# Solving by using Gauss law [closed]

Task: find the vector $\mathbf E$ in the center of the sphere with radius $R$, which has charge volume distribution $\rho$ , $$\rho = \mathbf a \cdot \mathbf r ,\qquad \mathbf a = \operatorname{const},$$ $\mathbf r$ = radius-vector from the center of the sphere.

I solved that task by routinely method. But can it be solved with using Gauss law?

My attempt to solve that task by using Gauss law. In first, I wrote, that $$\oint \mathbf E \cdot d\mathbf S = \int \nabla \cdot \mathbf E\ dV = 4\pi \int \rho dV = 4\pi a \int r cos(\mathbf a , \mathbf r) dV \qquad (.1).$$ Unfortunately, I don't know how to find $\mathbf E$ using written above: $|\mathbf E| \neq const$ into the surface of the sphere, so I can't write something like $$\oint \mathbf E \cdot d \mathbf S = E \int dS = ...$$ and equate it to (1).

Can you help me?

My routinely solution.

In the beginning

1. I can consider only surface of the sphere with surface denstity $\rho (S) = (\mathbf a \cdot \mathbf r)$.

2. I "moved" $\mathbf a$ to the center of the sphere by parallel transfer (look at the picture 1 below).

So, I can divide the sphere into thin rings with charge $$dQ = \rho (S)dS = \rho (S) 2\pi r Rd\varphi$$ (R - length from the ring to the center of the sphere, a - length from the ring plane to the center of the sphere, $\varphi$ - an angle between $\mathbf a , \mathbf R$, and $\mathbf r = \sqrt{R^{2} - a^{2}})$. Using that, $d|\mathbf E|$ creating by this ring in the center of the sphere (using the formula for the ring E = \frac{kaQ}{R}), can be write as $$dE = \frac{kadQ}{R^{3}} = \frac{ak\rho (S) 2\pi r R d\varphi}{R^{3}}.$$ According to the spherical symmetry, $$d\mathbf E_{center} = -\frac{\mathbf a}{a}dE,$$ so, according, that $$r = Rsin(\varphi ), \quad a = Rcos(\varphi ), \quad \rho (S) = (\mathbf a \cdot \mathbf r) = aRcos(\varphi ),$$ I get $$d\mathbf E = -\frac{k R cos(\varphi) R cos(\varphi ) 2\pi R sin(\varphi) R d\varphi}{R^{3}}\mathbf a = -2\pi k R cos^{2}(\varphi )sin(\varphi)\mathbf a \Rightarrow$$

$$\Rightarrow \mathbf E = 2\pi k \mathbf a \int \limits_{-1}^{1}cos^{2}(\varphi )d(cos(\varphi )) = \frac{4 \pi k R \mathbf a}{3}.$$ By the next step I accept, that R is a variable (takes into account the volume distribution of charge), so, with the new vector $\mathbf a$ (formally, it's changed only in a measure, $[|\mathbf a_{new}|] = \frac{[|\mathbf a_{old}|]}{meter}$), $$d\mathbf E = \frac{4 \pi k RdR}{3}\mathbf a \Rightarrow \mathbf E = \frac{2\pi k R^{2}}{3}\mathbf a.$$

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## closed as off-topic by Colin McFaul, Brandon Enright, Kyle Oman, Chris White, JamalSJun 17 '14 at 13:34

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This can't be solved by Gauss's law alone, of course, since it isn't symmetric by a rotation group, but I assume you mean solve it by Gauss's law and superposition. You can find the entire field this way, not just at the center.

Without loss of generality, choose the vector a in the y direction, and give a the magnitude $2\epsilon$ (the general solution is by rescaling), then the charge distribution is:

$$\rho(x,y,z) = 2\epsilon y = (x^2 + (y+\epsilon)^2 + z^2) - (x^2 + y^2 + z^2)$$

So the charge distribution is a superposition of two spherically symmetric shells displaced infinitesimally in the y-direction. The solution for a $\rho(r)=r^2$ is

$$E(r) = {1\over 5} r^3$$

By Gauss's law, and from this, you find the electric field components

$$E_x = {1\over 5} r^2 x$$ $$E_y = {1\over 5} r^2 y$$ $$E_z = {1\over 5} r^2 z$$

This is not the answer, because this is not quite the charge density you have. The sphere stops at radius R, and at this point, the superposition gives a spherical shell around the whole thing, which is dipole-polarized, and is absent in your problem.

Taking the y-derivative of the E-field above gives you the field of the charge density $2\epsilon y$ plus shell. You want just the charge density $\epsilon y$, no shell. But you see from taking the y-derivative of the field (superposing the two infinitesimally separated spheres) that it is zero at the origin, so the field of the charge density you have is equal and opposite to the field of the shell.

The field due to this spherical shell is found by superposing two uniform spheres of density $\rho = R^2$ displaced by $\epsilon$. The E inside a uniform sphere is

$$E_x = {R^2 \over 3} x$$ $$E_y = {R^2 \over 3} y$$ $$E_z = {R^2 \over 3} z$$

So that, superposing the two spheres displaced by $\epsilon$ (taking the y-derivative times $\epsilon$), you get the field from the shell

$$E_y = \epsilon {R^2\over 3}$$

And this is the answer, so that for the charge density coefficient of magnitude a (instead of $2\epsilon$), you find the field is

$$|E| = {a R^2 \over 6}$$

You multiply by $4\pi k$ to get your units and your answer.

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"...then the charge distribution is:..." How the term $e^{2}$ disappears? –  user8817 Sep 10 '12 at 14:23
"...The solution for a $\rho (r) = r^2$ is..." And how was earned that solution? –  user8817 Sep 10 '12 at 14:34
@Maxim_Ovchinnikov: For the first question, $\epsilon$ is infinitesimal as the name indicates, and for the second question, it's Gauss's law, $4\pi r^2 E(R) = \int_0^R 4\pi r^2 r^2 dr$. –  Ron Maimon Sep 10 '12 at 17:55
"...The E inside a uniform sphere is..." And how was earned that expression? Sorry for stupid questions. –  user8817 Sep 11 '12 at 17:59