Q. An infinite line charge produce a field of $7.182 \times 10^{8}\ N/C$ at a distance of $2\ cm.$ The linear charge density is
A) $7.27 \times 10^{- 4}\ C/m$
B) $7.98 \times 10^{- 4}\ C/m$
C) $7.11 \times 10^{- 4}\ C/m$
D) $7.04 \times 10^{- 4}\ C/m$
Explanation
(b) Relation for electric field is given by $E = \frac{\lambda}{2\pi\varepsilon_{0}r}$
$\lbrack Given:E = 7.182 \times 10^{8}\ N/C\rbrack$
$r = 2\ cm = 2 \times 10^{- 2}m$
$\frac{1}{4\pi\varepsilon_{0}} = 9 \times 10^{9} \Rightarrow \lambda = 2\pi\varepsilon_{0}rE = \frac{2 \times 2\pi\varepsilon_{0}rE}{2}$
$= \frac{1 \times 2 \times 10^{- 2} \times 7.182 \times 10^{8}}{2 \times 9 \times 10^{9}} = 7.98 \times 10^{- 4}C/m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.