Calculate change in radius for some change in electrostatic potential energy.

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.

• Can you clarify the initial question a bit more? I'm assuming we're trying to solve for a sphere radius, but what the parameters are to figure that out are fuzzy to me: "contains $90\%$ of energy". I don't know what that means. Are we envisioning another sphere of charge placed somewhere around this first sphere and considering the potential energy between them? – WB-man Apr 27 '17 at 17:53
• @WB-man There is a shell kept somewhere. It has some electric potential energy associated to it. Now we need to find some other shell which has same amount charge distributed over it but has 90% the energy that original sphere has. There is no interaction between those two shells. – A---B Apr 27 '17 at 18:05

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,

$U_{R_0}-U_{R_1}=90\%U_{R_0}\implies \dfrac{1}{R_0}-\dfrac{1}{R_1}=\dfrac{90\%}{R_0}\implies R_1=10R_0$

• $$\dfrac{1}{8\pi}\int_{R_0}^{+\infty}\int_\Omega\vert E\vert^2r^2d\Omega dr$$ What is omega here ? I am not similiar with this notation. Is it surface charge density ? – A---B Apr 28 '17 at 5:15
• It's the solid angle $d\Omega=\sin\theta d\theta d\phi$, $\int_\Omega d\Omega=4\pi$ – Rafa Budría Apr 28 '17 at 5:17
• You converted it to spherical coordinates ? – A---B Apr 28 '17 at 5:18
• Yes. The angular part is a constant and we can check the radius directly. – Rafa Budría Apr 28 '17 at 5:19
• $$\dfrac{1}{8\pi}\int_{R_0}^{+\infty}\color{red}{\int_\Omega\vert E\vert^2r^2d\Omega }dr$$ One last thing. As I understood, the first integral (red one) calculates the energy over the surface and the second integral calculates the energy over the whole space ? – A---B Apr 28 '17 at 5:23