# Divergence of the E field at a theoretical Point Charge

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## Main Question or Discussion Point

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?

## Answers and Replies

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Orodruin
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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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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?

Orodruin
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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 $\rho(\vec x) = Q\delta^{(3)}(\vec x)$.

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)?

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

No. Definitely not.
So we can conclude that its infinite at r=0? Thank you for clarifying.