Why is $\vec E$ constant for a uniform sphere? I know that a sphere is a desirable Gaussian surface because we can take $\vec E$ out of the integral of Gauss's Law.
However, it isn't apparent to me as to why $\vec E$ is constant for a sphere. If $E$ is a constant, it isn't a vector field, as it isn't a vector function. So, it would be something like $\vec E = (1,0,4)$, not varying with space. How is this the case for a sphere? If it is a constant $\vec E$, I would imagine it must be $0$, but I have trouble explaining why. All the electric field lines coming out of a sphere cancel each other, as if you drew lines off the sphere like drawing a Sun the way children do - the lines as vectors would cancel in vector addition. I can't see a justification for there to be a non-zero electric field if it's not a vector field.
Can someone make sense of all of this? 
 A: It is not the case that the electric field is constant around a sphere. It is the magnitude of the electric field that is constant. For a spherical Gaussian surface that surrounds a spherical charge distribution, the electric field on opposite sides of the sphere will point in opposite directions, but both fields will have the same strength. This allows us to do the following, starting from Gauss' law:
$$\oint\vec{E}\cdot d\vec{n} = \frac{q_{enc}}{\epsilon_0}$$
where $\vec{E}$ is the electric field, $d\vec{n}$ is the differential normal vector (whose length is equal to the infinitesimal area it's attached to), and $q_{enc}$ is the charge inside the surface, and $\epsilon_0$ is the permittivity of free space.
Since the electric field is always perpendicular to the spherical Gaussian surface, we can replace the integral with
$$\oint|\vec{E}|dn = \frac{q_{enc}}{\epsilon_0}$$
Since the magnitude of the elecric field is constant:
$$|\vec{E}|\oint dn = \frac{q_{enc}}{\epsilon_0}$$
The value of the integral is now the surface area, so
$$|\vec{E}|(4\pi r^2) = \frac{q_{enc}}{\epsilon_0}$$
And finally:
$$|\vec{E}| = \frac{1}{4\pi\epsilon_0}\frac{q_{enc}}{r^2}$$
A: With Gauss’ law, $\oint \vec E\cdot d\vec S$, you have
$\vec E\cdot d\vec S= \vert \vec E\vert dS\cos\theta$ so you want the magnitude of $\vec E$ to be constant so you can pull it out of the integral, and you want the trig factor $\cos\theta$, expressing the angle between $\vec E$ and $d\vec S$ at a point on the surface, to also be constant so that 
$$
\oint \vec E\cdot d\vec S= \vert \vec E\vert S
$$
(assuming $\theta=0$).  In particular, $\vec E$ can vary in direction on the surface, but hopefully always so that it has the same orientation w/r to $d\vec S$.
So to be clear: $\vec E$ is not constant, but its magnitude will be constant on a sphere of radius $R$ if the source charge density has spherical symmetry, and $\vec E$ will always be normal to the surface of that sphere.
A: Because net charge inside the sphere is 0 there will be no electric field inside the sphere unless and untill you put a charge inside it.
