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 R and the total charge 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,

role="math" localid="1657517331409" $\mathit{V}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}\mathit{\pi }{\mathit{R}}^{\mathbf{3}}$ …… (2)

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

Thus,

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

Here, R is the radius of the sphere.

Assume that, a point $\mathit{r}\mathbf{<}\mathit{R}$,

Consider the radius of the Gaussian sphere is $\mathit{r}$then its volume is expressed as,

$\mathit{V}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}\mathit{\pi }{\mathit{r}}^{\mathbf{3}}$ …… (4)

Now, Charge in shell is expressed as,

$\mathit{d}\mathit{Q}\mathbf{=}\mathit{\rho }\mathit{v}$

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

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

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

$\mathbf{\int }\boxed{\xcancel{}}\mathbf{}\mathit{E}\mathbf{·}\mathit{d}\mathit{a}\mathbf{=}\frac{\mathbf{dQ}}{{\mathbf{\epsilon }}_{\mathbf{0}}}$

Substitute $\frac{{\mathbf{Qr}}^{\mathbf{3}}}{{\mathbf{R}}^{\mathbf{3}}}$for $\mathit{d}\mathit{Q}$in $\frac{\mathbf{dQ}}{{\mathbf{\epsilon }}_{\mathbf{0}}}$for $\mathbf{\int }\boxed{\xcancel{}}\mathbf{}\mathit{E}\mathbf{·}\mathit{d}\mathit{a}$expression.

$$\begin{gathered} \mathbf{\int }\boxed{\xcancel{}}\mathbf{}\mathit{E}\mathbf{·}\mathit{d}\mathit{a}\mathbf{=}\frac{{\displaystyle \frac{{\mathbf{Qr}}^{\mathbf{3}}}{{\mathbf{R}}^{\mathbf{3}}}}}{{\mathbf{\epsilon }}_{\mathbf{0}}} \\ \mathit{E}\mathbf{\left(}\mathbf{4}{\mathbf{\pi r}}^{\mathbf{2}}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{Qr}}^{\mathbf{3}}}{{\mathbf{R}}^{\mathbf{3}}{\mathbf{\epsilon }}_{\mathbf{0}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{E}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{r}}{{\mathbf{R}}^{\mathbf{3}}} \end{gathered}$$

Thus, the electric filed inside the sphere is$\mathit{E}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{r}}{{\mathbf{R}}^{\mathbf{3}}}$.

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

$\mathit{f}\mathbf{=}\mathit{\rho }\mathit{E}$ …… (4)

Substitute the value $\frac{\mathbf{Q}}{{\displaystyle \frac{\mathbf{4}}{\mathbf{3}}{\mathbf{\pi R}}^{\mathbf{3}}}}$for $\mathit{\rho }$and role="math" localid="1657518360270" $\frac{\mathbf{Q}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{r}}{{\mathbf{R}}^{\mathbf{3}}}$for $\mathit{E}$in equation (4),

$$\begin{gathered} \mathit{f}\mathbf{=}\frac{\mathbf{Q}}{{\displaystyle \frac{\mathbf{4}}{\mathbf{3}}{\mathbf{\pi R}}^{\mathbf{3}}}}\mathbf{\left(}\frac{\mathbf{Q}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{r}}{{\mathbf{R}}^{}}\mathbf{\right)} \\ \mathbf{}\mathbf{=}\frac{\mathbf{3}}{{\mathbf{\epsilon }}_{\mathbf{0}}}{\mathbf{\left(}\frac{\mathbf{Q}}{\mathbf{4}{\mathbf{\pi R}}^{\mathbf{3}}}\mathbf{\right)}}^{\mathbf{2}}\mathit{r} \end{gathered}$$

Therefore, the force per unit volume acting on the sphere is $\mathit{f}\mathbf{=}\frac{\mathbf{3}}{{\mathbf{\epsilon }}_{\mathbf{0}}}{\mathbf{\left(}\frac{\mathbf{Q}}{\mathbf{4}{\mathbf{\pi R}}^{\mathbf{3}}}\mathbf{\right)}}^{\mathbf{2}}\mathit{r}$.

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

$\mathit{d}\mathit{\zeta }\mathbf{=}{\mathit{r}}^{\mathbf{2}}\mathbf{sin}\mathit{\theta }\mathit{d}\mathit{r}\mathit{d}\mathit{\theta }\mathit{d}\mathit{\varphi }$

By using the symmetry net force in the z on the$\mathit{d}\mathit{\zeta }$,

$\mathit{d}\mathit{\zeta }\mathbf{=}\mathit{f}\mathbf{}\mathbf{cos}\hat{\mathbf{\theta }}\mathit{z}$ …… (5)

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

$\mathbf{\int }\mathit{d}\mathit{F}\mathbf{=}\underset{\mathbf{0}}{\overset{\mathbf{R}}{\mathbf{\int }}}\underset{\mathbf{0}}{\overset{\mathbf{2}\mathbf{\pi }}{\mathbf{\int }}}\underset{\mathbf{0}}{\overset{\mathbf{\pi }\mathbf{/}\mathbf{2}}{\mathbf{\int }}}\mathbf{}\mathit{f}\mathbf{}\mathbf{cos}\mathit{\theta }\mathit{d}\mathit{\zeta }$ …… (6)

Substitute $\frac{\mathbf{3}}{{\mathbf{\epsilon }}_{\mathbf{0}}}{\mathbf{\left(}\frac{\mathbf{Q}}{\mathbf{4}{\mathbf{\pi R}}^{\mathbf{3}}}\mathbf{\right)}}^{\mathbf{2}}\mathit{r}$for $\mathit{f}$and ${\mathit{r}}^{\mathbf{2}}\mathbf{sin}\mathit{\theta }\mathit{d}\mathit{r}\mathit{d}\mathit{\theta }\mathit{d}\mathit{\varphi }$for $\mathit{d}\mathit{\zeta }$in equation (6).

${\mathit{f}}_{\mathbf{z}}\mathbf{=}\mathbf{\left(}\frac{\mathbf{3}}{{\mathbf{\epsilon }}_{\mathbf{0}}}{\mathbf{\left(}\frac{\mathbf{Q}}{\mathbf{4}{\mathbf{\pi R}}^{\mathbf{3}}}\mathbf{\right)}}^{\mathbf{2}}\mathbf{\right)}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{R}}{\mathit{r}}^{\mathbf{3}}\mathit{d}\mathit{r}{\mathbf{\int }}_{\mathbf{0}}^{\frac{\mathbf{\pi }}{\mathbf{2}}}\mathbf{sin}\mathit{\theta }\mathbf{cos}\mathit{\theta }\mathit{d}\mathit{\theta }{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{2}\mathbf{\pi }}\mathit{d}\mathit{\varphi }$ …… (7)

Let’s assume that, $\mathbf{sin}\mathit{\theta }\mathbf{=}\mathit{t}$then$\mathbf{cos}\mathit{\theta }\mathit{d}\mathit{\theta }\mathbf{=}\mathit{d}\mathit{t}$.

Substitute these values in equation (7), and simplify

$$\begin{gathered} {\mathit{f}}_{\mathbf{z}}\mathbf{=}\left(\frac{3}{{\epsilon }_0}{\left(\frac{Q}{4\pi R^3}\right)}^2\right){\mathbf{\int }}_{\mathbf{0}}^{\mathbf{R}}{\mathit{r}}^{\mathbf{3}}\mathit{d}\mathit{r}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{1}}\mathit{t}\mathit{d}\mathit{t}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{2}\mathbf{x}}\mathit{d}\mathit{\varphi } \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{3}}{{\mathbf{\epsilon }}_{\mathbf{0}}}{\left(\frac{Q}{4\pi R^3}\right)}^{\mathbf{2}}\left(\frac{R^4}{4}-0\right)\left(\frac{t^2}{2}-0\right)\left(2\pi -0\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{3}}{{\mathbf{\epsilon }}_{\mathbf{0}}}\left(\frac{Q^2}{16{\pi }^2{R}^6}\right)\left(\frac{R^4}{4}\right)\left(\frac{1^2}{2}\right)\left(2\pi \right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\left(\frac{3Q^2}{16R^6}\right) \end{gathered}$$

Hence, the force of the northern hemisphere is$\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{\left(}\frac{\mathbf{3}{\mathbf{Q}}^{\mathbf{2}}}{\mathbf{16}{\mathbf{R}}^{\mathbf{2}}}\mathbf{\right)}$.