Answer:[tex]E=2.078\times 10^{-10} J/cm^3[/tex]
Explanation:
Given
Diameter(D)=21.5 cm
distance between plate=1.75 mm
and we know capacitance is
[tex]C=\frac{\epsilon _0A}{d}[/tex]
and we know charge is
Q=CV
and charge density ([tex]\sigma [tex])[tex]=\frac{Q}{A}=\frac{\epsilon _0AV}{Ad}=\frac{\epsilon V}{d}[/tex]
Electric field(E)[tex]=\frac{\sigma }{\epsilon _0}[/tex]
Energy density[tex]=\frac{1}{2}\epsilon _0E^2[/tex]
Energy desity[tex]=\frac{4\pi \epsilon _0}{8\pi }\frac{V^2}{d^2}[/tex]
Energy desity[tex]=\frac{1}{8\pi \cdot 9\times 10^9}\times \left [ \frac{12\times 10^3}{1.75}\right ]^2[/tex]
[tex]E=2.078\times 10^{-10} J/cm^3[/tex]
