Question

If the linear charge density of a cylinder is $ 4 \,\mu cm^{-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} $

Answer 1

Answer: (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}} $
$=2 \times 10^{6} NC ^{-1}$