Respuesta :
Answer:
[tex]A=7.72\times 10^{-18}\ m^2[/tex]
Explanation:
Given:
- the voltage of the source, [tex]V=120\ V[/tex]
- distance of separation between the capacitor plates, [tex]d=2\times 10^{-3}\ J[/tex]
- energy of the capacitor, [tex]U=0.25\ J[/tex]
Assuming that there is free space between the plates, so the permittivity, [tex]\epsilon=9\times 10^{9}\ N.m^2.C^{-2}[/tex]
We have the energy stored in a capacitor as:
[tex]U=\frac{1}{2}\times \frac{\epsilon.A.V^2}{d}[/tex]
[tex]0.25=0.5\times \frac{9\times 10^{9}\times A\times 120^2}{2\times 10^{-3}}[/tex]
[tex]A=7.72\times 10^{-18}\ m^2[/tex]
Explanation:
Formula for energy stored in the capacitor is as follows.
E = [tex]\frac{1}{2}CV^{2}[/tex]
or, C = [tex]\frac{2E}{V^{2}}[/tex]
The given data is as follows.
E = 0.25 J, V = 120 V
Putting the given values into the above formula as follows.
C = [tex]\frac{2E}{V^{2}}[/tex]
= [tex]\frac{2 \times 0.25 J}{(120)^{2}}[/tex]
= [tex]3.47 \times 10^{-5}[/tex] F
Also, C = [tex]\frac{\epsilon_{o} A}{d}[/tex]
or, A = [tex]\frac{Cd}{\epsilon_{o}}[/tex]
Putting the given values into the above formula as follows.
A = [tex]\frac{Cd}{\epsilon_{o}}[/tex]
= [tex]\frac{3.47 \times 10^{-5} \times 2 \times 10^{-3}}{8.85 \times 10^{-12}}[/tex]
= 7846.82 [tex]m^{2}[/tex]
Thus, we can conclude that area of the capacitor plates is 7846.82 [tex]m^{2}[/tex].
