Stored Electrical Potential Energy Proof

Let’s imagine a cylindrical capacitor that has radii a and b as in the figure below. I am going to show that half the stored electric potential energy lies within a cylinder whose radius is r =\sqrt{ab} .

cylinder

In this figure the dotted line is the imagined Gaussian surface with length L and radius R , such that a<R<b . I need to create a Gaussian surface so that I can apply Gauss’ Laws. On the surface of the Gaussian surface the charge density is \mu = \frac{1}{2}\epsilon_{0}E where E is the Electric Field at points on the Gaussian Surface. Now if I take the formula for the electric field at any point due to an infinite line of charge with uniform linear charge density \lambda , namely the formula E=\frac{\lambda}{2\pi\epsilon_{0}R} , and plug it into the equation for the charge density on the Gaussian surface, I obtain \mu = \frac{1}{2}\epsilon_{0}(\frac{q}{2\pi\epsilon_{0}RL})^{2} , where \lambda=\frac{q}{L} . Now I will leave that formula for a moment to develop a volume integral that will sweep out the energy between the cylinders.

I will begin by stating that the charge density per unit volume is \mu = \frac{\mu}{V} . This implies that we must think of \mu as a function of the volume which implies that u(r) = \int_{a}^{R} \mu dV . This volume integral will give us the energy in the gaussian cylinder. Now I must find a substitution for the volume because the radius will change as I sweep out energy. This will be the volume of a cylinder which is V=\pi R^{2}L . But, I need this in differential notation because of the changing radius, which gives me dV=2 \pi RL dR . This substitution is important because now the variable of integration and its respective differential are apart of the same volume integral that I will be using. This will become obvious after the next step. Now, I take the energy density formula and the differential volume formula and plug them into the volume integral that I set up to obtain u(r) = \int_{a}^{R}\frac{q^{2}2\pi RL}{8{\pi^2}\epsilon_{0}{R^2}{L^2}}dR . After reducing this integral and factoring out the constants, I obtain u(r)= \frac{q^2}{4\pi\epsilon_{0}L}\int_{a}^{R}\frac{dR}{R} . After applying the fundamental theorem of Calculus, I obtain  the formula u(r) = \frac{q^2}{4\pi\epsilon_{0}L}ln\frac{R}{a} . This is not the total energy for the cylindrical capacitor and is only the energy within the radius of the Gaussian surface. The total energy for the cylinder is the same as this integral, except the limits of integration are different. This integral for the total energy is given by u(r) = \int_{a}^{b}\frac{q^2dR}{4\pi\epsilon_{0}LR} , which gives me \frac{q^2}{4\pi\epsilon_{0}L}ln\frac{b}{a} . What I now have to show is that half of the total stored electrical potential energy exists within a Gaussian radius that is exactly half of the  actual radius of the cylinder. This implies that \frac{ \mu_{R}}{\mu_{b}}=\frac{1}{2} . This now implies that I need to divide the formula for the total stored electrical potential energy into the formula for the stored electric potential energy within the Gaussian radius. This will look like \frac{ln\frac{R}{a}}{ln\frac{b}{a}} =\frac{1}{2} . If I solve for R, I obtain the final solution which is R =\sqrt{ab} . The proof is complete and now no further steps are necessary.