A charge Q is distributed uniformly within the material of a hollow sphere of inner and outer radii r₁ and r₂ (figure 30-E4). Find the electric field at a point P at a distance x away from the centre for r₁ < x < r. Draw a rough graph showing the electric field as a function of x for 0 < x < 2r₂ (figure 30-E4).

Amount of charge present on the hollow sphere = Q
Inner radii of the hollow sphere = $r_1$
Outer radii of the hollow sphere = $r_2$
Consider an imaginary sphere of radius x.
The charge on the sphere can be found by multiplying the volume charge density of the hollow spherical volume with the volume of the imaginary sphere of radius (x-$r_1$).

Charge per unit volume of the hollow sphere,
$\rho =\frac{Q}{{\displaystyle \frac{4}{3}}\pi \left(r_2^3-r_1^3\right)}$
Charge enclosed by this imaginary sphere of radius x:
q = ρ × Volume of the part consisting of charge


$$\begin{gathered} q=\frac{{\displaystyle \frac{4}{3}}\mathrm{\pi }\left(x^3-r_1^3\right)Q}{{\displaystyle \frac{4}{3}}\mathrm{\pi }\left(r_2^3-r_1^3\right)} \\ q=\frac{\left(x^3-r_1^3\right)}{\left(r_2^3-r_1^3\right)}Q \end{gathered}$$
According to Gauss's Law,

$\oint \mathrm{E}.ds=\frac{q}{{\in }_0}$

Here, the surface integral is carried out on the sphere of radius x and q is the charge enclosed by this sphere.





$$\begin{gathered} \mathrm{E}\oint ds=\frac{q}{{\in }_0} \\ E\left(4\mathrm{\pi }{x}^2\right)=\frac{q}{{\in }_0} \end{gathered}$$
$$\begin{gathered} E(4\mathrm{\pi }{x}^2)=\frac{\left(x^3-r_1^3\right)Q}{\left(r_2^3-r_1^3\right){\mathit{\in }}_0} \\ E\mathit{}=\frac{Q\left(x^3-r_1^3\right)}{4\mathrm{\pi }{\mathit{\in }}_0{\mathrm{x}}^2\left({\mathrm{r}}_2^3-{\mathrm{r}}_1^3\right)} \end{gathered}$$

The electric field is directly proportional to x for r₁ < x < r.

The electric field for r₂ < x < 2r_2,


$E=\frac{Q}{4\pi {\in }_0{x}^2}$

Thus, the graph can be drawn as: