Why can't electric field by a single charge be at an angle? I am having a very difficult time understanding this basic concept of why the direction of the electric field by a single isolated charge is radially outward or inward but not at an angle. I have been reading other sources on how this is due to spherical symmetry of charges, of how when you rotate the sphere, because charge is uniform at a certain radius, so it does not matter what angle the field points, only direction you have unique left is either radially outward or inward.
But I still don't understand this.Can anyone explain, possibly with a diagram, picture, of how at a certain point of space, electric fields can't be pointing at a angle due to symmetry of charge? Like are the electrical field pointing at angle canceling each other out? Can someone give some analogy of how symmetry makes electric field have unique direction only radially inward or outward?
 A: At an angle to what?  If you are imagining a point charge, it’s a coordinate singularity.  If you stand at the North Pole, every direction is “south.”  If you have an object which really has zero size, as you approach that point the only direction available is “away.”
Let’s ignore the zero-size effect by imagining you have some sphere of charge.  Here are two basic ways you can have a non-radial field:

*

*The field at the sphere’s equator might have some component tangent to the equator, like how the velocity vector for a city on Earth’s equator always points to the east.  This field would have nonzero line integral
$\oint \vec E\times\mathrm d\vec\ell$ around the equator, which you can actually accomplish by having a changing magnetic field $\mathrm d\vec B/\mathrm dt$ within the sphere.  This relationship actually survives into the limit of a zero-sized sphere: everywhere in space that $\mathrm d\vec B/\mathrm dt \neq 0$, the electric field has nonzero curl, $\vec\nabla\times\vec E \neq 0$.


*The field at the sphere’s equator might have a component tangent to the sphere’s surface, but pointing into the northern or southern hemisphere.  In that case you could subtract off the radial part of the field and leave yourself with just the dipole field. The dipole field also survives into the zero-size particle limit, but you have to imagine making the charge on your sphere larger as the diameter of the sphere gets smaller, so that the dipole moment is a constant.
In point of fact, modern quantum electrodynamics does predict some small electric dipole moment associated with pointlike charges.  The answer to “at an angle to what?” is the direction of the particle’s spin. The smallness of the permanent electric dipole moment, it turns out, is related to the lack of differences between matter and antimatter. I’ve written answers about this which specifically discuss the electron and the neutron.
A: Op has been getting several answers which may be correct but appear to be above the level of what OP was looking for. So I am taking a simpler approach.
Consider the following diagram where we observe a single charge on the left side and we are measuring the electric field at the point on the right. If the field was at an angle, there are components along the axis and perpendicular to it. Consider therefore such a perpendicular component as shown:

Now suppose you move behind and look at the setup from there. You would see the following:

And now, while you are behind, stand on your head. You will see the following:

But notice this is the same situation as in the first diagram - the charge is on the left but now the assumption there that there was a component perpendicular to the axis and pointing up leads to the contradiction that it would point down in the same situation. Thus, there can be no such component, only a component along the axis.
Note that if there were a complicated distribution of charge on the left instead of a single charge then we couldn't use such a symmetry argument as the third diagram would be different from the first one.
A: The experimental study of electricity  led to modeling the data with formulas that not only modeled the data, but were also predictive of new measurements. The attraction between two point charges . Coulombs law is the distillate of these observations, and it is called a "law" because  of that, extra axiom to pick mathematical solutions from the general mathematical formulas.
Your question , wanting the electric field to point at an angle between two point charges is not allowed  by Coulombs law. .

As the electric field between two charges is defined as the vector force divided by one of the charges , it has the same direction as the force, i.e. the line joining the two charges as stated above. (to measure an electric field there is always a test charge)
So within classical electrostatics there are no angular deviations of the field lines by definition of electric field.
A: The symmetries of the charge distribution say something about the symmetries of the resulting electric field. For example

*

*$\rho$ has rotational symmetry $\implies$ $\vec E$ only depends on $r$

*$\rho$ has mirror symmetry through some plane $\implies$ $\vec E(\vec r)$ is parallel to the plane for all points $\vec r$ on the plane

*$\rho$ has translation symmetry in some direction (for example $z$) $\implies$ $\vec E$ is independent of $z$.

The basic idea behind these statements is that since $\vec E$ can be uniquely determined from $\rho$ it shouldn't matter if we first apply the symmetry $\rho\rightarrow \rho'$ and than determine $\vec E'=f(\rho')$ or if we first determine $\vec E=f(\rho)$ and than apply the symmetry $\vec E\rightarrow \vec E'$. Since $\rho'=\rho$ (we said the charge distribution was symmetric under the transformation) we can reason that $\vec E'=\vec E$.
Let's see how to derive (2.) using this last fact. To make this easier we can decompose $\vec E$ in a tangential and a normal part: $\vec E=\vec E_\parallel+\vec E_\perp$.

From the image we can conclude that $\vec E'_\perp=-\vec E_\perp$ and $\vec E_\parallel'=\vec E_\parallel$. Since $\vec E'=\vec E$ we can conclude that $\vec E_\perp=-\vec E_\perp$ which has the only solution $\vec E_\perp=0$, or in other words: $\vec E$ is parallel to the plane.
Now why did I put all this effort in proving point (2.) when we are interested in spherical symmetry? Well for a spherically symmetric charge distribution we can draw a line connecting the origin to some other point $\vec r$. Now any plane that passes through this line is a plane of mirror symmetry. By choosing 2 nice planes we can deduce that $\vec E$ must be parallel to the line connecting the origin to $\vec r$, i.e. the electric field is in the radial direction.
As a bonus you can prove (1.) and (3.) using the fact that having $f(x+a)=f(x)$ for all $a$ implies $f(x) = \text{independent of }x$. To prove (1.) it might be useful to write $\vec E$ in spherical coordinates $\vec E(r,\theta,\phi)$.
A: Suppose you gave me a blank ball and closed your eyes, if I rotated and gave it back, you would not be able to say the difference. Now imagine the situation with the electric charge, imagine charge is kept at origin. If I were to rotate space about origin (you could imagine the charge was kept on a paper sheet and I decide to rotate the whole paper sheet), the situation in end would be indistinguishable from situation in the start.
This means that $\frac{\partial E_{\theta} }{\partial \theta}= 0$ i.e: the rate of change of component of the electric field in the angular direction is zero. This could mean either that at all points the $E_{theta}$ component is same or it is zero everywhere. Can you show that it must be infact zero everywhere?
A: The most basic answer lies purely in the mathematics of maxwells equations.
$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$
For a point charge $\rho = Q \delta^3(r)$
This is a single point of divergence.
A single point of divergence has spherical symettry, and thus the field must be symmetrical. Understanding the definition of divergence can show why this is the case
A: I may try to provide for a simple explanation if possible.
Electric field at a point can also be understood as the force vector per unit charge at that point.
That means if I have a test charge q at some point, the direction of force acting on q will be the same as that of electric field at that point.
Now coming to your question, you are referring to electric field lines, which are just a representation of how electric fields occur in a given scenario.
If you take a point charge Q and a test charge q, and bring q desirably closer to Q, the electric field at any point will always be in the direction of the force acting between them, which in turn is along the line joining the two charges.
If you take this charge q towards Q from any "angle" around it, the electric field will still be in the direction of the force, like before.
For a fixed distance r between the charges, if you take q around Q along the circle of radius r, the electric field direction will be along the radius(since the radius is on the line joining the two charges).
Since the electric fields at every point equidistant from a charge are equal in magnitude, and the intensity of electric field is characterized by the density of electric field lines around that point, we can say that equidistant points have equal density of electric field lines.
So coming back to our circle of radius r, since every point on the path of the circle is equidistant from the charge Q, the density of electric field lines across equal sections of the circumference of the circle are equal. Hence we can see that the electric field lines are evenly distributed throughout the circumference. Therefore the electric field is in a radial direction.
My explanation, if wrong, may need correction. Hope this helps.
A: Imagine a tap. Water is coming out of tap and falling to flat ground. This water is flowing towards far away point in all directions. How will water flow look like? I hope you can visualize that the water flow will be radially outward.
Think of positive charge as a "tap" of electric field. This electric field is flowing away from positive charge in all directions in 3D. If positive charge is "tap" (or source) then clearly negative charge will be "sink".
