Question

Two long parallel threads carry charge per unit length. The threads are separated by a distance $l$. The maximum field strength in the symmetry plane between them is given as $\frac{\lambda }{x\pi {\epsilon }_{0}l}$. Find $x$:

Answer 1

The electric field at a distance r due to long thread is $E_1=\frac{\lambda }{2\pi {ϵ}_{0}r}$

The field is symmetry due to both threads at middle of them. so $r=\sqrt{z^2+(l/2{)}^2}$

The net field at middle of them is $E=2E_1\sin \theta$

or $E=2\left(\frac{\lambda }{2\pi {ϵ}_{0}r}\right)\sin \theta =\frac{\lambda }{\pi {ϵ}_{0}r}\frac{z}{r}$

or $E=\frac{\lambda }{\pi {ϵ}_0}\frac{z}{z^2+\frac{l^2}{4}}$

For max E, $\frac{dE}{dz}=0$

$\Rightarrow [\frac{1}{z^2+\frac{l^2}{4}}-\frac{2z^2}{(z^2+\frac{l^2}{4}{)}^2}]=0$

or $2z^2=z^2+\frac{l^2}{4}$

or $z^2=\frac{l^2}{4}\Rightarrow z=\frac{l}{2}$

thus, $E_{max}=\frac{\lambda }{\pi {ϵ}_0}\frac{\left(l/2\right)}{(l/2{)}^2+\frac{l^2}{4}}=\frac{\lambda }{\pi {ϵ}_{0}l}$

so $x=1$