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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
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Verified Answer
The correct answer is:
$2 \times 10^{6} \mathrm{NC}^{-1}$
For a charged cylinder
$$
\begin{aligned}
E &=\frac{\lambda}{2 \pi \varepsilon_{0} r}=\frac{4 \times 10^{-6} \times 18 \times 10^{9}}{3.6 \times 10^{-2}} \\
&=2 \times 10^{6} \mathrm{NC}^{-1}
\end{aligned}
$$
$$
\begin{aligned}
E &=\frac{\lambda}{2 \pi \varepsilon_{0} r}=\frac{4 \times 10^{-6} \times 18 \times 10^{9}}{3.6 \times 10^{-2}} \\
&=2 \times 10^{6} \mathrm{NC}^{-1}
\end{aligned}
$$
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