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An infinitely long thin straight wire has uniform linear charge density of $\frac{1}{3} \mathrm{Cm}^{-1}$. Then, the magnitude of the electric intensity at a point $18 \mathrm{~cm}$ away is (given $\varepsilon_0=8.8 \times 10^{-12} \mathrm{C}^2 \mathrm{Nm}^{-2}$ )
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Verified Answer
The correct answer is:
$0.66 \times 10^{11} \mathrm{NC}^{-1}$
Charge density of long wire
$\begin{aligned}
\lambda & =\frac{1}{3} \mathrm{C}-\mathrm{m} \\
\text { and } \quad r & =18 \times 10^{-2} \mathrm{~m}
\end{aligned}$
From Gauss theorem
$\begin{gathered}
\oint \overrightarrow{\mathbf{E}} d \overrightarrow{\mathbf{S}}=\frac{q}{\varepsilon_0} \\
E \oint d S=\frac{q}{\varepsilon_0}
\end{gathered}$
or $\quad E \times 2 \pi r l=\frac{q}{\varepsilon_0}$
or $\quad E=\frac{q}{2 \pi \varepsilon_0 r l}=\frac{q / l}{2 \pi \varepsilon_0 r}$
$\begin{aligned} & =\frac{\lambda \times 2}{2 \pi \varepsilon_0 r \times 2}=\frac{\lambda \times 2}{4 \pi \varepsilon_0 r} \\ & =9 \times 10^9 \times \frac{1}{3} \times 2 \times \frac{1}{18 \times 10^{-2}} \\ & =\frac{1}{3} \times 10^{11}=0.33 \times 10^{11} \\ & =0.33 \times 10^{11} \mathrm{NC}^{-1}\end{aligned}$
$\begin{aligned}
\lambda & =\frac{1}{3} \mathrm{C}-\mathrm{m} \\
\text { and } \quad r & =18 \times 10^{-2} \mathrm{~m}
\end{aligned}$
From Gauss theorem
$\begin{gathered}
\oint \overrightarrow{\mathbf{E}} d \overrightarrow{\mathbf{S}}=\frac{q}{\varepsilon_0} \\
E \oint d S=\frac{q}{\varepsilon_0}
\end{gathered}$
or $\quad E \times 2 \pi r l=\frac{q}{\varepsilon_0}$
or $\quad E=\frac{q}{2 \pi \varepsilon_0 r l}=\frac{q / l}{2 \pi \varepsilon_0 r}$
$\begin{aligned} & =\frac{\lambda \times 2}{2 \pi \varepsilon_0 r \times 2}=\frac{\lambda \times 2}{4 \pi \varepsilon_0 r} \\ & =9 \times 10^9 \times \frac{1}{3} \times 2 \times \frac{1}{18 \times 10^{-2}} \\ & =\frac{1}{3} \times 10^{11}=0.33 \times 10^{11} \\ & =0.33 \times 10^{11} \mathrm{NC}^{-1}\end{aligned}$
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