Question

Electric field due to uniformly charged sphere.

Answer 1

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 $\varphi$ be the electric flux and $Q$ be the charge then :
 ${\epsilon }_0\varphi =Q_{enclosed}$
Also , electric flux=electric field X area of the enclosed surface : $\varphi =EA$
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 $q^{\prime }$ . Since volume density is same then-
$$\begin{gathered} {\displaystyle \frac{q^{\prime }}{{\displaystyle \frac{4}{3}}\pi r^3}}={\displaystyle \frac{q}{{\displaystyle \frac{4}{3}}\pi R^3}} \\ {q}^{\prime }=q{\displaystyle \frac{r^3}{R^3}} \end{gathered}$$
Applying Gauss Theorem here-
 $$\begin{gathered} \varphi =E4\pi r^2 \\ {\displaystyle \frac{Q_{enclosed}}{{\epsilon }_0}}=E4\pi r^2 \\ {\displaystyle \frac{q^{\prime }}{{\epsilon }_0}}=E4\pi r^2 \\ {\displaystyle \frac{q}{{\epsilon }_0}}\times {\displaystyle \frac{r^3}{R^3}}=E4\pi r^2 \\ E={\displaystyle \frac{1}{4\pi {\epsilon }_0}}\times {\displaystyle \frac{qr}{R^3}} \end{gathered}$$
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 .
$$\begin{gathered} E={\displaystyle \frac{1}{4\pi {\epsilon }_0}}\times {\displaystyle \frac{qr}{R^3}} \\ E={\displaystyle \frac{1}{4\pi {\epsilon }_0}}\times {\displaystyle \frac{qR}{R^3}} \\ E={\displaystyle \frac{1}{4\pi {\epsilon }_0}}\times {\displaystyle \frac{q}{R^2}} \end{gathered}$$
This is the electric field on the surface.
Case III: Outside the sphere $(r>R)$

We will apply Gauss theorem in this too.
$$\begin{gathered} \varphi =EA \\ {\displaystyle \frac{q}{{\epsilon }_0}}=E4\pi r^2 \\ E={\displaystyle \frac{1}{4\pi {\epsilon }_0}}\times {\displaystyle \frac{q}{r^2}} \end{gathered}$$
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 .