Q42P - Introduction to Electrodynamics

Question

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?

Step-by-Step Solution

Answer

The force of repulsion between the northern and the southern hemisphere is $\frac{Q^2}{32\pi {\epsilon }_0{R}^2}$

Step 1: Determine the expression for electrostatic forces.

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,${}^{\sigma }$ is the surface charge density,${}^{\mathrm{E}}$is normal to the surface of the sphere.

The below figure describes a metal sphere with a radius${}^{\mathrm{R}}$and a total charge${}^{\mathrm{Q}}$:

Here,${}^X$is the horizontal axis,${}^{\mathrm{Y}}$ is the vertical axis,${}^{\mathrm{E}}$ is the electric field vector and ${}^{\theta }$ is an angle between the vertical axes.

Step 2: Determine electric field

Write the expression for electric filed inside the sphere is,

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

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

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,

Step 3: Determine the expression for force

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,

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

Step 4: Determine the total force on the northern hemisphere

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

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 

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