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Question

Physics Question on Electric charges and fields

If the linear charge density of a cylinder is 4 μcm−14 \,\mu cm^{-1} then electric field intensity at point 3.6 cm3.6 \,cm from axis is

A

4×105 NC−14 \times 10^5\,NC^{-1}

B

2×106 NC−12 \times 10^6\,NC^{-1}

C

8×107 NC−18 \times 10^7\,NC^{-1}

D

12×107 NC−112 \times 10^7\,NC^{-1}

Answer

2×106 NC−12 \times 10^6\,NC^{-1}

Explanation

Solution

For a charged cylinder
E=λ2πε0rE =\frac{\lambda}{2 \pi \varepsilon_{0} r}
=4×10−6×18×1093.6×10−2=\frac{4 \times 10^{-6} \times 18 \times 10^{9}}{3.6 \times 10^{-2}}
=2×106NC−1=2 \times 10^{6} NC ^{-1}