Consider that the formula for the capacitance of a square parallel plate capacitor is:
[tex]C=\epsilon_o\frac{A}{d}=\epsilon_o\frac{L^2}{d}[/tex]where A=L^2 is the area of each plate, d is the separation between plates and
ε0 is the dielectric permitivity of vacuum ans is equal to 8.82*10^-12 F/m.
If you solve the previous expression for L and replace the given values for d and C, you obtain:
[tex]\begin{gathered} L=\sqrt[]{\frac{dC}{\epsilon_o}} \\ d=0.20mm=0.20\cdot10^{-3}m=2.0\cdot10^{-4}m \\ C=400pF=400\cdot10^{-12}F=4.00\cdot10^{-10}F \\ L=\sqrt[]{\frac{(2.0\cdot10^{-4}m)(4.00\cdot10^{-10}F)}{8.85\cdot10^{-10}\frac{F}{m}}} \\ L\approx0.0095m=0.95cm \end{gathered}[/tex]Hence, the proper value of L to construct the required capacitor is approximately 0.95cm
