Energy stored

This is one of the straightforward problem but it a little bit brute force.So I will skip some easy understand calculation.

Let start with finding the electric field.We'll find the field in two region one is the region in side a spherical shell of radius R and the other is the out side region.

$\displaystyle \vec{E} ; r<R$

$\displaystyle \vec{E_{tot}}=\vec{E_{+}}+\vec{E_{-}}$ and $\vec{E_{+}}=\dfrac{\rho \vec{r_{+}}}{3\epsilon_{0}} and \vec{E_{-}}=-\dfrac{\rho \vec{r_{-}}}{3\epsilon_{0}}$ where position vector point from center of each sphere to field point.

$\displaystyle \vec{E_{tot}}=\frac{\rho \vec{X}}{3\epsilon_{0}}$

Since the separation is very small compare to radius so we can treat our overlap cloud into one spherical cloud.

the energy inside this sphere is then

$\displaystyle E=\int \frac{1}{2}\epsilon_{0} \vec{E}^{2} dV=\frac{\epsilon_{0}}{2} \times \frac{\rho^{2}X^{2}}{9 \epsilon_{0}^{2} } \times \frac{4}{3} \pi R^{3}$

$\displaystyle E=\frac{2\rho^{2}R^{3}X^{2} \pi}{27 \epsilon_{0}}$

Now let's get outside the shell.

when we are outside the sphere we can treat our system as a dipole system without any violation of the original situation.

First we have to find the field from dipole and I have use polar coordinate and gradient in polar form to solve this.I'll skip how I get the field but this is the result.

$\displaystyle \vec{E_{out}}=\frac{p}{4\pi \epsilon_{0}r^{3}} \left ( 2\cos(\theta)\hat{r}+\sin(\theta)\hat{\theta} \right )$

Then the energy is

$\displaystyle \frac{\epsilon_{0}}{2} \left ( \iiint \vec{E}^{2} r^{2} \sin(\theta) d\phi d\theta dr \right )=\frac{p^{2}}{16\pi \epsilon_{0}} \left ( \int_{R}^{\infty}\frac{1}{r^{4}} dr \cdot \int_{\theta=0}^{\pi} (3\cos^{2}(\theta)+1 )d\cos(\theta) \right )$

$\displaystyle =\frac{p^{2}}{12\pi \epsilon_{0}R^{3}}=\frac{4\rho^{2} R^{3}X^{2}\pi}{27 \epsilon_{0}}$

Combine this two energy we get

$\displaystyle E=\frac{2\rho^{2}R^{3}X^{2} \pi}{9 \epsilon_{0}}$

$\displaystyle 2+2+3+2+9=\boxed{18}$

Quite an interesting problem.So we make use of spherical polar coordinates to solve.1)For the part inside the overlapping region.We could simply conclude that the field here is constant.This infact can be proved by superposing the fields due to the individual spheres.So as the file is constant we simple have $U(1)=(eE^2/2)(4/3)πR^3.2)$ .For an external point we could simply use a dipole approximation.since to calculate the field at an external point we may assume the charge at the centre and these are separated by small distances dipole approximation holds good.The vol can be written quite easily in spherical polar coordinates.Once its done we have to use $dU=dV(1/2)eE^2$ .Then integrate by setting limits of the polar angle from $0-π$ ,Azimuthal angle from $0-2π$ and radial distance fro R-infinite.This comes to be $U(2)=(q^2x^2/12πeR^3)$ and adding with $U(1)$ and expressing the charge in terms of vol. Density we get the and.Note that $e$ here refers to permittivity of vaccum.

Easy. E= [p(4/3)π(x/2)³]/4€π(x/2)²= px/3€ Therefore dU/dV=1/2( €E²) Substituting E=px/3€ Since feild is uniform U=1/2 €[px/3€]²(4/3)πR³ since x<<R=> r~R [2p²x²πR³/9€] Hence 2+2+2+3+9=18