Answer:
[tex]E = 6.38\times 10^6~N/C[/tex]
Explanation:
The question states that we can approximate the line as an infinite wire. In that case, the electric field can be found by Gauss' Law.
We should draw an imaginary cylindrical surface with an arbitrary height, h, around the wire. The radius of the cylinder should be equal to 65.1 cm.
Gauss' Law:
[tex]\int \vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}[/tex]
The integral in the left-hand side is not to be taken, because we know the area of the cylinder. The enclosed charge in the right-hand side is equal to the charge of the portion of the wire inside the imaginary surface.
The charge density of the wire is
[tex]\lambda = \frac{Q}{L} = \frac{230 \times 10^{-3}}{1000} = 2.3 \times 10^{-4}[/tex]
The charge enclosed by the imaginary surface is
[tex]Q_{enc} = \lambda h = 2.3\times 10^{-4}h[/tex]
Finally, Gauss' Law yields
[tex]E2\pi rh = \frac{\lambda h}{\epsilon_0}\\E = \frac{\lambda}{2\pi \epsilon_0r} = \frac{2.3 \times 10^{-4}}{2\pi\epsilon_0(65.1\times 10^{-2})} = 6.38\times 10^6~N/C[/tex]
