Answer:

[tex]\vec{E} = \frac{a}{2\pi \epsilon_0 R}\^R[/tex]

Explanation:

Since the wire is infinitely long, we will use Gauss’ Law:

[tex]\int\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}[/tex]

We will draw an imaginary cylindrical surface with height h around the wire. The electric flux through the imaginary surface will be equal to the net charge inside the surface.

In that case, the net charge inside the imaginary surface will be the portion of wire with height h. Then the charge of that portion will be equal to

[tex]Q_{enc} = ah[/tex]

The left-hand side of the Gauss’ Law is the flux through the imaginary surface. Since we choose our surface as a cylinder, of which we know the area, we do not have to take the surface integral.

[tex]\int\vec{E}d\vec{a} = E2\pi R h[/tex]

where R is the radius of the imaginary cylinder.

Finally, Gauss’ Law gives

[tex]E2\pi Rh = \frac{ah}{\epsilon_0}\\E = \frac{a}{2\pi \epsilon_0 R}[/tex]

The vector expression is

[tex]\vec{E} = \frac{a}{2\pi \epsilon_0 R}\^R[/tex]

As you can see, the electric field is independent from the height h, since that is merely an imaginary cylinder to apply Gauss’ Law. In the end, what matters is the charge density of the wire and the distance from the wire.