Conductor in a Vacuum Chamber

We know that the electric field between two points in space is given by the equation: E = $\frac{1}{4*π*ε0}$ $\frac{q}{(X-Xo)^2}$ If we consider the radius of the sphere to be the distance of effect of the electric field produced by the sphere's electric charge, the equation becomes: $\frac{1}{4*π*ε0}$ $\frac{q}{r^2}$

The energy density in a distance equal to the sphere's radius is given by the equation: u = $\frac{1}{2}$ ε0 E^2 = $\frac{q^2}{32*ε0*π^2*r^4}$

The diferential of electric energy accumulated in a distance of r plus a diferential distance dr from the surface of the sphere is: dU = 4 π r^2 dr u = $\frac{q^2*dr}{8*π*ε0*r^2}$

If we isolate dU and integrate both sides of the equation we will have our final equation: $\frac{q^2}{8*π*ε0*R}$

Notice that the variable of this final equation (R) is the radius of the spherical region of effect of the electric field.

By substituting the values of the electric charge that generates the electric field, and the radius of the sphere we come to the approximate result of 0.575 Joules of energy.