Question

A certain has rectangular plates$\mathbf{56}\mathbf{\text{cm}}$by $\mathbf{\text{24 cm}}$ and the gap width is $\mathbf{\text{20.0 mm}}$. What is its capacitance? We see that typical capacitances are very small when measured in farads. A role="math" localid="1662139654139" $\mathbf{\text{1F}}$capacitor is quite extraordinary. Apparently it has a very large area $\mathit{A}$(all wrapped up in a small package), and a vary small gap $\mathit{s}$.

Short answer

The capacitance is $6.0\text{nF}$.

Step-by-step solution

A concept:

A parallel plate capacitor is an arrangement of two metal plates connected in parallel and separated from each other by a certain distance. The gap between the plates is occupied by a dielectric medium.

Given data:

Area of the rectangular plates of a capacitor,

$$\begin{gathered} A=\left(56cm\right)\left(24cm\right) \\ =1344c{m}^2 \\ =\left(1344c{m}^2\right)\left(\frac{{10}^{-4}{m}^2}{1c{m}^2}\right) \\ =0.1344m^2 \end{gathered}$$

The width of the gap between the plates,

$\begin{array}{rcl}s& =& 0.20\text{mm}\\ & =& \left(0.20\text{mm}\right)\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\\ & =& 0.20\times {10}^{-3}\text{m}\end{array}$

Define capacitance:

The capacitance of a parallel plate capacitor given by a relation,

$C=\frac{{\epsilon }_{0}A}{s}$

Here, $s$is the width of the gap between the plates, $A$is the area of the capacitor’s plate, and ${\epsilon }_0$is the permittivity of free space having a value role="math" localid="1662140056064" $8.85\times {10}^{-12}\raisebox{1ex}{${\text{C}}^{\text{2}}$}\!\left/ \!\raisebox{-1ex}{$\text{N}·{\text{m}}^{\text{2}}$}\right.$.

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

role="math" localid="1662140204207" $\begin{array}{rcl}C& =& \frac{\left(8.85\times {10}^{-12}{\displaystyle \raisebox{1ex}{${\text{C}}^2$}\!\left/ \!\raisebox{-1ex}{$\text{N}·{\text{m}}^{\text{2}}$}\right.}\right)\left(\left(0.1344{\text{m}}^2\right)\right)}{0.20\times {10}^{-3}\text{m}}\\ & =& 5.947\times {10}^{-9}\text{F}\\ & =& 6.0\text{nF}\end{array}$

Hence, the required capacitance is $\begin{array}{rcl}& & 6.0\text{nF}\end{array}$.