Q112 / 150 · MHT CET 2009 · Physics
If the linear charge density of a cylinder is $4 \mu \mathrm{Cm}^{-1}$ then electric field intensity at point $3.6 \mathrm{~cm}$ from axis is
- A$4 \times 10^{5} \mathrm{NC}^{-1}$
- Correct answerB$2 \times 10^{6} \mathrm{NC}^{-1}$
- C$8 \times 10^{7} \mathrm{NC}^{-1}$
- D$12 \times 10^{7} \mathrm{NC}^{-1}$
Solution
For a charged cylinder
$$
\begin{aligned}
E &=\frac{\lambda}{2 \pi \varepsilon_{0} r}=\frac{4 \times 10^{-6} \times 18 \times 10^{9}}{3.6 \times 10^{-2}} \\
&=2 \times 10^{6} \mathrm{NC}^{-1}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.