# Integration with using Gauss law

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

My solution: I used Gauss theorem, $$\oint \mathbf E \cdot d\mathbf S = \int \nabla \cdot \mathbf E dV = 4\pi \int\rho\ dV = 4\pi a \int r\cos(\mathbf a , \mathbf r)\ dV .$$ 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?

I created a 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 to 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 by a dimension, $[|\mathbf a_{new}|] = \frac{[|\mathbf a_{old}|]}{meter}$), $$d\mathbf E = \frac{4 \pi k RdR}{3}\mathbf a \Rightarrow \mathbf E = \frac{2\pi k R^{2}}{3}\mathbf a.$$

• Two questions: 1) Why do you assume E is constant on the surface of the sphere? 2) Why are you integrating to the surface of the sphere when it is E at the center that you seek? – Robert Miller Sep 4 '12 at 18:38
• Oh yes, I made a mistake. I forgot about a scalar product. – John Taylor Sep 4 '12 at 18:45
• That charge density is interesting. What's the total charge? – James S. Cook Sep 4 '12 at 18:47
• So, this task cannot be solved by using Gauss law? – John Taylor Sep 4 '12 at 18:47
• "...I was reading that as a dot-product :)..." It was a dot product with $\mathbf a = const$ as a vector. Maybe I didn't understand your words. [:(]. – John Taylor Sep 4 '12 at 18:51