# Is the Divergence of a Vector Field Defined by a (Positive) Point Charge Negative?

I was wondering whether the divergence of a vector field which is defined by a (positive) point charge is positive, zero, or negative everywhere. It is assumed that the charge is at $$(0,0,0)$$. Intuitively, I would think that it would be negative since the field gets weaker the further you get from the charge, which would cause the vectors "pointing away" from a point to have less magnitude than the ones "pointing towards" it. But when I tried using a formula, I found that the divergence is 0 everywhere. Is my intuition wrong or my calculation?

• Are you talking about the divergence of the electric field of the point charge? This is just Gauss' law then. What formula did you use? You should provide precise details of your attempts in order to receive helpful responses here. Commented Nov 4, 2020 at 13:15
• @N.Steinle I used the following formula for the electric field: $$\displaystyle{\vec{E} = \frac{Q}{4\pi\epsilon_0}\frac{\vec{x}}{||\vec{x}||^3}}$$ Calculating the divergence from this gets me $0$, but $0$ doesn't feel right intuitively Commented Nov 4, 2020 at 13:34

The Gauss law specifies the divergence of the electric field in every circumstance as $$\nabla\cdot\mathbf E = \frac{1}{\epsilon_0}\rho,$$ where $$\rho=\rho(\mathbf r)$$ is the volumetric charge density. For a point charge $$q$$ at the origin, the charge density is given by a Dirac delta function: $$\rho(\mathbf r) = q \:\delta(\mathbf r).$$ As such, for $$\mathbf r \neq \mathbf 0$$, the charge density is zero and so is the divergence of the electric field.

I would think that it would be negative since the field gets weaker the further you get from the charge, which would cause the vectors "pointing away" from a point to have less magnitude than the ones "pointing towards" it,

is incorrect. Consider, as an example, a unit volume of cubical shape, with one face facing towards the unit charge. In this case, it is true that the electric field at this face is stronger than at the face at the back of the cube, so the electric flux going into the cube in the forwards face is more than what goes out of the back face. However, the walls on the side are not orthogonal to the point charge, and they subtend a nonzero solid angle from the origin, which means that there is a nonzero electric flux going out of the cube through the side. When you add all of this up, the net flux into the cube is exactly zero $$-$$ as guaranteed by the Gauss law.