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An infinite line charge produces a field of $9 \times 10^4 \mathrm{NC}^{-1}$ at a distance of $2 \mathrm{~cm}$. Its linear charge density is
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Verified Answer
The correct answer is:
$0.1 \mu \mathrm{C} \mathrm{m}{ }^{-1}$
Electric field produced by infinite line charge,
$$
\begin{aligned}
& E=9 \times 10^4 \quad \mathrm{NC}^{-1} \\
& r=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}
\end{aligned}
$$
Linear charge density, $\lambda=$ ?
We know that,
$$
\begin{aligned}
E & =\frac{\lambda}{2 \pi \varepsilon_0 r} \\
\Rightarrow \quad \lambda & =2 \pi \varepsilon_0 r E=\frac{4 \pi \varepsilon_0}{2} \cdot r E \\
& =\frac{1}{2 \times 9 \times 10^9} \times 2 \times 10^{-2} \times 9 \times 10^4 \\
& =10^{-7} \mathrm{Cm}^{-1}=0.1 \times 10^{-6} \mathrm{Cm}^{-1} \\
& =0.1 \mu \mathrm{Cm}^{-1}
\end{aligned}
$$
$$
\begin{aligned}
& E=9 \times 10^4 \quad \mathrm{NC}^{-1} \\
& r=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}
\end{aligned}
$$
Linear charge density, $\lambda=$ ?
We know that,
$$
\begin{aligned}
E & =\frac{\lambda}{2 \pi \varepsilon_0 r} \\
\Rightarrow \quad \lambda & =2 \pi \varepsilon_0 r E=\frac{4 \pi \varepsilon_0}{2} \cdot r E \\
& =\frac{1}{2 \times 9 \times 10^9} \times 2 \times 10^{-2} \times 9 \times 10^4 \\
& =10^{-7} \mathrm{Cm}^{-1}=0.1 \times 10^{-6} \mathrm{Cm}^{-1} \\
& =0.1 \mu \mathrm{Cm}^{-1}
\end{aligned}
$$
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