Explanation:
Formula for the capacitance of a parallel plate capacitor is as follows.
C = [tex]\epsilon_{o} \frac{A}{d}[/tex]
where, A = area = 1 [tex]cm^{2}[/tex] = [tex]1 \times 10^{-4} m^{2}[/tex]
d = diameter = 1 mm = [tex]10^{-3} m[/tex]
[tex]\epsilon_{o} = 8.854 \times 10^{-12} f/m[/tex]
Hence, putting the given values into the above formula as follows.
C = [tex]\epsilon_{o} \frac{A}{d}[/tex]
= [tex]8.854 \times 10^{-12} f/m \times \frac{10^{-4}}{10^{-3}}[/tex]
= [tex]8.854 \times 10^{-11} F[/tex]
Now, we will calculate the energy stored in the capacitor as follows.
U = [tex]\frac{1}{2}CV^{2}[/tex]
= [tex]\frac{1}{2} \times 8.854 \times 10^{-11} \times (120)^{2}[/tex]
= [tex]63748.8 \times 10^{-11} J[/tex]
or, = [tex]63.748 \times 10^{-8} J[/tex]
Thus, we can conclude that energy stored on each capacitor is [tex]63.748 \times 10^{-8} J[/tex].
