The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60...

Given
We are given the dielectric constant ${\displaystyle K=3.60}$. The dielectric strength is the electric field that the dielectric could tolerate before breakdown is occur and equals ${\displaystyle E=1.60\times {10}^7\frac{V}{m}}$. Also, we are given the capacitance of the capacitor ${\displaystyle C=1.25\times {10}^{-9}F}$ and the potential difference ${\displaystyle V=5500V}$
Required
We are asked to find the minimum area A of the plates.
Solution
The capacitor withstands till the electric field equals ${\displaystyle 1.60\times {10}^7\frac{V}{m}}$, after this a breakdown occurs. This electric field is related to the distance d between the two plates
${\displaystyle E=\frac{V}{d}}$
As shown, the electric field is inversely proportional to the distance d and E represents the maximum electric field between the two plates before the breakdown, therefore the distance d represents the minimum distance between the two electrodes before a breakdown. As we are given $E_{max}$ and V, we can calculate the minimum distance $d_{min}$​​​​​​​ as next
${\displaystyle d_{min}=\frac{V}{E_{max}}}$
${\displaystyle =\frac{5500V}{1.6\times {10}^7\frac{V}{m}}}$
${\displaystyle =3.44\times {10}^{4}m}$
The minmum area is related to the minimum distance as given by equation 24.19 in the next form
${\displaystyle A=\frac{Cd}{K{ϵ}_0}}$
Where ${\displaystyle {ϵ}_0}$, is the electric constant and equals ${\displaystyle 8.854\times {10}^{-12}\frac{C^2}{N}\cdot m^2}$ (See Appendix F) and C is the capacitance of the capcitor in the presence of the dielectric. Now we can plug our values for C, K, d and ${\displaystyle {ϵ}_0}$, into equation (1) to get the minimum area A
${\displaystyle A=\frac{Cd}{K{ϵ}_0}}$
${\displaystyle =\frac{(1.25\times {10}^{-9}F)(3.44\times {10}^{-4}m)}{\left(3.60\right)(8.854\times {10}^{-12}\frac{C^2}{N}\cdot m^2)}}$
${\displaystyle =0.0135m^2}$
${\displaystyle =135c{m}^2}$