Find the net force that the southern hemisphere of a uniformly charged solid sphere exerts on the northern hemisphere. Express your answer in terms of the radius$\mathit{R}$ and the total charge $\mathit{Q}$.

The charge per unit volume is called as volume charge density of the sphere. It is expressed as,

$\mathit{\rho }\mathbf{=}\frac{\mathbf{Q}}{\mathbf{V}}$ …… (1)

Here,$Q$is the charge of the solid sphere,$V$is the volume of the solid sphere.

The volume of the sphere is depends on the cube of the radius of the volume. It is expressed as,

$\mathbf{V}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}{\mathbf{\tau \tau R}}^{\mathbf{3}}$ …… (2)

Substitute the above value in$\rho =\frac{Q}{V}$and solve.

Thus,

$$\begin{gathered} \mathit{\rho }\mathbf{=}\frac{\mathbf{Q}}{{\displaystyle \frac{\mathbf{4}}{\mathbf{3}}}{\mathbf{\pi R}}^{\mathbf{3}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{3}\mathbf{Q}}{\mathbf{4}{\mathbf{\pi R}}^{\mathbf{3}}} \end{gathered}$$ …… (3)

Here,$R$ is the radius of the sphere.

Assume that, a point r <R ,

Consider the radius of the Gaussian sphere is $r$then its volume is expressed as,

$v=\frac{4}{3}{\mathrm{\pi r}}^3$ …… (4)

Now, Charge in shell is expressed as,

$dQ=\rho \nu$

Substitute the values derived from the equations (3) & (4) in the above equation,

$$\begin{gathered} dQ=\left(\frac{Q}{{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3}\right)\frac{4}{3}{\mathrm{\pi r}}^3 \\ =\frac{{\mathrm{Qr}}^3}{{\mathrm{r}}^3} \end{gathered}$$

By using Gauss’s law, the field from both the spheres can be obtained.

$\oint E\cdot da=\frac{dQ}{{\epsilon }_0}$

Substitute$\frac{Q{r}^3}{R^3}$ for$dQ$ in$\frac{dQ}{{\epsilon }_0}$ for$\oint E.da$ expression.

$$\begin{gathered} \hspace{0.33em}\hspace{0.33em}\oint E\cdot da=\frac{{\displaystyle \frac{Q{r}^3}{R^3}}}{{\epsilon }_0} \\ E\left(4{\mathrm{\pi r}}^2\right)=\frac{Q{r}^3}{R^3{\epsilon }_0} \\ E=\frac{Q}{4{\mathrm{\pi \epsilon }}_0}\frac{r}{R^3} \end{gathered}$$

Thus, the electric filed inside the sphere is$\mathrm{E}=\frac{\mathrm{Q}}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{r}}{{\mathrm{R}}^3}$.

Write the expression for the force per unit volume acting on the sphere.

$f=\rho E$ …… (4)

Substitute the valuerole="math" localid="1657363835277" $\frac{Q}{{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3}$for $\rho$and$\frac{Q}{{\displaystyle \frac{4}{3}\pi {\epsilon }_0}}\frac{r}{R^3}$for in equation (4),

role="math" localid="1657364140975" $$\begin{gathered} f=\frac{Q}{{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3}\left(\frac{Q}{4{\mathrm{\pi \epsilon }}_0}\frac{r}{R^3}\right) \\ =\frac{3}{{\epsilon }_0}{\left(\frac{Q}{4{\mathrm{\pi \epsilon }}_0}\right)}^{2}r \end{gathered}$$

Therefore, the force per unit volume acting on the sphere is$f=\frac{3}{{\epsilon }_0}{\left(\frac{Q}{4{\mathrm{\pi \epsilon }}_0}\right)}^{2}r$.

Now, consider the infinitesimal volume element in terms of spherical polar coordinates,

$d\tau =r^2\sin \theta drd\theta d\varphi$

By using the symmetry net force in the on the,

$f_z=f\cos \theta \stackrel{⏜}{z}$ …… (5)

Integrate the equation (5) over the range of surface area,

$\int dF=\underset{0}{\overset{R}{\int }}\underset{0}{\overset{2\mathrm{\pi }}{\int }}\underset{0}{\overset{\mathrm{\pi }/2}{\int }}f\cos \theta d\tau$ …… (6)

Substitute$\frac{3}{{\epsilon }_0}{\left(\frac{Q}{4{\mathrm{\pi R}}^3}\right)}^2$$r$for f and role="math" localid="1657364874652" $r^2\sin \theta drd\theta d\varphi$for $d\tau$in equation (6).

$f_z=\left(\frac{3}{{\epsilon }_0}{\left(\frac{Q}{4{\mathrm{\pi R}}^3}\right)}^2\right){\int }_0^R{r}^{3}dr{\int }_0^{\frac{\mathrm{\pi }}{2}}\sin \theta \cos \theta d\theta {\int }_0^{2\mathrm{\pi }}d\varphi$ …… (7)

Let’s assume that,$\sin \theta =t$then$\cos \theta d\theta =dt$.

Substitute these values in equation (7), and simplify

role="math" localid="1657365508813" $$\begin{gathered} f_z=\left(\frac{3}{{\epsilon }_0}{\left(\frac{Q}{4{\mathrm{\pi R}}^3}\right)}^2\right){\int }_0^R{r}^{3}dr{\int }_0^{1}tdt{\int }_0^{2\mathrm{\pi }}d\varphi \\ =\frac{3}{{\epsilon }_0}\frac{3}{{\epsilon }_0}{\left(\frac{Q}{4{\mathrm{\pi R}}^3}\right)}^2\left(\frac{R^4}{4}-0\right)\left(\frac{t^2}{2}-0\right)\left(2\mathrm{\pi }-0\right) \\ =\frac{3}{{\epsilon }_0}\left(\frac{Q^2}{16{\pi }^2{R}^{16}}\right)\left(\frac{R^4}{4}\right)\left(\frac{1^2}{2}\right)\left(2\mathrm{\pi }\right) \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{3Q^2}{16R^2}\right) \end{gathered}$$

Hence, the force of the northern hemisphere is $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{3Q^2}{16R^2}\right)$.