91Ó°ÊÓ

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

$\mathit{σ}\left(x.y.z\right)\mathbf{=}\frac{\mathbf{Q}}{\mathbf{4}\mathbf{π}\mathbf{a}\mathbf{b}\mathbf{c}}{\left(\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}\right)}^{\frac{\mathbf{1}}{\mathbf{2}}}$ …… (1)

Here, Q is the total charge.

The equations for ellipsoid is given as,

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

Then,

$\frac{{\mathbf{z}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}\mathbf{=}\mathbf{1}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{-}\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{b}}^{\mathbf{2}}}$ …… (2)

Now substitute the equation (2) in equation (1)

$$\begin{gathered} \mathit{σ}\left(x,y,z\right)\mathbf{=}\frac{\mathbf{Q}}{\mathbf{4}\mathbf{π}\mathbf{a}\mathbf{b}\mathbf{c}}{\left(\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{1}{c^2}\left(1-\frac{x^2}{a^2}-\frac{y^2}{b^2}\right)\right)}^{\frac{\mathbf{1}}{\mathbf{2}}} \\ \mathit{σ}\left(x,y,z\right)\mathbf{=}\frac{\mathbf{Q}}{\mathbf{4}\mathbf{π}\mathbf{a}\mathbf{b}\mathbf{c}}{\left(\frac{1}{c^2}\right)}^{\frac{\mathbf{-}\mathbf{1}}{\mathbf{2}}}{\left[\left(\frac{x^2}{a^4}\right)c^2+\left(\frac{y^2}{b^4}\right)c^2+\left(1-\frac{x^2}{a^2}-\frac{y^2}{b^2}\right)\right]}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}} \\ \mathit{σ}\left(x,y,z\right)\mathbf{=}\frac{\mathbf{Q}}{\mathbf{4}\mathbf{π}\mathbf{a}\mathbf{b}\mathbf{c}}{\left[\left(\frac{x^2}{a^4}\right)c^2+\left(\frac{y^2}{b^4}\right)c^2+\frac{1-x^2}{a^2}-\frac{y^2}{b^2}\right]}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}} \end{gathered}$$ …… (3)

a)

By using surface charge density of the ellipsoid, the net surface charge density of the circular disk,

The ellipsoid is reduced to the circular disk by setting the $\mathit{c}\mathbf{→}\mathbf{0}$in equation (3).

Therefore,

$\mathit{σ}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}\mathbf{Ï€²¹²ú}}{\mathbf{[}\mathbf{1}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{-}\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{b}}^{\mathbf{2}}}\mathbf{]}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}}$ …… (4)

Now consider the both side of the disk, so multiply by 2 to the equation (4).

Thus, the net surface charge density is,

$\mathit{σ}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{4}\mathbf{Ï€²¹²ú}}{\mathbf{[}\mathbf{1}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{-}\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{b}}^{\mathbf{2}}}\mathbf{]}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}}$ …… (5)

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

Then,

$$\begin{gathered} {\mathit{σ}}_{\mathbf{n}}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{4}\mathbf{Ï€²¹²ú}}{\mathbf{[}\mathbf{1}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{R}}^{\mathbf{2}}}\mathbf{]}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}} \\ {\mathit{σ}}_{\mathbf{n}}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{4}\mathbf{Ï€²¹²ú}}{\mathbf{[}\mathbf{1}\mathbf{-}\frac{{\mathbf{R}}^{\mathbf{2}}\mathbf{+}{\mathbf{r}}^{\mathbf{2}}}{{\mathbf{R}}^{\mathbf{2}}}\mathbf{]}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}} \\ {\mathit{σ}}_{\mathbf{n}}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{4}\mathbf{Ï€²¹²ú}}{\mathbf{[}{\mathbf{R}}^{\mathbf{2}}\mathbf{-}{\mathbf{r}}^{\mathbf{2}}\mathbf{]}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}} \\ {\mathit{σ}}_{\mathbf{n}}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{4}\mathbf{Ï€²¹²ú}}\frac{\mathbf{1}}{\sqrt{{\mathbf{R}}^{\mathbf{2}}\mathbf{-}{\mathbf{r}}^{\mathbf{2}}}} \end{gathered}$$

Hence, the net surface charge density on the circular disk is ${\mathit{σ}}_{\mathbf{n}}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{4}\mathbf{Ï€²¹²ú}}\frac{\mathbf{1}}{\sqrt{{\mathbf{R}}^{\mathbf{2}}\mathbf{-}{\mathbf{r}}^{\mathbf{2}}}}$.

The ${\mathit{σ}}_{\mathbf{n}}\mathbf{\left(}\mathbf{r}\mathbf{\right)}$is plotted as below.

b)

By using surface charge density of the ellipsoid, the net surface charge density on the ribbon,

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

Now consider the both side of the disk, so multiply by 2. Thus, the net surface charge density is,

$\mathit{σ}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}\mathbf{Ï€²¹²ú}}{\mathbf{[}\mathbf{1}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{-}\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{b}}^{\mathbf{2}}}\mathbf{]}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}}$ …… (6)

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

Now, let’s consider that, $\mathit{λ}$is the charge per unit length ribbon.

Then set $\frac{\mathbf{Q}}{\mathbf{b}}\mathbf{=}\mathit{λ}$and take the limit $\mathit{b}\mathbf{→}\mathbf{∞}$

$$\begin{gathered} \mathit{σ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\underset{\mathbf{b}\mathbf{→}\mathbf{∞}}{\mathbf{lim}}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)} \\ \mathit{σ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{Ï€²¹}}\frac{\mathbf{Q}}{\mathbf{b}}\underset{\mathbf{b}\mathbf{→}\mathbf{∞}}{\mathbf{lim}}{\mathbf{[}\mathbf{1}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{-}\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{b}}^{\mathbf{2}}}\mathbf{]}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}} \\ \mathit{σ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{Ï€²¹}}\frac{\mathbf{Q}}{\mathbf{b}}\underset{\mathbf{b}\mathbf{→}\mathbf{∞}}{\mathbf{lim}}{\mathbf{[}\mathbf{1}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{-}\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{b}}^{\mathbf{2}}}\mathbf{]}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}} \end{gathered}$$

Solve as further,

$$\begin{gathered} \mathit{σ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{λ}}{\mathbf{2}\mathbf{Ï€²¹}}{\mathbf{[}\mathbf{1}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{]}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}} \\ \mathit{σ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{λ}}{\mathbf{2}\mathbf{Ï€²¹}}{\mathbf{\left[}\frac{\mathbf{1}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{\right]}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}}{\mathbf{[}{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}\mathbf{]}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}} \\ \mathit{σ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{Ï€}}\frac{\mathbf{1}}{\sqrt{{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}} \end{gathered}$$

Therefore, the net charge density on the ribbon is $\mathit{σ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{π}}\frac{\mathbf{1}}{\sqrt{{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}}$.

The $\mathit{σ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$is plotted as,

c)

Now calculate the surface charge density of the conducting needle.

Let’s consider the equation of ellipsoid,

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

Assume that, $\mathit{b}\mathbf{=}\mathit{c}$and $\mathit{r}\mathbf{=}\sqrt{{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}}$

$$\begin{gathered} \frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{z}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}\mathbf{=}\mathbf{1} \\ \frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{c}}^{\mathbf{2}}}\mathbf{[}{\mathbf{y}}^{\mathbf{+}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}\mathbf{]}\mathbf{=}\mathbf{1} \\ \frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{c}}^{\mathbf{2}}}{\mathit{r}}^{\mathbf{2}}\mathbf{=}\mathbf{1} \end{gathered}$$

This is ellipsoid revolution.

Now the charge density of ellipsoid takes of the form,

$$\begin{gathered} \mathit{σ}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{4}{\mathbf{Ï€²¹c}}^{\mathbf{2}}}{\mathbf{[}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{4}}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{4}}}{{\mathbf{c}}^{\mathbf{4}}}\mathbf{+}\frac{{\mathbf{z}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{4}}}\mathbf{]}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}} \\ \mathit{σ}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{4}{\mathbf{Ï€²¹c}}^{\mathbf{2}}}{\mathbf{[}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{4}}}\mathbf{+}\frac{{\mathbf{r}}^{\mathbf{4}}}{{\mathbf{c}}^{\mathbf{4}}}\mathbf{]}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}} \\ \mathit{σ}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{4}{\mathbf{Ï€²¹c}}^{\mathbf{2}}}\frac{\mathbf{1}}{\sqrt{[{\displaystyle \frac{x^2}{a^4}}+{\displaystyle \frac{r^2}{c^4}}]}} \end{gathered}$$

Net charge per unit length is,

$\mathit{λ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{dQ}}{\mathbf{dx}}$

Now consider the ellipsoid revolution.

The equation for the charge of the ring of width $\mathit{d}\mathit{s}$is $\mathit{d}\mathit{q}\mathbf{=}\mathit{σ}\mathbf{2}\mathit{π}\mathit{r}\mathit{d}\mathit{s}$

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

$$\begin{gathered} \frac{\mathbf{2}\mathbf{x}}{{\mathbf{a}}^{\mathbf{2}}}\mathit{d}\mathit{x}\mathbf{+}\frac{\mathbf{2}\mathbf{r}}{{\mathbf{c}}^{\mathbf{2}}}\mathit{d}\mathit{r}\mathbf{=}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\frac{\mathbf{dr}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{-}{\mathbf{xc}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}\mathbf{r}} \end{gathered}$$

Solve as further,

$$\begin{gathered} \mathit{d}\mathit{s}\mathbf{=}\mathit{d}\mathit{x}\sqrt{\mathbf{1}\mathbf{+}{\left(\frac{-x{c}^2}{a^{2}r}\right)}^{\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{d}\mathit{x}\sqrt{\mathbf{1}\mathbf{+}\left(\frac{-x^2{c}^4}{a^4{r}^2}\right)} \\ \mathit{d}\mathit{s}\mathbf{=}\mathit{d}\mathit{x}\frac{{\mathbf{c}}^{\mathbf{2}}}{\mathbf{r}}\sqrt{\frac{{\mathbf{r}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{4}}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{4}}}} \end{gathered}$$

Then charge per unit length

$$\begin{gathered} \mathit{λ}\left(x\right)\mathbf{=}\frac{\mathbf{σ}\mathbf{2}\mathbf{π}\mathbf{r}\mathbf{d}\mathbf{s}}{\mathbf{d}\mathbf{x}} \\ \mathit{λ}\left(x\right)\mathbf{=}\frac{\mathbf{σ}\mathbf{2}\mathbf{d}\mathbf{x}{\displaystyle \frac{{\mathbf{c}}^{\mathbf{2}}}{\mathbf{r}}}\sqrt{{\displaystyle \frac{{\mathbf{r}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{4}}}}\mathbf{+}{\displaystyle \frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{4}}}}}}{\mathbf{d}\mathbf{x}} \\ \mathit{λ}\left(x\right)\mathbf{=}\frac{\mathbf{Q}\left(2\mathrm{π}r\right)}{\mathbf{4}\mathbf{π}\mathbf{a}{\mathbf{c}}^{\mathbf{2}}}\frac{\mathbf{1}}{\sqrt{{\displaystyle \frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{4}}}}\mathbf{+}{\displaystyle \frac{{\mathbf{r}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{4}}}}}}\frac{{\mathbf{c}}^{\mathbf{2}}}{\mathbf{r}}\sqrt{\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{4}}}\mathbf{+}\frac{{\mathbf{r}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{4}}}} \\ \mathit{λ}\left(x\right)\mathbf{=}\frac{\mathbf{Q}}{\mathbf{2}\mathbf{a}} \end{gathered}$$

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

is plotted as,