Question

A very long uniformly charged circular cylinder (radius R) has a surface charge density $\sigma$. A very long uniformly charged line charge (linear charge density $\lambda$) is placed along the cylinder axis. If electric field intensity vector outside the cylinder is zero, then $\lambda =-\frac{x}{4}\pi \sigma R.$ Find $x$.

Answer 1

Electric field at a point outside the cylinder (say 𝑃) would be due to two charge configurations, one due to linear charge and other due to charge on the cylinder.



We know that, E due to line charge
$\left(E_1\right)=\frac{\lambda }{2\pi {\epsilon }_{0}r}$ along x direction
Also, for cylinder, charge enclosed
$=\sigma \times \pi R^2$
If we consider a circular Gaussian surface of radius r passing through point 𝑃, then by Gauss law
$Flux=(\overrightarrow{E}.\overrightarrow{A})=\frac{Q_{enclosed}}{{\epsilon }_0}$
$\Rightarrow E_2\left(\pi r^2\right)=\frac{\sigma \times \pi R^2}{{ϵ}_0}$
Therefore,
$E_2=\frac{\sigma R}{r{\epsilon }_0}$ along x direction
Resultant eletric field (E) s given by :
$\overrightarrow{E}=\overrightarrow{E_1}+\overrightarrow{E_2}$
$=\frac{\lambda }{2\pi {\epsilon }_{0}r}\hat{i}+\frac{\sigma R}{{\epsilon }_{0}r}\hat{i}$
It is given that field at point P is zero, then
For $\overrightarrow{E}$ to be zero,
$\frac{\lambda }{2\pi {\epsilon }_{0}r}+\frac{\sigma R}{{\epsilon }_{0}R}=0$
$\frac{\lambda }{2\pi {\epsilon }_{0}r}=-\frac{\sigma R}{{\epsilon }_{0}R}=0$
$\lambda =-2\pi \sigma R$
$\Rightarrow x=8$
Hence, the value of x is 8.