Answer :
Answer:
(A) Total energy will be equal to [tex]0.044\times 10^{-5}J[/tex]
(b) Energy density will be equal to [tex]0.0175J/m^3[/tex]
Explanation:
We have given diameter of the plate d = 2 cm = 0.02 m
So area of the plate [tex]A=\pi r^2=3.14\times 0.02^2=0.001256m^2[/tex]
Distance between the plates d = 0.50 mm = [tex]0.50\times 10^{-3}m[/tex]
Permitivity of free space [tex]\epsilon _0=8.85\times 10^{-12}F/m[/tex]
Potential difference V =200 volt
Capacitance between the plate is equal to [tex]C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-12}\times 0.001256}{0.50\times 10^{-3}}=0.022\times 10^{-9}F[/tex]
(a) Total energy stored in the capacitor is equal to
[tex]E=\frac{1}{2}CV^2[/tex]
[tex]E=\frac{1}{2}\times 0.022\times 10^{-9}\times 200^2=0.044\times 10^{-5}J[/tex]
(b) Volume will be equal to [tex]V=Ad[/tex], here A is area and d is distance between plates
[tex]V=0.001256\times 0.02=2.512\times 10^{-5}m^3[/tex]
So energy density [tex]=\frac{Energy}{volume}=\frac{0.044\times 10^{-5}}{2.512\times 10^{-5}}=0.0175J/m^3[/tex]