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An infinitely long thin straight wire has uniform charge density of $\frac{1}{4} \times 10^{-2} \mathrm{~cm}^{-1}$.
What is the magnitude of electric field at a distance $20 \mathrm{~cm}$ from the axis of the wire?
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What is the magnitude of electric field at a distance $20 \mathrm{~cm}$ from the axis of the wire?
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Verified Answer
The correct answer is:
$2.25 \times 10^{10} \mathrm{NC}^{-1}$
Given, charge density of uniformly long thin wire,
$\begin{aligned}
\lambda &=\frac{1}{4} \times 10^{-2} \mathrm{~cm}^{-1} \\
&=\frac{1}{4} \times 10^{-2} \times 10^{2} \mathrm{~m}^{-1}=\frac{1}{4} \mathrm{~m}^{-1} \\
r &=20 \mathrm{~cm}=0.2 \mathrm{~m}
\end{aligned}$
Electric field at a distance $r$ from the axis of wire is given as
$\begin{aligned}
E &=\frac{\lambda}{2 \pi \varepsilon_{0} r}=\frac{2 \lambda}{4 \pi \varepsilon_{0} r} \\
&=\frac{9 \times 10^{9} \times 2 \times \frac{1}{4}}{0.2} \\
&=2.25 \times 10^{10} \mathrm{NC}^{-1}
\end{aligned}$
$\begin{aligned}
\lambda &=\frac{1}{4} \times 10^{-2} \mathrm{~cm}^{-1} \\
&=\frac{1}{4} \times 10^{-2} \times 10^{2} \mathrm{~m}^{-1}=\frac{1}{4} \mathrm{~m}^{-1} \\
r &=20 \mathrm{~cm}=0.2 \mathrm{~m}
\end{aligned}$
Electric field at a distance $r$ from the axis of wire is given as
$\begin{aligned}
E &=\frac{\lambda}{2 \pi \varepsilon_{0} r}=\frac{2 \lambda}{4 \pi \varepsilon_{0} r} \\
&=\frac{9 \times 10^{9} \times 2 \times \frac{1}{4}}{0.2} \\
&=2.25 \times 10^{10} \mathrm{NC}^{-1}
\end{aligned}$
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