91Ó°ÊÓ

Determine the expression for the electric field with respect to the displacement as follows:

$\mathbf{Δ³Õ}\mathbf{=}\mathbf{-}∫\mathbf{E}â‹\dots \mathbf{ds}$

Consider the electric field is Eand the surface integral is ds.

Consider the expression for the force by the coulomb’s law as:

$\mathbf{F}\mathbf{=}\frac{{\mathbf{kq}}_1{\mathbf{q}}_2}{{\mathbf{r}}^2}$

Here, the two charges are$q_1$and $q_2$. While the distance between the two charge is $r$.

Consider the diagram for the conducting sphere of radius of R and the charge placed at the distance afrom it as shown below.

Here, the distance b is equal to the half of the chord of the circle.

Note that the point charge is placed outside the conducting surface and is at the distance afrom the centre. The image charge$q^{\text{'}}$ is at the lien that joins the ray on the ray emerging from the original charge q. In order to increase the potential of the sphere from the zero to${\mathrm{V}}_0$ place the second image charge at the centre the sphere.

The equation for the charge to be used is as follows:

$q^{\text{'}\text{'}}=4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}R{V}_0$

Consider the sphere is neutral when the charge enclosed is 0 and is given as:

$\begin{array}{c}{q}^{\text{'}\text{'}}+q^{\text{'}}=0\\ q^{\text{'}\text{'}}=−q^{\text{'}}\end{array}$

Solve for the force of attraction as:

role="math" localid="1658897960538"

Therefore, the force of attraction is$=\frac{q^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{\left(\frac{R}{a}\right)}^3\left(\frac{2a^2−R^2}{{(a^2−R^2)}^2}\right)$