A certain capacitor has rectangular plates $\mathbf{\text{59 cm}}$by $\mathbf{\text{33cm}}$, and the gap width is role="math" localid="1662140429408" $\mathbf{\text{0.27mm}}$. If the gap is filled with a material; whose dielectric constant is $\mathbf{2}\mathbf{.}\mathbf{9}$, what is the capacitance of this capacitor?

The dielectric constant of a medium is defined as the ratio of the capacitance of a capacitor with a dielectric as a medium to its capacitance with a vacuum between its plates.

Area of the rectangular plates of a capacitor,

$\begin{array}{rcl}A& =& \left(56\text{cm}\right)\left(\text{33 cm}\right)\\ & =& 1947{\text{cm}}^2\\ & =& \left(1947{\text{cm}}^2\right)\left(\frac{{10}^{-4}{\text{m}}^2}{1{\text{cm}}^2}\right)\\ & =& 0.1947{\text{m}}^2\end{array}$

The width of the gap between the plates,

$\begin{array}{rcl}s& =& 0.27\text{mm}\\ & =& \left(0.27\text{mm}\right)\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\\ & =& 0.27\times {10}^{-3}\text{m}\end{array}$

Dielectric constant, $k=2.9$

The capacitance of the capacitor is given by relation as below.

$C=\frac{k{\epsilon }_{0}A}{s}$ ..... (1)

Here, $s$is the width of the gap between the plates, $A$ is the area of the capacitor’s plate, $k$is the dielectric constant, and ${\epsilon }_0$is the permittivity of free space having a value $\begin{array}{rcl}& & 8.85\times {10}^{-12}\raisebox{1ex}{${\text{C}}^{\text{2}}$}\!\left/ \!\raisebox{-1ex}{$\text{N}·{\text{m}}^{\text{2}}$}\right.\end{array}$.

Using equation (1), the capacitance of the capacitor can be calculated as,

$\begin{array}{rcl}C& =& \frac{\left(2.9\right)\left(8.85\times {10}^{-12}{\displaystyle \raisebox{1ex}{${\text{C}}^{\text{2}}$}\!\left/ \!\raisebox{-1ex}{$\text{N}·{\text{m}}^{\text{2}}$}\right.}\right)\left(0.1947{\text{m}}^2\right)}{0.27\times {10}^{-3}\text{m}}\\ & =& 1.85\times {10}^{-8}\text{F}\\ & =& 18.5\times {10}^{-9}\text{F}\\ & =& 18.5\text{nF}\end{array}$.

Hence, the capacitance of the capacitor is $18.5\text{nF}$.