# Using symmetry in Gauss' Law

I have to find electric field at any inside point due to a uniformly charged solid sphere

I do it in following steps

$$\to$$ First I choose a spherical gaussian surface passing through required point concentric with the charged sphere.

$$\to$$ Next by symmetry I say that electric field will be same in magnitude at all points as all the points are equivalent to given charge distribution.

$$\to$$ By symmetry I say that the electric field will be directed radially.

But problem that arises is that how can I show that this field will be directed radially outwards i.e. what is the argument to prove that electric field will not be radially inwards

• I erased my answer, I was solving the problem outside, and you wanted it inside. My bad Sep 20 at 22:28
• @Wolphramjonny Didn't you read question properly? Sep 20 at 22:35

The direction of the field is given by the sign of the charge. It can be either outward or inward .

Your arguments are not in the right order:

First, you prove by symmetry, that the field is radial and depends only on the distance to the center O : $$\vec{E}=E(r)\vec{e_r}$$. Note that $$E(r)$$ is an algebraic quantity: you don't need to know its sign.

Then, you choose a Gauss surface adapted to this symmetry: here, a sphere centered on O. With this choice of Gauss surface, the electric flux is written very simply $$\Phi=4\pi r^2E(r)$$ and the field is determined completely $$E(r)=Q_{int}/4\pi \varepsilon r^2$$.

Gauss's theorem gives you the field with its sign : the radial component is positive if $$Q_{int}$$ is positive.

Let $$\,r\,$$ be the radius of a spherical surface with uniform surface charge density $$\,\sigma\boldsymbol >0\,$$ so with charge $$\,q\boldsymbol=4\,\pi\,r^2\sigma\boldsymbol >0$$. Then the field inside this spherical surface is zero while the field outside is the same as if the charge $$\,q\boldsymbol >0\,$$ is concentrated on the center of the sphere so directed outwards.

For the case of a point in the interior of a solid sphere (ball) with uniform volume charge density $$\,\rho\boldsymbol >0\,$$ think that the field due to the sum (integral) of the spherical shells beyond the reference point have zero contribution while the field due to the sum (integral) of the spherical shells below the reference point have a contribution directed outwards as if the charge of the solid sphere below the reference point is concentrated on the center. Of cource if $$\,\rho\boldsymbol <0\,$$ then the field is directed inwards.

By an other argument : if a closed smooth surface like a sphere of radius $$\,R\,$$ is oriented with its normal $$\,\mathbf n\,$$ directed outwards then by Gauss Law for the electric flux $$\,\Phi\boldsymbol=Q/\epsilon_0\,$$ where $$\,Q\,$$ the enclosed electric charge. But $$\,\Phi\boldsymbol=4\,\pi\,R^2 E\boldsymbol=Q/\epsilon_0\,$$ where $$\,E\,$$ the algebraic projection of $$\,\mathbf E\,$$ on $$\,\mathbf n\,$$ at every point on the surface $$\,E\boldsymbol=\mathbf E\boldsymbol\cdot\mathbf n$$. So $$\,E\,$$ must have the sign of $$\,Q$$.

For the electric field due to a point charge, by Coulomb's law, the field direction points from the charge point to the field point along the straight line, which is radially outwards.