Question

If the linear charge density of a cylinder is $4\mu C{m}^{-1}$ then electric field intensity at point $3.6cm$ from axis is.

• A. $4\times {10}^{5}N{C}^{-1}$
• B. $2\times {10}^{6}N{C}^{-1}$
• C. $8\times {10}^{7}N{C}^{-1}$
• D. $12\times {10}^{7}N{C}^{-1}$

Answer 1

Answer: (B)

$2\times {10}^{6}N{C}^{-1}$

For a charged cylinder $E=\frac{\lambda }{2\pi {\epsilon }_{0}r}=\frac{4\times {10}^{-6}\times 18\times {10}^9}{3.6\times {10}^{-2}}$
$=2\times {10}^{6}N{C}^{-1}$