Question

A section of an infinite rod of charge having linear charge density λ$\lambda$ which is constant for all points on the line.
Electric field E at a distance r from the line is

• A. $\frac{\lambda }{\pi {\epsilon }_{0}r}$
• B. $\frac{\lambda }{2\pi {\epsilon }_{0}r}$
• C. $\frac{\lambda }{4\pi {\epsilon }_{0}r}$
• D. $\frac{2\lambda }{\pi {\epsilon }_{0}r}$

Answer 1

The correct option is B $\frac{\lambda }{2\pi {\epsilon }_{0}r}$

From symmetry, $\overrightarrow{E}$ due to a uniform linear charged can only be radially directed. As a Gaussian surface, we can choose a circular cylinder of radius r and length. L, closed at each end by plane caps normal to the axis.
$\epsilon {}_0\oint \overrightarrow{E}.d\overrightarrow{s}=q_{in}{\epsilon }_0\left[\int \overrightarrow{E}.d\overrightarrow{s}+\int \overrightarrow{E}.d\overrightarrow{s}+\right]=q_{in}$
Cylindrical Plane Surface
${\epsilon }_{0}E\left(2\pi rl\right)+E.dscos{90}^{\circ }=\lambda I$
$E=\frac{\lambda I}{{\epsilon }_0\left(2\pi rl\right)}=\frac{\lambda }{2\pi {\epsilon }_{0}r}$
The direction of is radially outwards for a line of positive charge.