Q. If the linear charge density of a cylinder is $4\mu\complement m^{- 1}$ then electric field intensity at point 3.6 cm from axis is
A) $4 \times 10^{5}\ {NC}^{- 1}$
B) $2 \times 10^{6}{NC}^{- 1}$
C) $8 \times 10^{7}\ {NC}^{- 1}$
D) $12 \times 10^{7}{NC}^{- 1}$
Explanation
(b) For a charged cylinder
$E = \frac{\lambda}{2\pi\varepsilon_{0}r} = \frac{4 \times 10^{- 6} \times 18 \times 10^{9}}{3.6 \times 10^{- 2}}$
$\ \ \ \ = 2 \times 10^{6}NC^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.