# Divergence of the E field at a theoretical Point Charge

• blizzardof96
So we can conclude that its infinite at r=0? Thank you for clarifying.In summary, the divergence at a theoretical point charge is either infinite or equal to 4pi.

#### blizzardof96

I've been thinking about this problem and would like some clarification regarding the value of the divergence at a theoretical point charge.

My logic so far:
Because the integral over all space(in spherical coordinates) around the point charge is finite(4pi), then the divergence at r=0 must be nonzero. If we can say that the dirac-delta function is a good physical model for the point charge then the Divergence is infinite at r=0 and zero elsewhere.

A property of the dirac-delta function is that when you integrate it picks out the value of the function at the location of the spike. With the integral value being 4pi, it would make sense that divergence is equal to 4pi at the origin.

So back to my question. Would the divergence be infinite at r=0 or equal to 4pi?

In general, you have that the field of a point charge is given by
$$\vec E = \frac{Q\vec e_r}{4\pi \epsilon_0 r^2}$$
and so the flux through a sphere of radius ##r## is ##Q/\epsilon_0## (as it should be for a point charge ##Q##). This means that
$$\int \nabla \cdot \vec E \, dV = \frac{Q}{\epsilon_0}.$$
Hence, ##\nabla \cdot \vec E = Q \delta^{(3)}(\vec x)/\epsilon_0##. This is perfectly reasonable, since ##\nabla\cdot \vec E = \rho(\vec x)/\epsilon_0## generally and the charge density is ##\rho(\vec x) = Q\delta^{(3)}(\vec x)##.

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Orodruin said:
In general, you have that the field of a point charge is given by
$$\vec E = \frac{Q\vec e_r}{4\pi \epsilon_0 r^2}$$
and so the flux through a sphere of radius ##r## is ##Q/\epsilon_0## (as it should be for a point charge ##Q##). This means that
$$\int \nabla \cdot \vec E \, dV = \frac{Q}{\epsilon_0}.$$
Hence, ##\nabla \cdot \vec E = Q \delta^{(3)}{\vec x}/\epsilon_0##. This is perfectly reasonable, since ##\nabla\cdot \vec E = \rho(\vec x)/\epsilon_0## generally and the charge density is ##\rho(\vec x) = Q\delta^{(3)}(\vec x)##.

Thank you for your response. Why do we assume that the charge density is equal to the dirac-delta function times Q?

blizzardof96 said:
Thank you for your response. Why do we assume that the charge density is equal to the dirac-delta function times Q?
The charge density of a point charge ##Q## at the origin by definition has the following property
$$\int_V \rho(\vec x) dV = \begin{cases}Q, & \vec x = 0 \in V \\ 0, & \vec x = 0 \notin V\end{cases}.$$
The only distribution that satisfies this is ##\rho(\vec x) = Q\delta^{(3)}(\vec x)##.

Orodruin said:
The charge density of a point charge ##Q## at the origin by definition has the following property
$$\int_V \rho(\vec x) dV = \begin{cases}Q, & \vec x = 0 \in V \\ 0, & \vec x = 0 \notin V\end{cases}.$$
The only distribution that satisfies this is ##\rho(\vec x) = Q\delta^{(3)}(\vec x)##.

That makes sense. Can we then say that the charge density and therefore the divergence is finite at the origin(and equal to 4pi)?

blizzardof96 said:
Can we then say that the charge density and therefore the divergence is finite at the origin(and equal to 4pi)?
No. Definitely not.

Orodruin said:
No. Definitely not.

So we can conclude that its infinite at r=0? Thank you for clarifying.

## 1. What is the significance of the divergence of the electric field at a theoretical point charge?

The divergence of the electric field at a theoretical point charge is a measure of the strength of the electric field at that point. It indicates the amount of electric flux flowing out of or into the point charge, and therefore, provides information about the distribution of electric charge in the surrounding space.

## 2. How is the divergence of the electric field at a theoretical point charge calculated?

The divergence of the electric field is calculated using the Gauss's law, which states that the electric flux through a closed surface surrounding a point charge is proportional to the charge enclosed by that surface. The divergence of the electric field is equal to the charge density at that point divided by the permittivity of the surrounding medium.

## 3. What does a positive or negative divergence of the electric field at a theoretical point charge indicate?

A positive divergence of the electric field indicates that the electric field is flowing outwards from the point charge, while a negative divergence indicates that the electric field is flowing inwards towards the point charge. This can be interpreted as a source or sink of electric field, respectively.

## 4. Can the divergence of the electric field at a theoretical point charge be zero?

Yes, the divergence of the electric field at a point charge can be zero if the charge is located at the center of a spherical surface, as the flux through the surface is equal in all directions and cancels out.

## 5. How does the divergence of the electric field at a theoretical point charge relate to the Coulomb's law?

The divergence of the electric field at a point charge is directly related to the charge density, which is a key component in Coulomb's law. The electric field strength at a point charge is inversely proportional to the square of the distance from the charge, which is reflected in the calculation of the divergence of the electric field.