Why is the electric field due to a point charge and a uniformly charged spherical insulator outside of the insulator the same? Assume that the point charge and the spherical insulator have identical charges. I am just wondering why this is the case. I know that from Gauss' Law, the charged enclosed by a Gaussian sphere of a radius r (with r being greater than the radius of the spherical insulator with uniform charge distribution) is the same, so the E field is identical for both.
However, this doesn't make intuitive sense to me. Since the sum of the charges on the spherical insulator are closer to the Gaussian sphere, shouldn't the E field due to the spherical insulator be bigger than for the point charge?
 A: If the sphere is of radius $R$ and has the total charge $Q$, the surface charge density is $\frac{Q}{4\pi R^2}$. So you see as the sphere gets bigger, the charge concentration on it get less.
For a fixed point outside, assuming that the sphere radius is increasing linearly with time, the electric field magnitude due to getting closer increases like $x^2$, while the charge density increases with $\frac1{x^2}$. So these two exactly cancel each other.
So if the total charge on the sphere is remained constant, the radius of it is of no importance.
Now let $R$ shrink toward zero. We have a point charge of $Q$. And we had seen that the radius isn't important. So there's no way that the field magnitudes not be the same.
From symmetry, we know that the field directions are radially outward for all radii, So summing these two conclusions up in one proposition, we can say that the field due to a point charge and a uniformly charged, non-conducting (or even conducting) sphere of any radius, as long as the total charge, $Q$, is kept constant, are the same.
QED.

EDIT:
For the generality of the proof, to even cover solid balls, you can say that a ball is the superposition of infinite thin layers of charge $dq$, in a way that the total charge is $Q$. 
this case, again make no difference, since, according to the proof, you can squeeze any spherical layer into a point charge of the same charge amount and the electric field will not be altered (of course the total energy of the arrangement WILL differ but we're not currently talking about it).
According to this, you can squeeze your charged ball into a point charge of the same total charge, and the only important thing is that:
$$\int_{\mathbb{ball}} dq=Q$$
It means, as long as the total charge is $Q$, And we have a radially symmetric ball (so that we CAN think of it as the superposition of layers), the electric field outside is the same as that of a point charge of the same charge amount.
A: It's unclear whether you have a thin shell insulator of radius R and total charge Q with uniform surface charge ($\sigma=\frac{Q}{4\pi R^2}$) or a solid sphere with uniform volume charge ($\rho=\frac{3Q}{4\pi R^3}$), but the argument is the same.
The net electric field outside the spherical shape, at distance $R_0>R$ from the center of the sphere, is the vector sum of tiny electric fields, $d\vec{E}$, due to each point charge on/in the sphere. That tiny charge depends on the radius, $dQ=\sigma( dA)$ or $dQ=\rho (dV)$, and the tiny fields are $$d\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{dQ}{r^2}\hat{r}$$
where $r$ is the distance from the small charge to the point where you are calculating the field (point of interest), and $\hat{r}$ is a unit vector pointing from the small charge toward the point of interest.  For some of the tiny charges $r<R$ and for others $r\ge R$. We can calculate $r$ using the law of cosines to get
$$r^2=R^2+R_0^2-2RR_0\cos(\theta),$$ where $\theta$ is the polar angle with the polar axis along the sphere-center-to-point-of-interest line.
When you add up all these small electric fields due to small charges, the vector components which aren't radial all sum to zero. Also, the $r$ values are changing but $R$ value in the $dQ$ isn't, so some tiny charges are more influential than others. That is, each dQ has a small contribution to the total because the total charge isn't all in one location.
That's why the spherically distributed charge looks to the outside world to be the same as a point charge.  
Apocryphally, this same question, except regarding the influence of Earth's mass, was Newton's motivation behind inventing the summing technique which we know as calculus. Before him, the concept of infinitesimally small things was abhorrent to mainstream mathematicians (apologies to Liebnitz).
A: If you draw the electric field lines (due to the uniformly charged sphere) outside the sphere you will find that they leave the sphere perpendicularly.
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Now if you put a point charge at the place of the sphere's centre (and then removing the sphere) you will observe that the electric field lines are identical to the electric field lines due to the charged sphere.

So it is safe to assume that a point charge lies within a uniformly charged insulating (or conducting) sphere and acts as a source of E. field, as long as the electric field lines outside the sphere are concerned, inside the sphere the electric field lines are not identical to that of point charge at the centre of a INSULATING sphere. 
In an conducting sphere the electric field lines are absent inside because all the charge gets distributed at the outer layer of the sphere in order to achieve a sate of minimum repulsion and maximum possible stability. This distribution is not possible in insulating spheres. So electric field exists inside insulating charged spheres. 
Hope it Helps..
