# I Differential form of Gauss' law: All three terms the same value?

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1. Oct 16, 2016

### greypilgrim

Hi.

Is the Maxwell equation
$$\nabla\cdot\vec{E}=\frac{\rho}{\varepsilon_0}$$
even true in the stronger form
$$\frac{\partial E_i}{\partial x_i}=\frac{\rho}{3\cdot\varepsilon_0}\enspace ?$$
I guess not, since I haven't found a source suggesting this. But shouldn't the isotropic electric field of a point charge change the total electric field in all directions by the same amount?

2. Oct 17, 2016

### vanhees71

No, why should that hold? It's better to think about the divergence in terms of its coordinate-independent definition,
$$\vec{\nabla} \cdot \vec{E}(\vec{x})=\lim_{\Delta V \rightarrow \{\vec{x} \} }\int_{\partial \Delta V} \mathrm{d}^2 \vec{f} \cdot \vec{E}.$$

3. Oct 17, 2016

### greypilgrim

My idea was somehow as follows: Since the electric field of a point charge is isotropic and everything is linear, shouldn't it be possible to separate its field into three component fields where every one of those fields only has field vectors in one coordinate direction? And since everything can be thought of as made up by point charges (delta distribution), this would translate to any charge distributions and fields.

4. Oct 19, 2016

### DarkBabylon

While you could seperate the electric field into components, what you are actually getting from this equation is a scalar that tells you the flux density by volume.
It will be the same regardless of dimension, plus the charge itself can't be seperated into vectors as it is a scalar as well. You don't get a third of electric field in each axis, the coordinates share components using trigonometry.

I would suggest going through linear algebra and vector analysis before tackling electromagnetism, I think there is a salad here.

5. Oct 19, 2016

### vanhees71

Take he Coulomb field as an example,
$$\vec{E}=\frac{q}{4 \pi} \frac{\vec{x}}{r^3}.$$
Then
$$\partial_x E_x=\frac{q}{4 \pi} \frac{-2x^2+y^2+z^3}{r^5},$$
but
$$\vec{\nabla} \cdot \vec{E}=0, \quad \vec{x} \neq 0.$$

6. Oct 21, 2016

### Khashishi

The field of a dipole is not isotropic.

7. Oct 22, 2016

### vanhees71

The OP was taking about the Coulomb not a dipole field. Of course, also for the dipole field you have $\vec{\nabla} \cdot \vec{E}=0$ except at the place of the dipole, where the field is singular.