Explanation:
Given data :
Area of the plate = 0.13×0.13
= [tex]16.9[/tex]×[tex]10^{-3}[/tex] m²
Distance between the plates = 0.40 mm = 400×[tex]10^{-6}[/tex] m
The critical strength of the field in case of air is 3.0 MV/m.
Lets find the voltage in case of air as a dielectric.
V = 3×[tex]10^{6}[/tex]×400×[tex]10^{-6}[/tex]
V = 1200 V
Now we find capacitance,
C = εA/d
C = (8.854×[tex]10^{-12}[/tex])(16.9×[tex]10^{-3}[/tex])/(400×[tex]10^{-6}[/tex])
= 374×[tex]10^{-12}[/tex] F = 374pF
Let's find energy,
[tex]E = \frac{1}{2} CV^{2}[/tex]
E = (1/2)(374×[tex]10^{-12}[/tex])(1200)²
E = 269.28 × [tex]10^{-6}[/tex] J
