Answer to Question #113493 in Electricity and Magnetism for ALI

Question

Positive charge Q is distributed uniformly along the positive y-axis between y = 0 and y = a. A point P lies on the positive x-axis, a distance x from the origin (see figure)
(a) Calculate the x and y-components of the electric field produced by the charge distribution Q at points on the positive x-axis.
(b) Show that if x≫a
E_x ≅Q/(4πε_0 x^2 )
E_y ≅- Qa/(8πε_0 x^3 )
Explain why this result is obtained

Answer 1

As per the given question,

here i suppose $r_o=x$

a)

let the linear charge density $=\lambda$

Let the length of the rod is l and the total charge distributed on the rod is Q.

$dE=\dfrac{dQ}{4\pi \epsilon_o r^2}$

$dE=\dfrac{\lambda da}{4\pi \epsilon_o x^2}$

$\cos\theta=\dfrac{r_o}{r}$

$r=r_o\sec\theta$

$dE=\dfrac{\lambda da}{4\pi \epsilon_o r_o^2 \sec^2\theta}$

$\tan \theta=\dfrac{a}{r_o}$

$da=r_o sec^2\theta d\theta$

$dE=\dfrac{\lambda da}{4\pi \epsilon_o r_o^2\sec^\theta}$

$\Rightarrow dE=\dfrac{\lambda r_o \sec^2\theta d\theta}{4\pi \epsilon_o r_o^2 \sec^2\theta}$

Now, taking x component and y component

$dE=\dfrac{\lambda d\theta}{4\pi\epsilon_o r_o}(\cos\theta \hat{i}+\sin\theta \hat{j})$

Now, integrating both side

$\int _0^EE=\dfrac{\lambda}{4\pi \epsilon_o r_o}(\int_o^\theta \cos\theta d\theta \hat{i}+\int_o^\theta \sin\theta d\theta\hat{j})$

$E=\dfrac{\lambda}{4\pi \epsilon_o r_o}(\sin \theta \hat{i}-(\cos\theta+1)\hat{j})$

now we can write sin and cos in the form of x,

$E=\dfrac{\lambda}{4\pi \epsilon_o x}(\dfrac{x}{\sqrt{x^2+a^2}} \hat{i}-(\dfrac{a}{\sqrt{x^2+a^2}}+1)\hat{j})$

Now, if x>>a

so $a^2=0$

$E=\dfrac{\lambda }{4\pi \epsilon_ox^2}$

if x=a

$E=\dfrac{\lambda a}{8\pi \epsilon_ox^3}$