Question

Consider the electric field 𝐸 = 𝑎𝑟

𝑟𝜌𝑙

4𝜋𝜖0

∫

𝑑𝑧

′

(𝑟

2+𝑧′

2)

3/2

+∞

−∞

due to the uniform line charge density ρl

= 12 µC/m of an infinitely long straight conducting wire at a distance r. Find out the numeric value of 

r. Remember that prime variables corresponds to the source quantity.

Answer 1

Given,

Electric field $E=ar$

$E=\frac{r\rho l}{4\pi \epsilon_o}\int_{-\infty}^{\infty} \frac{dz}{(r^2+z^2)^{3/2}}$

$=\frac{ar\rho l}{4\pi \epsilon_o}\times [\frac{z}{r^2(z^2+r^2)}]_{-\infty}^{\infty}$

$=\frac{ar\rho l}{4\pi \epsilon_o}\times (\frac{2}{r^2})$

So,

$ar=\frac{r\rho l}{4\pi \epsilon_o}\times (\frac{2}{r^2})$

$\Rightarrow$ $r=\sqrt{\frac{2\rho l}{4\pi \epsilon_o a}}$