A sphere of radius $R$ has a charge $Q$ distributed uniformly over its surface. How large will a sphere that contains $90\ \%$ of energy stored in electrostatic field of this charge distribution.
Since, $$U = {1\over 8\pi}\int_{\text{Entire} \\ \text{Field}} \vec E\cdot \vec E\ \ dv$$
I assume that a sphere with charge only surface would behave like a spherical shell with charge, then $E = 4\pi \sigma$, where $\sigma$ is the surface charge density.
From the integral I got,
$$8\pi\Delta U = \vec E\cdot \vec E \times \Delta v$$
For which I get, $-8\pi(0.1)U = 16\pi^2\sigma^2 \Delta v$.
Everything seems fine here except that I don't the value of $U$. I can't solve two unknown in one equation.
I also tried just to use $\displaystyle U = {Q^2 \over R}$.
$$\begin{align} \Delta U &= -(0.1)U_i \\{Q^2 \over R^\prime} - {Q^2 \over R} &= -(0.1){Q^2 \over R} \\ \frac{10}9 R &= R^\prime \end{align}$$.
The given answer is $10R$.
I think I am missing something very silly. Any hints are welcomed.
I was too with this puzzling problem. We've been assuming that there are two spherical distributions of charge to compare, but no, there is only one and not comparing anything. In a classical language it can be explained so: Consider all imaginary spheres surrounding the material one. By other side, we can think of the field as carrying some amount of energy at every point where it is not zero. With this idea, every region of the space contains some amount of energy and the entire space the total energy of the field. So, each of those spheres contains some amount of the total energy. We want to find the sphere that contains a 90% of the total energy.
First, for a some region, $U =\displaystyle\dfrac{1}{8\pi}\int_{\text{Region}} \vec E\cdot \vec E\ \ dv$.
For the distribution, being $R_0$ the radius of the spheric distribution, $E=\vert\vec E\vert=\dfrac{Q}{r^2}$ and $E=0$ in the sphere's interior.
$\displaystyle U_{R_0}=U_{total}=\dfrac{1}{8\pi}\int_{R_0}^{+\infty}\int_\Omega\vert\vec E\vert^2r^2d\Omega dr=\dfrac{Q^2}{2}\int_{R_0}^{+\infty}\dfrac{dr}{r^2}=$
$\displaystyle U_{R_0}=\dfrac{Q^2}{2}\dfrac{1}{R_0}$
Now, we calculate the energy for the region out of the sphere of radius $R_1$
$U_{R_1}=\dfrac{Q^2}{2}\dfrac{1}{R_1}$
At last,
$U_{R_0}-U_{R_1}=90\%U_{R_0}\implies \dfrac{1}{R_0}-\dfrac{1}{R_1}=\dfrac{90\%}{R_0}\implies R_1=10R_0$