魅影直播

Write the expression for potential using law of cosines.

$\mathbf{\text{痴(谤,胃)=}}\frac{\mathbf{1}}{\mathbf{4}\mathbf{蟺}{\mathbf{蔚}}_{\mathbf{0}}}\left[\frac{q}{\sqrt{r^2+a^2鈭�2racos鈦�胃}}鈭�\frac{q}{\sqrt{R^2+(ra/R{)}^2鈭�2racos鈦�胃}}\right]$ 鈥︹�� (1)

Here, role="math" localid="1657372493931" $\mathrm{r}$and $\mathrm{胃}$are the spherical polar coordinates, role="math" localid="1657372543438" $\mathrm{q}$is the charge, ${\mathrm{蔚}}_0$is the permittivity for the free space and $\mathrm{R}$is the radius of the sphere.

a) From the figure given below we can write the equation as,

$\mathrm{蟽}\left(\mathrm{胃}\right)=鈭�{\mathrm{蔚}}_0\frac{\mathrm{q}}{4{\mathrm{蟺蔚}}_0}{\left(\begin{array}{l}鈭�\frac{1}{2}{\left({\mathrm{r}}^2+{\mathrm{a}}^2鈭�2\mathrm{racos}鈦�\mathrm{胃}\right)}^{鈭�\frac{3}{2}}(2\mathrm{r}鈭�2\mathrm{acos}鈦�\mathrm{胃})+\\ \frac{1}{2}{\left({\mathrm{R}}^2+{\left(\frac{\mathrm{ra}}{\mathrm{R}}\right)}^2鈭�2\mathrm{racos}鈦�\mathrm{胃}\right)}^{鈭�\frac{3}{2}}\left(\frac{{\mathrm{a}}^2}{{\mathrm{R}}^2}2\mathrm{r}鈭�2\mathrm{acos}鈦�\mathrm{胃}\right)\end{array}\right)}_{\mathrm{r}=\mathrm{R}}$鈥︹�� (2)

Write the equation for $\mathrm{r}$

$\mathrm{r}=\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2鈭�2\mathrm{racos}鈦�\mathrm{胃}}$

${\mathrm{r}}_1=\sqrt{{\mathrm{r}}^2+{\mathrm{b}}^2鈭�2\mathrm{rbcos}鈦�\mathrm{胃}}$

As we know that,

$\mathrm{b}=\frac{{\mathrm{R}}^2}{\mathrm{a}}$

${\mathrm{q}}^1=\frac{鈭�\mathrm{R}}{\mathrm{a}}\mathrm{q}$ 鈥︹��. (3)

$\frac{{\mathrm{q}}^1}{{\mathrm{r}}_1}=\frac{鈭�\mathrm{R}}{\mathrm{a}}\frac{\mathrm{q}}{{\mathrm{r}}_1}$

Substitute $\sqrt{{\mathrm{r}}^2+{\mathrm{b}}^2鈭�2\mathrm{rbcos}鈦�\mathrm{胃}}$for ${\mathrm{r}}_1$in equation (3).

$\frac{{\mathrm{q}}^1}{{\mathrm{r}}_1}=\frac{鈭�\mathrm{R}}{\mathrm{a}}\frac{\mathrm{q}}{\sqrt{{\mathrm{r}}^2+{\mathrm{b}}^2鈭�2\mathrm{rbcos}鈦�\mathrm{胃}}}$ 鈥︹�� (4)

Now, write the expression for the potential for this configuration.

$\mathrm{V}\left(\mathrm{r}\right)=\frac{1}{4{\mathrm{蟺蔚}}_0}\left(\frac{\mathrm{q}}{\mathrm{r}}+\frac{{\mathrm{q}}^1}{{\mathrm{r}}_1}\right)$鈥︹��. (5)

Substitute $\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2鈭�2\mathrm{racos}鈦�\mathrm{胃}}$for $\mathrm{r},\sqrt{{\mathrm{r}}^2+{\mathrm{b}}^2鈭�2\mathrm{rbcos}鈦�\mathrm{胃}}$for${\mathrm{r}}_1$and$\frac{鈭�\mathrm{R}}{\mathrm{a}}\mathrm{q}$for${\mathrm{q}}^1$

$\mathrm{V}\left(\mathrm{r}\right)=\frac{1}{4{\mathrm{蟺蔚}}_0}\left(\frac{\mathrm{q}}{\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2鈭�2\mathrm{racos}鈦�\mathrm{胃}}}+\frac{{\mathrm{a}}_{\mathrm{q}}}{\sqrt{{\mathrm{r}}^2+{\mathrm{b}}^2鈭�2\mathrm{rbcos}鈦�\mathrm{胃}}}\right)$ 鈥︹赌�(6)

Substitute $\frac{{\mathrm{R}}^2}{\mathrm{a}}$for $\mathrm{b}$.

$\mathrm{V}\left(\mathrm{r}\right)=\frac{1}{4{\mathrm{蟺蔚}}_0}\left(\frac{\mathrm{q}}{\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2鈭�2\mathrm{racos}鈦�\mathrm{胃}}}+\frac{\frac{鈭�\mathrm{R}}{\mathrm{a}}\mathrm{q}}{\sqrt{{\mathrm{r}}^2+{\left(\frac{{\mathrm{R}}^2}{\mathrm{a}}\right)}^2鈭�2\mathrm{r}\left(\frac{{\mathrm{R}}^2}{\mathrm{a}}\right)\cos 鈦�\mathrm{胃}}}\right)$

$\mathrm{V}\left(\mathrm{r}\right)鈭�\frac{\mathrm{q}}{4{\mathrm{蟺蔚}}_0}\left(\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2鈭�2\mathrm{racos}鈦�\mathrm{胃}}}鈭�\frac{1}{\left.\frac{\mathrm{a}}{\mathrm{R}}\sqrt{{\mathrm{r}}^2+{\left(\frac{{\mathrm{R}}^2}{\mathrm{a}}\right)}^2鈭�2\mathrm{r}\left(\frac{{\mathrm{R}}^2}{\mathrm{a}}\right)\cos 鈦�\mathrm{胃}}\right)}\right.$

$\mathrm{V}\left(\mathrm{r}\right)=\frac{\mathrm{q}}{4{\mathrm{蟺蔚}}_0}\left(\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^{2}2\mathrm{racos}鈦�\mathrm{胃}}}鈭�\frac{1}{\left.\sqrt{{\left(\frac{\mathrm{a}}{\mathrm{R}}\right)}^2{\mathrm{r}}^2+{\left(\frac{\mathrm{a}}{\mathrm{R}}\right)}^2{\left(\frac{{\mathrm{R}}^2}{\mathrm{a}}\right)}^2鈭�2\mathrm{r}{\left(\frac{\mathrm{a}}{\mathrm{R}}\right)}^2\left(\frac{{\mathrm{R}}^2}{\mathrm{a}}\right)\cos 鈦�\mathrm{胃}}\right)}\right.$

$\mathrm{V}\left(\mathrm{r}\right)=\frac{\mathrm{q}}{4{\mathrm{蟺蔚}}_0}\left(\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2鈭�2\mathrm{racos}鈦�\mathrm{胃}}}鈭�\frac{1}{\sqrt{{\left.\frac{\mathrm{a}}{\mathrm{R}}\right)}^2{\mathrm{r}}^2+(\mathrm{R}{)}^2鈭�2\mathrm{racos}鈦�\mathrm{胃}}}\right)$

Rearranging the statement,

$\mathrm{V}\left(\mathrm{r}\right)=\frac{\mathrm{q}}{4{\mathrm{蟺蔚}}_0}\left(\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2鈭�2\mathrm{racos}鈦�\mathrm{胃}}}鈭�\frac{1}{\sqrt{{\mathrm{R}}^2+{\left(\frac{\mathrm{ar}}{\mathrm{R}}\right)}^2鈭�2\mathrm{racos}鈦�\mathrm{胃}}}\right)$

The potential is zero, when $\mathrm{r}=\mathrm{R}$.

$\mathrm{V}\left(\mathrm{r}\right)=\frac{1}{4{\mathrm{蟺蔚}}_0}\left(\frac{\mathrm{q}}{\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2鈭�2\mathrm{racos}鈦�0}}+\frac{\frac{鈭�\mathrm{R}}{\mathrm{a}}\mathrm{q}}{\sqrt{{\mathrm{r}}^2+{\mathrm{b}}^2鈭�2\mathrm{rbcos}鈦�0}}\right)$ 鈥︹�� (7)

$=0$

Hence, the potential is zero, when $\mathrm{r}=\mathrm{R}$.

b)

Write the expression for the induced surface charge density.

$\mathrm{蟽}\left(\mathrm{胃}\right)={\mathrm{蔚}}_0\frac{\mathrm{鈭�}\mathrm{V}}{\widehat{\mathrm{o}}\mathrm{n}}$ 鈥︹��. (8)

But $\frac{\mathrm{鈭�}\mathrm{V}}{\mathrm{鈭�}\mathrm{n}}=\frac{\mathrm{鈭�}\mathrm{V}}{\mathrm{鈭�}\mathrm{r}}$at $\mathrm{r}=\mathrm{R}$

Substitute $\frac{\mathrm{q}}{4{\mathrm{蟺蔚}}_0}\left(\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2鈭�2\mathrm{racos}鈦�\mathrm{胃}}}鈭�\frac{1}{\sqrt{{\mathrm{R}}^2+{\left(\begin{array}{c}\mathrm{ar}\\ \mathrm{R}\end{array}\right)}^2鈭�2\mathrm{racos}鈦�\mathrm{胃}}}\right)$ for $\mathrm{V}$,

$\mathrm{蟽}\left(\mathrm{胃}\right)=鈭�{\mathrm{蔚}}_0\frac{\mathrm{鈭�}}{\mathrm{鈭�}\mathrm{n}}\left(\frac{\mathrm{q}}{4{\mathrm{蟺蔚}}_0}\left(\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2鈭�2\mathrm{racos}鈦�\mathrm{胃}}}鈭�\frac{1}{\sqrt{{\mathrm{R}}^2+{\left(\frac{\mathrm{ar}}{\mathrm{R}}\right)}^2鈭�2\mathrm{racos}鈦�\mathrm{胃}}}\right)\right)$

$=鈭�{\mathrm{蔚}}_0\frac{\mathrm{q}}{4{\mathrm{蟺蔚}}_0}{\left(\begin{array}{l}鈭�\frac{1}{2}{\left({\mathrm{r}}^2+{\mathrm{a}}^2鈭�2\mathrm{racos}鈦�\mathrm{胃}\right)}^{鈭�\frac{3}{2}}(2\mathrm{r}鈭�2\mathrm{acos}鈦�\mathrm{胃})\\ +\frac{1}{2}{\left({\mathrm{R}}^2+{\left(\frac{\mathrm{ra}}{\mathrm{R}}\right)}^2鈭�2\mathrm{racos}鈦�\mathrm{胃}\right)}^{鈭�\frac{3}{2}}\left(\frac{{\mathrm{a}}^2}{{\mathrm{R}}^2}2\mathrm{r}鈭�2\mathrm{acos}鈦�\mathrm{胃}\right)\end{array}\right)}_{\mathrm{r}鈭�\mathrm{R}}$

$=\frac{鈭�\mathrm{q}}{4\mathrm{蟺}}\left(\begin{array}{l}鈭�{\left({\mathrm{R}}^2+{\mathrm{a}}^2鈭�2\mathrm{Racos}鈦�\mathrm{胃}\right)}^{\frac{3}{2}}(\mathrm{R}鈭�\mathrm{acos}鈦�\mathrm{胃})\\ +{\left({\mathrm{R}}^2+{\mathrm{a}}^2鈭�2\mathrm{Racos}鈦�\mathrm{胃}\right)}^{\frac{3}{2}}\left(\frac{{\mathrm{a}}^2}{\mathrm{R}}鈭�\mathrm{acos}鈦�\mathrm{胃}\right)\end{array}\right)$

Write induced surface charge formula.

${\mathrm{q}}_{\mathrm{irvi}}=鈭�\mathrm{蟽诲补}$ 鈥︹�� (9)

Substitute $\frac{\mathrm{q}}{4\mathrm{蟺搁}}{\left({\mathrm{R}}^2+{\mathrm{a}}^2鈭�2\mathrm{Racos}鈦�\mathrm{胃}\right)}^{鈭�\frac{3}{2}}\left({\mathrm{R}}^2鈭�{\mathrm{a}}^2\right)$for $\mathrm{蟽}$in equation (9).

${\mathrm{q}}_{\mathrm{in}}={鈭�}_0^{2\mathrm{蟺}}鈥�{鈭�}_0^{\mathrm{蟺}}鈥�\frac{\mathrm{q}}{4\mathrm{蟺搁}}{\left({\mathrm{R}}^2+{\mathrm{a}}^2鈭�2\mathrm{Racos}鈦�\mathrm{胃}\right)}^{鈭�\frac{3}{2}}\left({\mathrm{R}}^2鈭�{\mathrm{a}}^2\right){\mathrm{R}}^2\sin 鈦�\mathrm{胃诲胃诲蠒}$

$=\frac{\mathrm{q}}{4\mathrm{蟺搁}}\left({\mathrm{R}}^2鈭�{\mathrm{a}}^2\right)2{\mathrm{蟺搁}}^2{鈭�}_0^2鈥�{\left({\mathrm{R}}^2+{\mathrm{a}}^2鈭�2\mathrm{Racos}鈦�\mathrm{胃}\right)}^{鈭�\frac{3}{2}}\sin 鈦�\mathrm{胃诲胃}$

$=\frac{\mathrm{q}}{4\mathrm{蟺搁}}\left({\mathrm{R}}^2鈭�{\mathrm{a}}^2\right)2{\mathrm{蟺搁}}^2{\left(\frac{鈭�1}{\mathrm{Ra}}{\left({\mathrm{R}}^2+{\mathrm{a}}^2鈭�2\mathrm{Racos}鈦�\mathrm{胃}\right)}^{鈭�\frac{1}{2}}\right)}_0^{\mathrm{蟺}}$

$=\frac{\mathrm{q}}{2\mathrm{a}}\left({\mathrm{R}}^2鈭�{\mathrm{a}}^2\right)(鈭�1)\left({\left({\mathrm{R}}^2+{\mathrm{a}}^2鈭�2\mathrm{Racos}鈦�\left(\mathrm{蟺}\right)鈭�\left({\mathrm{R}}^2+{\mathrm{a}}^2鈭�2\mathrm{Racos}鈦�\left(0\right)\right)\right)}^{鈭�\frac{1}{2}}\right)$

Further solving,

${\mathrm{q}}_{\mathrm{in}}=\frac{\mathrm{q}}{2\mathrm{a}}\left({\mathrm{R}}^2鈭�{\mathrm{a}}^2\right)(鈭�1){\left({\mathrm{R}}^2+{\mathrm{a}}^2鈭�2\mathrm{Ra}(鈭�1)鈭�{\mathrm{R}}^2+{\mathrm{a}}^2鈭�2\mathrm{Ra}\left(1\right)\right)}^{\frac{鈭�1}{2}}$

$=\frac{\mathrm{q}}{2\mathrm{a}}\left({\mathrm{a}}^2鈭�{\mathrm{R}}^2\right){\left(\left({\mathrm{R}}^2+{\mathrm{a}}^2+2\mathrm{Ra}\right)鈭�\left({\mathrm{R}}^2+{\mathrm{a}}^2鈭�2\mathrm{Ra}\right)\right)}^{\frac{鈭�1}{2}}$

$=\frac{\mathrm{q}}{2\mathrm{a}}\left({\mathrm{a}}^2鈭�{\mathrm{R}}^2\right)\left(\frac{1}{\sqrt{{\mathrm{R}}^2+{\mathrm{a}}^2+2\mathrm{Ra}}}鈭�\frac{1}{\sqrt{{\mathrm{R}}^2+{\mathrm{a}}^2鈭�2\mathrm{Ra}}}\right)$ 鈥︹�� (10)

For$\mathrm{a}>\mathrm{R}$ ,

$\sqrt{{\mathrm{R}}^2+{\mathrm{a}}^{2}2\mathrm{Ra}}=\sqrt{(\mathrm{arR}{)}^2}$

$=\mathrm{a}-\mathrm{R}$

Substitute $\mathrm{a}-\mathrm{R}$for $\sqrt{{\mathrm{R}}^2+{\mathrm{a}}^2鈭�2\mathrm{Ra}}$and $\mathrm{a}+\mathrm{R}$for $\sqrt{{\mathrm{R}}^2+{\mathrm{a}}^2+2\mathrm{Ra}}$ in equation (10).

Therefore,

${\mathrm{q}}_{\mathrm{in}}=\frac{\mathrm{q}}{2\mathrm{a}}\left({\mathrm{a}}^2鈭�{\mathrm{R}}^2\right)\left(\frac{1}{\mathrm{a}+\mathrm{R}}鈭�\frac{1}{\mathrm{a}鈭�\mathrm{R}}\right)$

$=\frac{\mathrm{q}}{2\mathrm{a}}\left[(\mathrm{a}+\mathrm{R})(\mathrm{a}鈭�\mathrm{R})\left(\frac{1}{\mathrm{a}+\mathrm{R}}鈭�\frac{1}{\mathrm{a}鈭�\mathrm{R}}\right)\right]$

$=\frac{\mathrm{q}}{2\mathrm{a}}\left[\left(\frac{(\mathrm{a}+\mathrm{R})(\mathrm{a}鈭�\mathrm{R})}{\mathrm{a}鈭�\mathrm{R}}鈭�\frac{(\mathrm{a}+\mathrm{R})(\mathrm{a}鈭�\mathrm{R})}{\mathrm{a}鈭�\mathrm{R}}\right)\right]$

$=\frac{\mathrm{q}}{2\mathrm{a}}\left[(\mathrm{a}鈭�\mathrm{R})鈭�(\mathrm{a}+\mathrm{R}{)}_{鈭�}^{鈭�}\right.$

Solve as further,

${\mathrm{q}}_{\mathrm{in}}=\frac{\mathrm{q}}{2\mathrm{a}}(鈭�2\mathrm{R})$

$=\frac{\mathrm{q}}{\mathrm{a}}\mathrm{R}$

$={\mathrm{q}}^1$

Hence, the induced surface charge is ${\mathrm{q}}^1$.

c)

Write the expression for the force of image charge ${\mathrm{q}}^1$on $\mathrm{q}$.

$\mathrm{F}=\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{{\mathrm{qq}}^1}{(\mathrm{a}鈭�\mathrm{b}{)}^2}$鈥︹�� (11)

Substitute $\frac{鈭�\mathrm{R}}{\mathrm{a}}\mathrm{q}$for ${\mathrm{q}}^1$and $\frac{{\mathrm{R}}^2}{\mathrm{a}}$for $\mathrm{b}$in equation (11).

$\mathrm{F}=\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{\mathrm{q}\left(\frac{鈭�\mathrm{R}}{\mathrm{a}}\mathrm{q}\right)}{{\left(\begin{array}{c}\mathrm{a}鈭�{\mathrm{R}}^2\\ \mathrm{a}\end{array}\right)}^2}$

$=\frac{1}{4{\mathrm{蟺锄}}_0}\frac{{\mathrm{q}}^2\mathrm{Ra}}{{\left({\mathrm{a}}^2鈭�{\mathrm{R}}^2\right)}^2}$

Thus, the force of image charge ${\mathrm{q}}^1$on $\mathrm{q}$is $\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{{\mathrm{q}}^2\mathrm{Ra}}{{\left({\mathrm{a}}^2鈭�{\mathrm{R}}^2\right)}^2}$.

Write the expression for work done.

$\mathrm{W}={鈭�}_{\mathrm{a}}^{\mathrm{A}}鈥�\mathrm{Fda}$ 鈥︹�� (12)

Substitute $\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{{\mathrm{q}}^2\mathrm{Ra}}{{\left({\mathrm{a}}^2鈭�{\mathrm{R}}^2\right)}^2}$for $\mathrm{F}$in equation (12)

$\mathrm{W}={鈭�}_{\mathrm{a}}^{\mathrm{a}}鈥�鈭�\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{{\mathrm{q}}^2\mathrm{Ra}}{{\left({\mathrm{a}}^2鈭�{\mathrm{R}}^2\right)}^2}\mathrm{da}$

$=鈭�\frac{{\mathrm{q}}^2\mathrm{R}}{4{\mathrm{蟺蔚}}_0}{\left(\frac{鈭�1}{2\left({\mathrm{a}}^2鈭�{\mathrm{R}}^2\right)}\right)}^{\mathrm{a}}$

$=\frac{{\mathrm{a}}^2\mathrm{R}}{8{\mathrm{蟺蔚}}_0\left({\mathrm{a}}^2鈭�{\mathrm{R}}^2\right)}$

Therefore, the energy of this configuration is$\frac{{\mathrm{a}}^2\mathrm{R}}{8{\mathrm{蟺蔚}}_0\left({\mathrm{a}}^2鈭�{\mathrm{R}}^2\right)}$