# Why is Gauss' Law for Electric Fields still applicable when $\vec{E}=kr^3\hat{r}$?

Problem 2.9 from Griffith's "Introduction to Electrodynamics"

Suppose the electric field in some region is found to be $$\vec{E}=kr^5\hat{r}$$, in spherical coordinates ($$k$$ is some constant). Find the charge density $$\rho(\vec{r})$$ and the total charge enclosed by a spherical surface of radius $$R$$.

I obtained the charge density with $$\nabla\cdot\vec{E}=\frac{1}{\epsilon_0}\rho(\vec{r})$$ which gave the correct answer ($$5\epsilon_okr^2$$)

Similarly, the total charge can be calculated using $$\oint_S\vec{E}\cdot\hat{n}dS=\frac{q_{enc}}{\epsilon_0} \to q_{enc}=4\pi\epsilon_0kR^5$$

Why is it that the two formulae above (differential, integral, Gauss' Law) still hold even when $$E$$ no longer falls off as $$\frac{1}{r^2}$$?

Isn't Gauss' Law for electric fields $$=\frac{q_{enc}}{\epsilon_0}$$derived based on the assumption that the $$r^2$$ (one from $$d\vec{S}_{spherical}$$ and the other from $$\vec{E}$$) cancel, otherwise the flux would depend on the radius of the sphere and we would need to know the relation bewteen $$k$$ and $$q_{enc}$$?

• $1/r^2$ is for a point charge. Now you have many many point charges inside the sphere. The charge distribution renders the field to the $r^5$.
– ytlu
Jan 15, 2021 at 17:38

Isn't Gauss' Law for electric fields $$=\frac{q_{enc}}{\epsilon_0}$$derived based on the assumption that the $$r^2$$ (one from $$d\vec{S}_{spherical}$$ and the other from $$\vec{E}$$) cancel, otherwise the flux would depend on the radius of the sphere and we would need to know the relation between $$k$$ and $$q_{enc}$$?

It is easiest to see that $$\Phi_E = \oint \vec{E} \cdot d\vec{a} = \frac{q_{enc}}{\epsilon_0}$$ for spherical volumes surrounding point charges, and that's generally how it's justified to students. But in fact, it holds for for all volumes and charge configurations (not just spherical volumes surrounding point charges). And if you buy that, you can then show that $$\vec{\nabla} \cdot \vec{E} = \rho/\epsilon_0$$; in other words, the two statements are equivalent. The way to show this (informally) is as follows:

Imagine a cuboid spanning the coordinates $$[x, x+\Delta x]$$, $$[y, y+\Delta y]$$, and $$[z, z+\Delta z]$$. We will assume that $$\Delta x$$, $$\Delta y$$, and $$\Delta z$$ are "small" in some appropriate sense. The net flux through the faces whose normals are in the $$x$$-direction is approximately $$\Phi_{E,x} \approx [E_x(x+\Delta x, y, z) - E_x(x,y,z)]\Delta y \Delta z = \frac{E_x(x+\Delta x, y, z) - E_x(x,y,z)}{\Delta x} \Delta x \Delta y \Delta z \approx \frac{\partial E_x}{\partial x} \Delta V$$ where $$\Delta V \equiv \Delta x \Delta y \Delta z$$. Note that the fluxes have opposite signs because the face at $$x$$ has normal $$-\hat{\imath}$$ while the face at $$x +\Delta x$$ has normal $$+\hat{\imath}$$. Similarly, the net fluxes through the pairs of faces facing in the $$y$$ and $$z$$ directions are $$\Phi_{E,y} \approx \frac{\partial E_y}{\partial y} \Delta V \qquad \Phi_{E,z} \approx \frac{\partial E_z}{\partial z} \Delta V$$ respectively. Thus, the total flux through these surfaces is approximately $$\Phi_E = \left( \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z} \right) \Delta V = (\vec{\nabla} \cdot \vec{E}) \Delta V.$$ And if Gauss's Law holds for this volume, which holds a charge $$\Delta q$$, then we have $$(\vec{\nabla} \cdot \vec{E}) \Delta V \approx \Phi_E = \frac{\Delta q}{\epsilon_0} \quad \Rightarrow \quad \vec{\nabla} \cdot \vec{E} \approx \frac{1}{\epsilon_0} \frac{\Delta q}{\Delta V} \approx \frac{\rho}{\epsilon_0}.$$ All of the "approximately equals signs" above become exact equalities in the limit that $$\Delta x, \Delta y, \Delta z \to 0$$.

As it is pointed out in the comment that $$1/r^2$$ is for a point charge. And a charge distribution can make of any kind of field like in this case that goes like $$r^5$$.

As an example, consider potential due to a change distribution $$\rho(x',y',z')$$. If you recall the multipole expansion that looks like $$\phi_A=\frac{1}{4\pi\epsilon_0}\left[\frac{K_0}{r}+\frac{K_1}{r^2}+\frac{K_2}{r^3}+\cdots\right]$$ So you see, You get different powers of $$r$$ here.

But Even so, the field and the charge distribution is not physicals both tend to infinite as $$r\rightarrow \infty$$. That's the reason then asked for charge only inside the sphere. The physical field goes to zero as you go to infinite.