# Differentiating the electric field in Gauss's law, I get zero charge density. Can anyone help me where am I going wrong?

$$\mathbf E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2} \mathbf e_r$$ $$\nabla \cdot\mathbf E = \frac{\rho_V}{\epsilon}$$ \begin{align} \implies \nabla\cdot\mathbf E & = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{Q}{4\pi\epsilon_0r^2}\right) \\ & = \frac{1}{r^2}0 \\ & = 0 = \rho_V/\epsilon \\ \implies \rho_V &= 0 \longrightarrow ? \end{align}

(From the original image)

Where am I going wrong on this? But at origin it is not?

• Is it zero every where, even at origin – The seeker May 26 '17 at 17:48
• Ask yourself the following question: what does $\rho(r)$ represent physically? Given your system of charges, can you guess what $\rho(r)$ should look like? Does it match with your answer? – Philip May 26 '17 at 17:53
• The formula $\nabla\cdot \mathbf F = \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 \frac{\partial}{\partial r} F_r)$ is only valid for $r>0$ (and of course under $SO(3)$ symmetry). – md2perpe May 26 '17 at 17:53
• I have transcribed your image into LaTeX; please use this format in future posts. – Emilio Pisanty May 26 '17 at 17:54
• Possible duplicates: physics.stackexchange.com/q/75557/2451 , physics.stackexchange.com/q/9255/2451 and links therein. – Qmechanic May 26 '17 at 18:03