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I might have a bug in my brain, but I just can't figure this out. Please help. According to the first of the Maxwell Equations, we have $$ \nabla\cdot\vec{E}=\frac{\rho}{\epsilon_0}. $$ And we have for the electric field of a point charge $$ \vec{E} = k\frac{\vec{r}}{r^3}, $$ where $k=\frac{q}{4\pi\epsilon_0}$, $\vec{r}=[x_1,x_2,x_3]$ and $r=\sqrt{x_1^2+x_2^2+x_3^2}$. Now I want to find the divergence of this electric field $\vec{E}$, so let's derive the i-th component w.r.t. $x_i$: $$ \frac{d}{dx_i}\vec{E}_i = k\frac{r^3 - 3r^2r'x_i}{r^6} = k\frac{1}{r^3}-3k\frac{rx_i^2}{r^6} = k\frac{1}{r^3}-3k\frac{x_i^2}{r^5} $$ where I used $r'=\frac{dr}{dx}=\frac{x}{r}$. Now let's add up the tree derivatives.

$$ \nabla\cdot\vec{E}=\sum_{i=1}^{3} \frac{d}{dx_i}\vec{E}_i = \frac{3k}{r^3} - 3k\frac{x_1^2+x_2^2+x_3^2}{r^5} = \frac{3k}{r^3} - \frac{3k}{r^3} = 0. $$

So the divergence of the electric field of a point charge would always be zero, when it actually should be equal to $\frac{\rho}{\epsilon_0}$. Please help me figuring out, what I am doing wrong or where I have a misconception. I'm thankful for every kind help :)

tobi-v
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2 Answers2

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Because this a point charge, the charge is 0 everywhere except the origin. This is consistent (everywhere except the origin) with your calculation that found the divergence to be 0.

What you didn't consider is that your expression for the field,

$$\vec{E}=k\frac{\vec{r}}{r^3}$$

is not defined at the origin, and therefore you can't use it to find the divergence at the origin.

The Photon
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You missed one point while calculation. Let me make calculation more easier by going towards spherical polar coordinates.

Since $\vec E$ = $E_r \hat r + E_\theta \hat \theta + E_\phi \hat \phi$ = $\frac{k}{r^2}\hat r$ For spherical polar coordinates $\vec{\nabla} \cdot \vec E = \frac{1}{r^2}\frac{\partial}{\partial r}(r^2 E_r)+0+0=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2 \frac{k}{r^2})=0$ (only if $r\neq 0$ because $\frac{1}{r^2}$ is not defined at r=0. So at r= 0 divergence of Electric field is infinite and elsewhere it is zero, which can explain the result $\vec \nabla \cdot \vec{E}=\frac{\rho}{\epsilon_0}$.

Chauhan
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