1) [tex]1.11\cdot 10^{-7} J[/tex]
The capacitance of a parallel-plate capacitor is given by:
[tex]C=\frac{\epsilon_0 A}{d}[/tex]
where
[tex]\epsilon_0[/tex] is the vacuum permittivity
A is the area of each plate
d is the distance between the plates
Here, the radius of each plate is
[tex]r=\frac{2.0 cm}{2}=1.0 cm=0.01 m[/tex]
so the area is
[tex]A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2[/tex]
While the separation between the plates is
[tex]d=0.50 mm=5\cdot 10^{-4} m[/tex]
So the capacitance is
[tex]C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F[/tex]
And now we can find the energy stored,which is given by:
[tex]U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J[/tex]
2) 0.71 J/m^3
The magnitude of the electric field is given by
[tex]E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m[/tex]
and the energy density of the electric field is given by
[tex]u=\frac{1}{2}\epsilon_0 E^2[/tex]
and using
[tex]E=4\cdot 10^5 V/m[/tex], we find
[tex]u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3[/tex]
