For a uniform line charge of 8 nc/m lying along the z axis, the electric field at a radius of 3 m from the uniform line is given by :
47.9 aρ V/m
The field of an infinite line charge falls as 1/ρ and points radially outward :
\(\vec{E}=\dfrac{\rho_{L}}{2\pi\varepsilon_{0}\rho}\,\vec{a}_{\rho}\)
Substituting the given values, with \(\rho_{L}=8\times10^{-9}\) C/m and \(\rho=3\) m:
\(2\pi\varepsilon_{0}\rho=2\pi\times8.854\times10^{-12}\times3=1.669\times10^{-10}\)
\(E=\dfrac{8\times10^{-9}}{1.669\times10^{-10}}=47.9\ \text{V/m}\)
so \(\vec{E}=47.9\,\vec{a}_{\rho}\) V/m — option 3.
A faster route using the standard constant. Since \(1/2\pi\varepsilon_{0}=17.98\times10^{9}\), the formula becomes
\(E=\dfrac{17.98\times10^{9}\times8\times10^{-9}}{3}=\dfrac{143.8}{3}=47.9\ \text{V/m}\)
Where the formula comes from. Apply Gauss's law to a cylinder of radius \(\rho\) and length L coaxial with the line. By symmetry E is radial and constant over the curved surface, and the flat ends contribute nothing since E is parallel to them:
\(E\left(2\pi\rho L\right)=\dfrac{\rho_{L}L}{\varepsilon_{0}}\quad\Rightarrow\quad E=\dfrac{\rho_{L}}{2\pi\varepsilon_{0}\rho}\)
The L cancels, which is what makes the result independent of how much of the line is enclosed.
Why the exponent is 1 and not 2. A point charge spreads its flux over a sphere of area \(4\pi r^{2}\), giving \(1/r^{2}\). An infinite line spreads it over a cylinder, whose area grows only as \(2\pi\rho L\) — one power of the distance — so the field falls as \(1/\rho\). By the same argument an infinite sheet gives a field that does not fall off at all.
Options 1 and 3 differ only by transposed digits, 49.7 against 47.9, so the arithmetic has to be carried through rather than estimated. Option 2 is roughly double the correct value, which is what results from omitting the factor of 2 in the denominator — that is, from using the point-charge constant \(1/4\pi\varepsilon_{0}\) in the wrong place.
Hence, E = 47.9 aρ V/m.
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