91影视

Consider the figure for the given condition as shown in figure below:

From the figure write the equation as:

Consider the equations as:

$$\begin{gathered} r=\sqrt{r^2+a^2-2ra\cos 胃} \\ {r}_1=\sqrt{r^2+b^2-2rb\cos 胃} \end{gathered}$$

Since, $b=\frac{R^2}{a}$, $q^1=\frac{-R}{a}q$and $\frac{q^1}{r_1}=\frac{-R}{a}\frac{q}{r_1}$.

Write the equation as:

$\frac{q^1}{r_1}=\frac{-R}{a}\frac{q}{\sqrt{r^2+b^2-2rb\cos 胃}}$

Consider the equation for the potential for the condition as follows:

$V\left(r\right)=\frac{1}{4蟺{蔚}_0}\left(\frac{q}{r}+\frac{q^1}{r_1}\right)$

Substitute the values and rewrite the equation for the potential as:

$\begin{array}{rcl}V\left(r\right)& =& \frac{1}{4蟺{蔚}_0}\left(\frac{q}{\sqrt{r^2+a^2-2ra\cos 胃}}+\frac{-R}{a}\frac{q}{\sqrt{r^2+{\left(\frac{R^2}{a}\right)}^2-2r\left(\frac{R^2}{a}\right)\cos 胃}}\right)\\ & =& \frac{q}{4蟺{蔚}_0}\left(\frac{1}{\sqrt{r^2+a^2-2ra\cos 胃}}-\frac{1}{\sqrt{r^2+a^2-2ra\cos 胃}}\right)\\ & =& 0\\ & & \end{array}$

Therefore, it is proved that the equation of the potential is 0 for $R=r$.

Consider the formula for the induced surface charge density as:

$蟽\left(胃\right)=-{蔚}_0\frac{鈭�V}{鈭�n}$

Since, $\frac{鈭�V}{鈭�n}=\frac{鈭�V}{鈭�r}$for the value of requal R.

Substitute the values and solve as:

$\begin{array}{rcl}蟽\left(胃\right)& =& -{蔚}_0\frac{鈭�}{鈭�n}\left(\frac{q}{4蟺{蔚}_0}\left(\frac{1}{\sqrt{r^2+a^2-2ra\cos 胃}}-\frac{1}{\sqrt{R^2+{\left(\frac{ar}{R}\right)}^2-2ra\cos 胃}}\right)\right)\\ & =& -{蔚}_0\frac{鈭�}{鈭�n}{\left(\frac{q}{4蟺{蔚}_0}{\left(r^2+a^2-2ra\cos 胃\right)}^{\frac{3}{2}}\left(2r-2a\cos 胃\right)+\frac{1}{2}{\left(R^2+{\left(\frac{ar}{R}\right)}^2-2ra\cos 胃\right)}^{\frac{3}{2}}\left({\left(\frac{a}{R}\right)}^{2}2r-2a\cos 胃\right)\right)}_{r=R}\\ & =& \frac{q}{4蟺{蔚}_0}{\left(R^2+a^2-2Ra\cos 胃\right)}^{\frac{-3}{2}}\left(R^2-a^2\right)\\ & & \end{array}$

Consider the expression for the induced surface charge as:

$q_{ind}=鈭�蟽da$

Substitute the values and solve as:

$\begin{array}{rcl}{q}_{ind}& =& {鈭�}_0^{2蟺}{鈭�}_0^{蟺}\frac{q}{4蟺R}{\left(R^2+a^2-2Ra\cos 胃\right)}^{\frac{-3}{2}}\left(R^2-a^2\right)R^2\sin 胃d胃d蠒\\ & =& \frac{q}{4蟺R}\left(R^2-a^2\right)2蟺R^2{\left(-\frac{1}{Ra}{\left(R^2+a^2-2Ra\cos 胃\right)}^{\frac{-1}{2}}\right)}_0^{蟺}\\ & =& \frac{q}{2a}\left(a^2-R^2\right)\left(\frac{1}{\sqrt{R^2+a^2+2Ra}}-\frac{1}{\sqrt{R^2+a^2-2Ra}}\right)\\ & & \end{array}$

Consider for $a>R$rewrite the equation as:

$\begin{array}{rcl}{q}_{ind}& =& \frac{q}{2a}\left(a^2-R^2\right)\left(\frac{1}{a+R}-\frac{1}{a-R}\right)\\ & =& -\frac{q}{a}R\\ & =& q^1\\ & & \end{array}$

Therefore, the induced charge on the surface is q¹.

Consider the formula for the force of image charge q¹on the charge q is obtained as:

$F=\frac{1}{4蟺{蔚}_0}\frac{q{q}^1}{{\left(a-b\right)}^2}$

Substitute the values and solve for the force as:

$\begin{array}{rcl}F& =& \frac{1}{4蟺{蔚}_0}\frac{q\left(-\frac{R}{a}q\right)}{{\left(a-\frac{R^2}{a}\right)}^2}\\ & =& \frac{1}{4蟺{蔚}_0}\frac{q^{2}Ra}{{\left(a^2-R^2\right)}^2}\\ & & \end{array}$

Solve for the work done as follows:

$\begin{array}{rcl}W& =& {鈭�}_{鈭�}^a\frac{1}{4蟺{蔚}_0}\frac{q^{2}Ra}{{\left(a^2-R^2\right)}^2}da\\ & =& -\frac{q^{2}R}{4蟺{蔚}_0}{\left(\frac{-1}{2\left(a^2-R^2\right)}\right)}_{鈭�}^a\\ & =& \frac{q^{2}R}{8蟺{蔚}_0\left(a^2-R^2\right)}\\ & & \end{array}$

Therefore, the energy of the configuration is $\frac{{\mathrm{q}}^2\mathrm{R}}{8{\mathrm{蟺蔚}}_0({\mathrm{a}}^2-{\mathrm{R}}^2)}$.