left(1 \mathrm{pF}=10^{-12} \mathrm{~F}\right)$. 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 $6 ?$

Solution:

Capacitance between the parallel plates of the capacitor, C = 8 pF

Initially, distance between the parallel plates was and it was filled with air. Dielectric constant of air, k = 1

Capacitance, C, is given by the formula,

$C=\frac{k \in_{0} A}{d}$

$=\frac{\epsilon_{0} A}{d}$   ...(i)

Where,

A = Area of each plate

$E_{0}=$ Permittivity of free space

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

Dielectric constant of the substance filled in between the plates, $k^{\prime}=6$

Hence, capacitance of the capacitor becomes

$C^{\prime}=\frac{k^{\prime} \in_{0} A}{d}=\frac{6 \in_{0} A}{\frac{d}{2}}$   ...(ii)

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

$C^{\prime}=2 \times 6 C$

$=12 C$

$=12 \times 8=96 \mathrm{pF}$

Therefore, the capacitance between the plates is 96 pF.