# What does the charge density signify in the differential form of Gauss law?

Let us say that I have uniformly charged sphere of total charge Q and radius R. The electric field due to this charge distribution at r=R/2 is given by, $$E = \frac{kQ}{2R^2}\hat{r}$$(This was derived from the integral form of Gauss Law). Suppose that I have this expression and I'm required to find the charge distribution, I'll use the differential form, which says that $$\nabla.E = \frac{\rho}{\epsilon_0}$$ $\nabla.E$ will give me $\frac{\delta^3(r)Q}{2\epsilon_0 }$ and this equals $\frac{\rho}{\epsilon_0}$. When I integrate this charge distribution over all space I'll get Q/2 and not Q, that is,$$\iiint\rho dV = \iiint\frac{\delta^3(r) Q}{2} = \frac{Q}{2}$$ So, does the charge density in the differential form of Gauss law correspond to the charge enclosed by the initial Gaussian surface which was used to derive the Electric field? Why is the remaining part of the charge distribution missing? Am i doing things right in the first place? Thanks in advance.

## 1 Answer

Uniform volume charge density $$\rho_{\rm v}=\dfrac{Q}{V}=\dfrac{3Q}{4\pi R^{3}} \tag{01}$$ Gauss Law on a ball of radius $\:r<R$ (spherical symmetry) $$E\cdot 4\pi r^{2}=\dfrac{\rho_{\rm v}}{\epsilon_{0}}\cdot \dfrac{4\pi}{3}r^{3}=\dfrac{Q}{\epsilon_{0}}\cdot \dfrac{r^{3}}{R^{3}} \tag{02}$$ so $$E(r)=\dfrac{1}{4\pi\epsilon_{0}} \dfrac{Q}{R^{3}}r=k \dfrac{Q}{R^{3}}r \tag{03}$$ and in vector form $$\mathbf{E}(\mathbf{r})=\dfrac{1}{4\pi\epsilon_{0}} \dfrac{Q}{R^{3}}\,\mathbf{r} \tag{04}$$ Now $$\boldsymbol{\nabla}\boldsymbol{\cdot}\mathbf{E}=\dfrac{1}{4\pi\epsilon_{0}} \dfrac{Q}{R^{3}}\,\underbrace{\boldsymbol{\nabla}\boldsymbol{\cdot}\mathbf{r}}_{3}=\dfrac{3Q}{4\pi\epsilon_{0} R^{3}}=\dfrac{\rho_{\rm v}}{\epsilon_{0}} \tag{05}$$

• Thank you, found my mistake. The electrostatic field equation should be a function. But in my case, it is the value of the field at a specific point. Feb 26, 2018 at 9:41
• Precisely. That's your mistake. Feb 26, 2018 at 9:42