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Given a charge distribution $\rho(\vec{r})$ where $\vec{r}$ is the position vector and that $\rho$ is a function of only $|x|$, Why is it that the corresponding electric field $E$ is necessarily of the form $(E(x),0,0)$ and $E(x)$ is antisymmetric ?

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Is this a homework or self-study (or similar) problem? If not then we can remove the homework tag. – David Z Feb 12 '12 at 2:47
up vote 2 down vote accepted

This is because of the symmetry of the problem. Using Coulomb's law for each point of the charge distribution (summing over each point)

$$ E(\vec{r}) = \frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\vec{s})(\vec{r}-\vec{s})}{|\vec{r}-\vec{s}|^3} d\vec{s}$$

Since the charge only depends on $|x|$, you can view this as a infinite sheet of charge place at any value of $x$. Thus, if you consider a test charge at $r_1 = (x_1,y_1,z_1)$, the contribution from $y>y_1$ is opposite to that from $y<y_1$, and therefore cancels. The same holds in the $z$ direction. Thus, the components of the field in $y$ and $z$ are zero.

The antisymmetry of $E$, i.e. $E(x) = - E(-x)$, comes from the fact that the charge distribution depends on $|x|$. Indeed, considering 1D, $\rho(-x)\cdot (-x) = -\rho(x) \cdot x = - [\rho(x)\cdot x]$ because of $\rho = f(|x|)$.

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Thanks, Karolos, but isn't $|x|$ symmetric since $|x|=|-x|$? – flux Feb 11 '12 at 15:57
You're right. Sorry for the shortcut. I've edited my answer. – Karolos Feb 11 '12 at 16:11
Thanks again. Is there a way that uses Gauss's law? – flux Feb 11 '12 at 16:23
For this purpose, and many others, both laws are equivalent. Coulomb's law can be obtained from integrating Gauss' law $\nabla\cdot E(\vec{r}) = \rho(\vec{r})/\varepsilon_0$ on both sides. – Karolos Feb 11 '12 at 16:26

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