Question

A charge is distributed uniformly along the z axis from $z=-a$to$z=+a$. Show that the electric potential at a point r is given by

$V\left(r,\theta \right)=\frac{Q}{4\pi {\epsilon }_0}\frac{1}{r}\left[1+\frac{1}{3}{\left(\frac{a}{r}\right)}^2{P}_2\left(cos\theta \right)+\frac{1}{5}{\left(\frac{a}{r}\right)}^4{P}_4\left(cos\theta \right)+...\right]$

for $r>a$.

Short answer

The electrical potential at point r is$V\left(r,\theta \right)=\frac{Q}{4\pi {\epsilon }_0}\frac{1}{r}\left[1+\frac{1}{3}{\left(\frac{a}{r}\right)}^2{P}_2\left(cos\theta \right)+\frac{1}{5}{\left(\frac{a}{r}\right)}^4{P}_4\left(cos\theta \right)+...\right]$.

Step-by-step solution

Define function

Write the expression for charge for the small line segment.

$\rho d\tau$or $\lambda dz$ …… (1)

Here,P is the volume charge density and is the linear charge density.

Here, the charge Q is uniformly distributed along the z-axis, from$z=-atoz=+a$

$\lambda =\frac{Q}{2a}$ …… (2)

Multiply with on both sides of the above equation.

$\lambda dz=\frac{Q}{2a}dz$ ……. (3)

Determine potential

$V\left(r\right)=\frac{1}{4\pi {\epsilon }_0}\sum _{n=0}^{\infty }\frac{1}{r^{n+1}}\underset{-a}{\overset{+a}{\int }}{z}^n{P}_n\left(\cos \theta \right)\frac{Q}{2a}dz$...... (4)

Now, take the following equation from equation (4)

$$\begin{gathered} \underset{-a}{\overset{+a}{\int }}{z^{n}dz=\left[\frac{z^{n+1}}{n+1}\right]}_{-a}^{+a} \\ =\frac{2a^{n+1}}{n+1} \end{gathered}$$ ……. (5)

If the n is even, then equation (5) can be as follows,

$$\begin{gathered} \underset{-a}{\overset{+a}{\int }}{z}^{n}dz=\frac{2a^{n+1}}{n+1} \\ =0 \end{gathered}$$

Determine potential

Substitute the equation (4) in (5)

$\begin{array}{rcl}V& =& \frac{1}{4\pi {\epsilon }_0}\sum _{0,2,4}^{\infty }\frac{1}{r^{n+1}}\left(\frac{Q}{2a}\frac{2a^{n+1}}{n+1}\right)P_n\cos \theta \\ V\left(r\right)& =& \frac{1}{4\pi {\epsilon }_0}\frac{Q}{r}\sum _{0,2,4}^{\infty }\left(\frac{1}{n+1}{\left(\frac{a}{r}\right)}^n{P}_n\cos \theta \right)\\ V\left(r,\theta \right)& =& \frac{1}{4\pi {\epsilon }_0}\frac{Q}{r}\left[1+{\left(\frac{a}{r}\right)}^0{P}_0\cos \theta +\frac{1}{3}{\left(\frac{a}{r}\right)}^2{P}_2\cos \theta +\frac{1}{5}{\left(\frac{a}{r}\right)}^4{P}_4\cos \theta +....\right]\\ & =& \frac{Q}{4\pi {\epsilon }_{0}r}\left[1+\frac{1}{3}{\left(\frac{a}{r}\right)}^2{P}_2\cos \theta +\frac{1}{5}{\left(\frac{a}{r}\right)}^4{P}_4\cos \theta +....\right]\\ & & \end{array}$

Therefore, the electric potential at a point is the proved. That is,

$\frac{Q}{4\pi {\epsilon }_{0}r}\left[1+\frac{1}{3}{\left(\frac{a}{r}\right)}^2{P}_2\cos \theta +\frac{1}{5}{\left(\frac{a}{r}\right)}^4{P}_4\cos \theta +....\right]$