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**Hint:**This is the case of solid non-conducting spheres. We will have three cases associated with it . They are : electric fields inside the sphere, on the surface, outside the sphere .

Apply the gauss theorem to find the electric field at the three different places.

**Complete step by step solution:**

Consider a charged solid sphere of radius $R$ and charge $q$ which is uniformly distributed over the sphere. We will use Gauss Theorem to calculate electric fields. If $\phi $ be the electric flux and $Q$ be the charge then :

${\varepsilon _0}\phi = {Q_{enclosed}}$

Also , electric flux=electric field X area of the enclosed surface : $\phi = EA$

Case I- Inside the sphere $(r < R)$

The charge distribution is uniform . Volume density will be the same. Let the charge enclosed by a circle of radius $r$ be $q'$ . Since volume density is same then-

$

\dfrac{{q'}}{{\dfrac{4}{3}\pi {r^3}}} = \dfrac{q}{{\dfrac{4}{3}\pi {R^3}}} \\

q' = q\dfrac{{{r^3}}}{{{R^3}}} \\

$

Applying Gauss Theorem here-

$

\phi = E4\pi {r^2} \\

\dfrac{{{Q_{enclosed}}}}{{{\varepsilon _0}}} = E4\pi {r^2} \\

\dfrac{{q'}}{{{\varepsilon _0}}} = E4\pi {r^2} \\

\dfrac{q}{{{\varepsilon _0}}} \times \dfrac{{{r^3}}}{{{R^3}}} = E4\pi {r^2} \\

E = \dfrac{1}{{4\pi {\varepsilon _0}}} \times \dfrac{{qr}}{{{R^3}}} \\

$

This is the electric field inside the charged sphere .

Case II: On the surface $(r = R)$

In the above case we have calculated the electric field inside the sphere. In that formula we will put $(r = R)$ , so evaluate the electric field on the surface of the sphere .

$

E = \dfrac{1}{{4\pi {\varepsilon _0}}} \times \dfrac{{qr}}{{{R^3}}} \\

E = \dfrac{1}{{4\pi {\varepsilon _0}}} \times \dfrac{{qR}}{{{R^3}}} \\

E = \dfrac{1}{{4\pi {\varepsilon _0}}} \times \dfrac{q}{{{R^2}}} \\

$

This is the electric field on the surface.

Case III: Outside the sphere $(r > R)$

We will apply Gauss theorem in this too.

$

\phi = EA \\

\dfrac{q}{{{\varepsilon _0}}} = E4\pi {r^2} \\

E = \dfrac{1}{{4\pi {\varepsilon _0}}} \times \dfrac{q}{{{r^2}}} \\

$

This is the electric field outside the sphere.

If we plot these variations on a graph we will get the following graph:

**Note:**Since this is a solid sphere , it has charge inside it as well and that is why the electric field is non zero. In case of a hollow spherical shell, the electric field inside the shell is zero .

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