Question about Gauss' law and dipoles I've got a question about dipoles and Gauss' law:

Consider an electric dipole $p=ql$ in free space, centred at the origin of a Cartesian reference frame. Using Gauss’s Law, discuss if the electric field $E(r)$ can be such that the scalar product $E(R) \cdot R$ is everywhere positive, everywhere negative, or everywhere null for all points $R$ located on a spherical surface of radius $R \gg l$.

I don't understand how it can be anything other than null as the total charge enclosed by the Gaussian surface will be $0$.
 A: For simplicity I treat this as a two-dimensional problem.
The diagram below shows a dipole of length $l$ and a Gaussian surface of radius R.

At point $A$ the dot product $\vec E_{R1}\cdot \vec R$ is positive and at point $B$ the dot product $\vec E_{R2}\cdot \vec R$ is negative where $\vec E_{R1}$ and $\vec E_{R2}$ are the net electric fields at points $A$ and $B$ respectively.
Note that if points $A$ and $B$ are mirror images about a plane perpendicular to the line joining the two charges then by symmetry the electric flux through a small element of the surface of the sphere of radius $R$ around point $A$ is equal but opposite to the electric flux around point $B$.
This is true for all elements which make up the sphere except at points like $D$ and $F$ where the net electric field is at right angles to the surface and hence the radius vector $\vec R$, so $\vec E \cdot \vec R$ is zero.
Now as the radius of the Gaussian surface becomes larger and larger compared to the separation of the charges $l$, ie $R\gg l$, then the angle $\theta$ gets smaller and smaller and the net electric field at a point on the Gaussian surface becomes smaller and smaller.
This results in the dot product $\vec E \cdot \vec R$ getting smaller and smaller but never becoming zero except at ponts like $D$ and $F$.
A: Gauss's Law states that
$$ \oint \mathbf{E} \cdot \mathbf{dA} = \frac{Q_{\text{enclosed}}}{\epsilon_0} $$
As you mentioned, since $Q_{\text{enclosed}} = 0$ , $\mathbf{E} \cdot \mathbf{R}$ cannot be everywhere positive or negative since that would make the integral $\oint \mathbf{E} \cdot \mathbf{dA}$ nonzero. Your intuition is correct.
In addition, even though the electric field due to a dipole can be very small, it will be nonzero for all points in space. Therefore, $\mathbf{E} \cdot \mathbf{R}$ cannot be everywhere positive, everywhere negative, or everywhere null.
