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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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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