We know that the charge on a conductor goes to the surface, but just

how it distributes itself there is not easy to determine. One famous example in which the surface charge density can be calculated explicitly is the ellipsoid:

$\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{b}}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{z}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}\mathbf{=}\mathbf{1}$

In this case15

$\mathit{\sigma }\mathit{=}\frac{\mathit{Q}}{\mathit{4}\mathit{\pi }\mathit{a}\mathit{b}\mathit{c}}{\left(\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}=1\right)}^{\mathit{-}\mathit{1}\mathit{/}\mathit{2}}$

(2.57) where is the total charge. By choosing appropriate values for a,band c. obtain (from Eq. 2.57):

(a) the net (both sides) surface charge a (r)density on a circular disk of radius R;(b) the net surface charge density a (x) on an infinite conducting "ribbon" in the xyplane, which straddles theyaxis from x=-ato x=a(let A be the total charge per unit length of ribbon);

(c) the net charge per unit length $\mathit{\lambda }\mathit{\left(}\mathit{x}\mathit{\right)}$ on a conducting "needle," running from x= -ato x= a . In each case, sketch the graph of your result.

Write the equation for the surface charge density on an ellipsoid.

$\mathit{\sigma }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathit{=}\frac{\mathbf{Q}}{\mathbf{4}\mathbf{\pi }\mathbf{a}\mathbf{b}\mathbf{c}}{\mathbf{(}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{4}}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{b}}^{\mathbf{4}}}\mathbf{+}\frac{{\mathbf{z}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{4}}}\mathbf{)}}^{\frac{\mathbf{1}}{\mathbf{2}}}$ ........(1)

Here, is the total charge.

The equation for ellipsoid is as follows:

$\frac{{\mathit{x}}^{\mathit{2}}}{{\mathit{a}}^{\mathit{2}}}\mathit{+}\frac{{\mathit{y}}^{\mathit{2}}}{{\mathit{b}}^{\mathit{2}}}\mathit{+}\frac{{\mathit{z}}^{\mathit{2}}}{{\mathit{c}}^{\mathit{2}}}\mathit{=}\mathit{1}$

Then,

$\frac{{\mathit{z}}^{\mathit{2}}}{{\mathit{c}}^{\mathit{2}}}\mathit{=}\mathit{1}\frac{{\mathit{x}}^{\mathit{2}}}{{\mathit{a}}^{\mathit{2}}}\mathit{+}\frac{{\mathit{y}}^{\mathit{2}}}{{\mathit{b}}^{\mathit{2}}}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}$ ...........(2)

Now, put equation (2) in equation (1)

$\sigma \left(x,y,z\right)\mathit{=}\frac{\mathit{Q}}{\mathit{4}\mathit{\pi }\mathit{a}\mathit{b}\mathit{c}}{\left(\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{1^2}{c^2}\left(1-\frac{x^2}{a^2}-\frac{y^2}{b^2}\right)\right)}^{\frac{\mathit{1}}{\mathit{2}}}$

localid="1657350509463" $\sigma \mathit{(}x\mathit{,}y\mathit{,}z\mathit{)}\mathit{=}\frac{Q}{\mathit{4}\pi abc}{\left(\frac{1}{{\mathrm{c}}^2}\right)}^{\frac{-1}{2}}{\mathit{[}\left(\frac{x^2}{a^4}\right){\mathit{c}}^{\mathit{2}}\mathit{+}\left(\frac{y^2}{b^4}\right){\mathit{c}}^{\mathit{2}}\mathit{+}\left(1-\frac{x^2}{a^2}-\frac{y^2}{b^2}\right)\mathit{]}}^{-\frac{1}{2}}$ ........(3)

localid="1657349308362" $\sigma \mathit{(}x\mathit{,}y\mathit{,}z\mathit{)}\mathit{=}\frac{Q}{4\pi ab}{\mathit{[}\mathit{\left(}\frac{{\mathit{x}}^{\mathit{2}}}{{\mathit{a}}^{\mathit{4}}}\mathit{\right)}{c}^{\mathit{2}}\mathit{+}\mathit{\left(}\frac{{\mathit{y}}^{\mathit{2}}}{{\mathit{b}}^{\mathit{4}}}\mathit{\right)}{c}^{\mathit{2}}\mathit{+}\frac{\mathit{1}\mathit{-}{x}^{\mathit{2}}}{a^{\mathit{2}}}\mathit{-}\frac{y^{\mathit{2}}}{b^{\mathit{2}}}\mathit{]}}^{-\frac{1}{2}}$

a)

The net surface charge density of the circular disk can be calculated using the surface charge density of the ellipsoid.

The ellipsoid is reduced to the circular disk by setting the in equation (3).

Therefore,

$\sigma \left(x,y\right)=\frac{q}{4\mathrm{\pi ab}}{\left[1-\frac{x^2}{a^2}-\frac{y^2}{b^2}\right]}^{-\frac{1}{2}}$ ........(5)

For circular disk, $a=b=R$, and $r=\sqrt{x^2+y^2}$

Then,

$\sigma \mathit{\left(}r\mathit{\right)}\mathit{=}\frac{Q}{\mathit{2}\pi R^{\mathit{2}}}{\mathit{[}\mathit{1}\mathit{-}\frac{x^{\mathit{2}}\mathit{+}{y}^{\mathit{2}}}{R^{\mathit{2}}}\mathit{]}}^{-\frac{1}{2}}$

$\sigma \mathit{\left(}r\mathit{\right)}\mathit{=}\frac{Q}{\mathit{2}\pi R^{\mathit{2}}}{\mathit{[}\mathit{1}\mathit{-}\frac{R^{\mathit{2}}\mathit{-}{r}^{\mathit{2}}}{R^{\mathit{2}}}\mathit{]}}^{-\frac{1}{2}}$

$\sigma \mathit{\left(}r\mathit{\right)}\mathit{=}\frac{Q}{\mathit{2}\pi R^{\mathit{2}}}{\mathit{[}{R}^{\mathit{2}}\mathit{-}{r}^{\mathit{2}}\mathit{]}}^{-\frac{1}{2}}$

$\sigma \mathit{\left(}r\mathit{\right)}\mathit{=}\frac{Q}{\mathit{2}\pi R^{\mathit{2}}}\mathit{}\frac{\mathit{1}}{\sqrt{R^{\mathit{2}}\mathit{-}{r}^{\mathit{2}}}}$

Hence, the net surface charge density on the circular disk is $\sigma \mathit{\left(}r\mathit{\right)}\mathit{=}\frac{Q}{\mathit{2}\pi R^{\mathit{2}}}\mathit{}\frac{\mathit{1}}{\sqrt{R^{\mathit{2}}\mathit{-}{r}^{\mathit{2}}}}$.

The ${\sigma }_n\left(r\right)$is plotted below.

b)

The net surface charge density on the ribbon is calculated using the surface charge density of the ellipsoid.

The ellipsoid is reduced to the ribbon by setting $\mathrm{c}\to 0$the in equation (3).

Now, consider both the disk. Multiply them by 2. Thus, the net surface charge density is as follows:

localid="1657353507416" $\sigma \mathit{(}x\mathit{,}y\mathit{)}\mathit{=}\frac{q}{\mathit{4}\pi ab}{\mathit{[}\mathit{1}\mathit{-}\frac{x^{\mathit{2}}}{a^{\mathit{2}}}\mathit{-}\frac{y^{\mathit{2}}}{b^{\mathit{2}}}\mathit{]}}^{-\frac{1}{2}}$ ..........(6)

The infinite conducting ribbon straddles they-axis from x= -a to x=a.

Now, let’s consider that,localid="1657353903340" $\lambda$is the charge per unit length ribbon.

Then, set localid="1657353511750" $\frac{Q}{b}=\lambda$and take the limit localid="1657353517400" $b\to \infty$.

localid="1657353524783" $\sigma \left(x\right)=\underset{b\to \infty }{lim}{\sigma }_n\left(x,y\right)$

localid="1657353527832" $\sigma \left(x\right)=\frac{1}{2\mathrm{\pi a}}\frac{Q}{b}\underset{b\to \infty }{lim}{\left[1-\frac{x^2}{a^2}-\frac{y^2}{b^2}\right]}^{-\frac{1}{2}}$

localid="1657353532308" $\sigma \left(x\right)=\frac{\lambda }{2\mathrm{\pi a}}\frac{Q}{b}\underset{b\to \infty }{lim}{\left[1-\frac{x^2}{a^2}-\frac{y^2}{b^2}\right]}^{-\frac{1}{2}}$

Solve further.

localid="1657353536442" $\sigma \left(x\right)=\frac{\lambda }{2\mathrm{\pi a}}{\left[1-\frac{x^2}{a^2}\right]}^{-\frac{1}{2}}$

localid="1657353540375" $\sigma \left(x\right)=\frac{\lambda }{2\mathrm{\pi a}}{\left[\frac{1}{a^2}\right]}^{-\frac{1}{2}}{\left[a^2-x^2\right]}^{\frac{1}{2}}$

$\sigma \left(x\right)$is plotted as follows.

c)

Now, calculate the surface charge density of the conducting needle.

Let’s consider the equation of ellipsoid.,

$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$

Assume that b =cand $r=\sqrt{y^2+z^2}$.

$\frac{x^2}{a^2}+\frac{y^2}{c^2}+\frac{z^2}{c^2}=1$

$\frac{x^2}{a^2}+\frac{1}{c^2}\left[y^2+z^2\right]=1$

$\frac{x^2}{a^2}+\frac{1}{c^2}{r}^2=1$

This is an ellipsoid revolution.

Now, the charge density of the ellipsoid takes the following form:

$\sigma \mathit{(}\mathit{x}\mathit{,}\mathit{y}\mathit{,}\mathit{z}\mathit{)}\mathit{=}\frac{Q}{\mathit{4}\pi a{c}^{\mathit{2}}}{\mathit{[}\frac{{\mathit{x}}^{\mathit{2}}}{{\mathit{a}}^{\mathit{4}}}\mathit{+}\frac{{\mathit{y}}^{\mathit{4}}}{{\mathit{c}}^{\mathit{4}}}\mathit{+}\frac{{\mathit{z}}^{\mathit{2}}}{{\mathit{c}}^{\mathit{4}}}\mathit{]}}^{-\frac{1}{2}}$

$\sigma \mathit{(}\mathit{x}\mathit{,}\mathit{y}\mathit{,}\mathit{z}\mathit{)}\mathit{=}\frac{Q}{\mathit{4}\pi a{c}^{\mathit{2}}}{\mathit{[}\frac{{\mathit{x}}^{\mathit{2}}}{{\mathit{a}}^{\mathit{4}}}\mathit{+}\frac{{\mathit{r}}^{\mathit{2}}}{{\mathit{c}}^{\mathit{4}}}\mathit{]}}^{\mathit{-}\frac{\mathit{1}}{\mathit{2}}}$

$\sigma \mathit{(}x\mathit{,}y\mathit{,}z\mathit{)}\mathit{=}\frac{Q}{\mathit{4}\pi a{c}^{\mathit{2}}}\mathit{}\frac{\mathit{1}}{\sqrt{[{\displaystyle \frac{x^2}{a^4}}+{\displaystyle \frac{r^2}{c^4}}]}}$

The net charge per unit length is as follows:

$\lambda \left(x\right)=\frac{dQ}{dx}$

Now, consider the ellipsoid revolution.

The equation for the charge of the ring of width dsis dq = $\sigma 2\pi rds$.

Here,$ds=\sqrt{d{x}^2+d{r}^2}=dx\sqrt{1+{\left(\frac{dr}{dx}\right)}^2}$

$\frac{2x}{a^2}dx+\frac{2r}{c^2}dr=0$

$\frac{dr}{dx}=\frac{-x{c}^2}{a^{2}r}$

Solve further

localid="1657360021391" $$\begin{gathered} ds=dx\sqrt{1+{\left(\frac{-xc}{a^{2}r}\right)}^2} \\ =dx\sqrt{1+{\left(\frac{-x^2{c}^4}{a^4{r}^2}\right)}^2} \\ ds=dx\frac{c^2}{r}\sqrt{\frac{r^2}{c^4}+\frac{x^2}{a^4}} \end{gathered}$$

Then charge per unit length

$\lambda \left(x\right)=\frac{\sigma 2\pi rds}{dx}$

$\lambda \left(x\right)=\frac{\sigma 2\pi rdx{\displaystyle \frac{c^2}{r}}\sqrt{{\displaystyle \frac{r^2}{c^4}}+{\displaystyle \frac{x^2}{a^4}}}}{dx}$

$\lambda \left(x\right)=\frac{Q\left(2\mathrm{\pi r}\right)}{4{\mathrm{\pi ac}}^2}\frac{1}{\sqrt{{\displaystyle \frac{x^2}{a^4}}}+{\displaystyle \frac{r^2}{c^4}}}\frac{c^2}{r}\sqrt{\frac{x^2}{a^4}}+\frac{r^2}{c^4}$

localid="1657360436597" $\lambda \left(x\right)=\frac{Q}{2a}$

Therefore, the charge per unit length on the needle is$\lambda \left(x\right)=\frac{Q}{2a}$

$\lambda \left(x\right)$is plotted as follows