A metal sphere of radius${}^{\mathbf{R}}$carries a total charge${}^Q$.What is the force

of repulsion between the "northern" hemisphere and the "southern" hemisphere?

Write the expression for the electrostatic force is,

$\mathbf{F}\mathbf{=}\frac{\mathbf{k}{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}$

Here,${}^{q_1}$is the first charge, ${}^{q_2}$is the second charge,${}^k$is the Coulombs constant and${}^r$is the separation between the two charges.

The electric field can be expressed as,

$\mathbf{E}\mathbf{=}\frac{\mathbf{kq}}{{\mathbf{r}}^{\mathbf{2}}}$

Here, ${}^q$is charge.

The force acting on the sphere's surface when it has a surface charge density is as follows:

$\mathbf{F}\mathbf{=}\mathbf{\sigma E}$

Here,localid="1654326663841" ${}^{\sigma }$is the surface charge density,localid="1654326681149" ${}^{\mathrm{E}}$is normal to the surface of the sphere.

The below figure describes a metal sphere with a radiuslocalid="1654326701219" ${}^{\mathrm{R}}$and a total chargelocalid="1654326716375" ${}^{\mathrm{Q}}$:

Here,localid="1654326727026" ${}^X$is the horizontal axis,localid="1654326740031" ${}^{\mathrm{Y}}$is the vertical axis,localid="1654326828096" ${}^{\mathrm{E}}$is the electric field vector and localid="1654325912944" ${}^{\theta }$is an angle between the vertical axes.

Write the expression for electric filed inside the sphere is,

$$\begin{gathered} {\mathrm{E}}_{\mathrm{in}}=\frac{\mathrm{k}\left(0.0\mathrm{C}\right)}{{\mathrm{r}}^2} \\ =0.00N/C \end{gathered}$$

Therefore, the electric filed inside the sphere is $E_{in}=0.00N/C$

Now, consider that a charge ${}^Q$on the sphere's surface, the electric field outside the sphere may be calculated by replacing${}^Q$for${}^q$in the equation.${}^{E=\frac{kq}{r^2}}$

$E_{out}=\frac{kQ}{R^2}$

The sphere's electric field is ${}^E$and the hemisphere is half of the sphere, the hemisphere's electric field is given as, ${}^{\frac{1}{2}E}$.

Thus, the expression will be,

$E_{hemi}=\frac{1}{2}E$

Therefore, the electric field of the hemisphere is obtained by substituting ${}^{\frac{kQ}{R^2}}$for $E$in the equation.${}^{E_{hemi}=\frac{1}{2}E}$

$$\begin{gathered} E_{hemi}=\frac{1}{2}\left(\frac{kQ}{R^2}\right) \\ =\frac{kQ}{2R^2} \end{gathered}$$

The direction of the electric field will be radially outward because it is normal to the sphere's surface. That is,

The force is defined as the product of the surface charge density and the electric field, it is given by.

${\mathrm{F}}_{\mathrm{Z}}=\mathrm{\sigma }{\overrightarrow{\mathrm{E}}}_{\mathrm{hemi}}$

On the basis of charge and sphere radius write the expression for the surface charge density:

$\sigma =\frac{Q}{4\pi R^2}$

Also write the expression for force in the ${}^Z$direction,

$d{F}_Z=F_Z\cos \theta$

$$\begin{gathered} F_z=\iint \left(\frac{Q}{4\pi R^2}\right)\left(\frac{1}{4\pi {\epsilon }_0}\right)\frac{Q}{2R^2}R\cos \theta dA \\ ={\int }_0^{\frac{\pi }{2}}\left(\left(\frac{Q}{4\pi R^2}\right)\left(\frac{1}{4\pi {\epsilon }_0}\right)\frac{Q}{2R^2}\right)R^2\cos \theta \sin \theta d\theta {\int }_0^{2\pi }d\varphi \\ =2\pi \left(\frac{1}{2}\right){\left(\frac{Q}{4\pi }\right)}^2\left(\frac{1}{{\epsilon }_0{R}^2}\right){\int }_0^{\frac{\pi }{2}}\cos \theta \sin \theta d\theta \\ =\frac{Q^2}{16\pi {\epsilon }_0{R}^2}\left(\frac{1}{2}\right) \end{gathered}$$

Solve further as,

$F_Z=\frac{Q^2}{32\pi {\epsilon }_0{R}^2}$

Therefore, the force of repulsion between the northern and the southern hemisphere is

$\frac{Q^2}{32\pi {\epsilon }_0{R}^2}$