2.57P - Introduction to Electrodynamics

Question

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 

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

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

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

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

Step-by-Step Solution

Answer

Answer

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

2. The net charge density on the ribbon is $\mathit{\sigma }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi }}\frac{\mathbf{1}}{\sqrt{{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}}$.

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

Step 1: Define functions

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

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

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

Step 2: Determine net surface charge density

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 $\mathit{c}\mathbf{\to }\mathbf{0}$ in equation (3).

Therefore, 

$\mathit{\sigma }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}\mathbf{\pi ab}}{\mathbf{[}\mathbf{1}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{-}\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{b}}^{\mathbf{2}}}\mathbf{]}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}}$                            …… (4)

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

Thus, the net surface charge density is,

$\mathit{\sigma }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{4}\mathbf{\pi ab}}{\mathbf{[}\mathbf{1}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{-}\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{b}}^{\mathbf{2}}}\mathbf{]}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}}$                            …… (5)

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

Then, 

$$\begin{gathered} {\mathit{\sigma }}_{\mathbf{n}}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{4}\mathbf{\pi ab}}{\mathbf{[}\mathbf{1}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{R}}^{\mathbf{2}}}\mathbf{]}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}} \\ {\mathit{\sigma }}_{\mathbf{n}}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{4}\mathbf{\pi ab}}{\mathbf{[}\mathbf{1}\mathbf{-}\frac{{\mathbf{R}}^{\mathbf{2}}\mathbf{+}{\mathbf{r}}^{\mathbf{2}}}{{\mathbf{R}}^{\mathbf{2}}}\mathbf{]}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}} \\ {\mathit{\sigma }}_{\mathbf{n}}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{4}\mathbf{\pi ab}}{\mathbf{[}{\mathbf{R}}^{\mathbf{2}}\mathbf{-}{\mathbf{r}}^{\mathbf{2}}\mathbf{]}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}} \\ {\mathit{\sigma }}_{\mathbf{n}}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{4}\mathbf{\pi ab}}\frac{\mathbf{1}}{\sqrt{{\mathbf{R}}^{\mathbf{2}}\mathbf{-}{\mathbf{r}}^{\mathbf{2}}}} \end{gathered}$$

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

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

Step 3: Determine surface charge density on the ribbon

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 $\mathit{c}\mathbf{\to }\mathbf{0}$ in equation (3).

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

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

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

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

Then set $\frac{\mathbf{Q}}{\mathbf{b}}\mathbf{=}\mathit{\lambda }$ and take the limit $\mathit{b}\mathbf{\to }\mathbf{\infty }$

$$\begin{gathered} \mathit{\sigma }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\underset{\mathbf{b}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)} \\ \mathit{\sigma }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi a}}\frac{\mathbf{Q}}{\mathbf{b}}\underset{\mathbf{b}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}{\mathbf{[}\mathbf{1}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{-}\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{b}}^{\mathbf{2}}}\mathbf{]}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}} \\ \mathit{\sigma }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi a}}\frac{\mathbf{Q}}{\mathbf{b}}\underset{\mathbf{b}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}{\mathbf{[}\mathbf{1}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{-}\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{b}}^{\mathbf{2}}}\mathbf{]}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}} \end{gathered}$$

Solve as further,

$$\begin{gathered} \mathit{\sigma }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{2}\mathbf{\pi a}}{\mathbf{[}\mathbf{1}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{]}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}} \\ \mathit{\sigma }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{2}\mathbf{\pi a}}{\mathbf{\left[}\frac{\mathbf{1}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{\right]}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}}{\mathbf{[}{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}\mathbf{]}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}} \\ \mathit{\sigma }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi }}\frac{\mathbf{1}}{\sqrt{{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}} \end{gathered}$$

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

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

Step 4: Determine surface charge density of conducting the needle

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,

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

Net charge per unit length is,

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

Now consider the ellipsoid revolution.

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

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 

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

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

 is plotted as,