Difficulty in differential form of gauss law

• eng.phy
In summary: I interpret the expression without the dot. So, I'm pretty sure that the dot is not implied. But, like I said, I'm no expert in the customs of electromagnetics.In summary, the conversation discusses the concept of divergence and charge density in relation to Gauss's law for empty space. It is explained that for an empty space, the divergence of the electric field intensity is equal to zero, and that the dot between ∇ and E is important in this context. The conversation also includes a mathematical explanation for the derivation of this form of the law.
eng.phy
hi all,
im new at electromagnetics and vector calculus, so facing these trivial problems. The differential form of gauss's law for empty space states ∇E = ρ/ε .The right side refers to charge density at the point of calculation of divergence? If a point charge(or a small uniformly charged sphere) is placed at (0,0) then the divergence at position vector r comes out positive( k/r). But charge density at that point is 0?

A sphere of radius r and total charge Q, uniformly distributed, will have a non-zero charge density everywhere inside the sphere. The total charge at any point is zero - since a point has zero volume.

Don't forget to keep thinking "divergence of the electric field".

To model a point charge, one simply sets the charge density to be the dirac delta function, which is like an infinite density point function.

Thanks for replying... I am still not clear. Suppose to avoid point charge, I take a uniformly charged sphere(Q) of radius R, centerd at (0,0). Electric field distribution will be same as that of point charge Q outside the sphere. Now can we say that charge density at any position r in space is given by ρ = Q/4∏εr$^{2}$ for r<R and 0 for r>R. So for any point outside the sphere, the charge density is 0 right? But divergence of electric field is not. Basically I find difficulty in the divergence theorem which is used to derive this form of gauss law. Let's take a closed volume surface in the above space, away from the charged sphere. EM states that flux of E through the surface = total charged enclosed by surface(scaled) = integral of divergence of electric field over volume of surface. The first 2 terms should be 0(outside charge) but since divergence is positive at all points in surface(k/r), how is the sum 0? Thanks in advance

I take a uniformly charged sphere(Q) of radius R, centerd at (0,0) ... Now can we say that charge density at any position r in space is given by ρ = Q/4∏εr2 for r<R and 0 for r>R
The density is the total charge Q divided by the total volume V=4πR3/3 inside the sphere. To avoid the peicewise function we can write:

$$\rho(r)=\frac{3Q}{4\pi R^3}h(R-r)$$... where h(r) is the Heaviside Step function.

Lets take a closed volume surface in the above space, away from the charged sphere. EM states that flux of E through the surface = total charged enclosed by surface(scaled) = integral of divergence of electric field over volume of surface. The first 2 terms should be 0(outside charge) but since divergence is positive at all points in surface(k/r), how is the sum 0?

You appear to be conflating two distinct things.
If there is charge outside the volume, then there is non-zero passing through any place inside the volume - and that flux can have a divergence. Since the same amount of flux leaves the volume as enters it, then the net flux through the volume is zero.

It's like a train station - nobody stays there ... over time the same number of people leave as enter. However, there can still be a crowd in there.

Yes, Simon Bridge has the right idea that the total flux through a Gaussian surface which does not contain the sphere, is equal to zero. In other words, the divergence of the electric field outside the sphere is zero. This may seem weird, because the field lines are 'spreading out' in space, but the divergence is still zero. This is because the field lines are spreading out in such a way that the divergence is zero.

I slightly mangled that description didn't I? I shouldn't answer these things before I have my coffee!

Thanx! that helped :)

The easiest mathematical way to show it is to take a volume element in spherical polar coordinates. The four side-wall normals are perpendicular to the field (so no flux), and the inner wall has a smaller area than the outer wall by a factor of r^2, which is exactly the factor by which the electric field of a point charge is reduced from the inner wall to the outer wall. The flux entering the inner wall must equal the flux exiting the outer wall for every value of r>0, and so the divergence is zero.

eng.phy said:
hi all,
im new at electromagnetics and vector calculus, so facing these trivial problems. The differential form of gauss's law for empty space states ∇E = ρ/ε .The right side refers to charge density at the point of calculation of divergence? If a point charge(or a small uniformly charged sphere) is placed at (0,0) then the divergence at position vector r comes out positive( k/r). But charge density at that point is 0?

This is not the differential form of gauss's law fror empty space. For empty space, the divergence of the electric field intensity is equal to zero. This is the differential from of gauss's law for a region in which a distributed charge is present with charge density ρ.

Chestermiller said:
This is not the differential form of gauss's law fror empty space.
It's not even a divergence as written - there is an all-important dot missing between ∇ and E, something not picked up on.

The dot is often implied, isn't it?

Simon Bridge said:
The dot is often implied, isn't it?
I'm not an expert in the mathematical customs of people working in electrostatics, but I have to agree with Q-reeus. It really struck me unusual that no responder mentioned this (I didn't mention it in my response because I wanted to keep focused on just the one point I was trying to make). Also, no one mentioned that the gradient operator and the electric field intensity were not represented in boldface to signify vectors. When I see the gradient operator acting on a vector without a dot in-between, I interpret this as the gradient of the vector, which is a second order tensor. Of course, I'm not sure what the practical application of this second order tensor might be.

I'm not sure what you mean by the gradient of a vector, but said second order tensor would marry the divergence and curl into a single entity, and doing so can be useful.

eng.phy said:
So for any point outside the sphere, the charge density is 0 right? But divergence of electric field is not
This is not so. The divergence of the field outside the charged sphere is zero.
Just calculate the Div(1/r^2) in spherical (easier) or Cartesian coordinates.

Ya sorry about the loose notations, meant divergence only... silly mistake :P

You cannot calculate the divergence of a scalar. The correct formula for the divergence of the Coulomb field is
$$\vec{\nabla} \cdot \left (\frac{\vec{r}}{4 \pi |\vec{r}|^3} \right )=\delta^{(3)}(\vec{r}).$$
In terms of the electrostatic potential
$$\Phi(\vec{r})=\frac{1}{4 \pi |\vec{r}|}$$
this gives the important formula
$$-\Delta \left (\frac{1}{4 \pi |\vec{r}|} \right )=\delta^{(3)}(\vec{r}).$$
To prove this take an arbitrary test function to show
$$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{r} \; f(\vec{r}) \vec{\nabla} \cdot \left (\frac{\vec{r}}{4 \pi |\vec{r}|^3} \right )=f(0).$$

What is the differential form of Gauss's law?

The differential form of Gauss's law is a mathematical equation that describes the relationship between the electric field and the distribution of electric charges. It states that the electric flux through a closed surface is proportional to the total electric charge enclosed by that surface.

Why is the differential form of Gauss's law useful?

The differential form of Gauss's law is useful because it allows us to calculate the electric field at any point in space, given the distribution of electric charges. This makes it easier to solve complex problems involving electric charges and fields.

How is the differential form of Gauss's law different from the integral form?

The differential form and integral form of Gauss's law are equivalent and describe the same physical phenomenon. The main difference is that the differential form uses calculus to express the relationship between the electric field and the charge distribution, while the integral form uses a summation over the electric flux through a closed surface.

What are the limitations of the differential form of Gauss's law?

The differential form of Gauss's law only applies to static electric fields and does not take into account the effects of changing magnetic fields. It also assumes that the electric field is continuous and differentiable at all points in space, which may not always be the case in real-world scenarios.

How is the differential form of Gauss's law derived?

The differential form of Gauss's law is derived from the integral form by applying the divergence theorem and simplifying the resulting equation using vector calculus. This results in the differential form, which is a more convenient and powerful way to express Gauss's law in certain situations.

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