Answer:
E(r) = λ/2πrε0
Explanation:
If we consider an infinitely long line of charge with the charge per unit length being λ, we can take advantage of the cylindrical symmetry of this situation.
By symmetry, i mean that the electric fields all point radially away from the line of charge and thus there is no component parallel to the line of charge.
Niw, let's use a cylinder (with an arbitrary radius (r) and length (l)) centred on the line of charge as our Gaussian surface.
Doing that will mean that the electric field would be perpendicular to the curved surface of the cylinder. Therefore, the angle between the electric field and area vector is equal to zero and cos θ = cos 0 = 1
Now, the top and bottom surfaces of the cylinder will lie parallel to the electric field. Therefore, the angle between the area vector and the electric field would be 90° and cos θ = cos 90 = 0
Now, we know that according to Gauss Law,
Electric Flux, Φ = E•dA
Thus,
Total Φ = Φ_curved + Φ_top + Φ_bottom
Thus,
Φ = ∫E•dA cos 0 + ∫E•dA cos 90° + ∫E•dA cos 90°
We now have ;
Φ = ∫E . dA × 1
Since we are dealing with the radial component, the curved surface would be equidistant from the line of charge and the electric field in the surface will be the same magnitude throughout.
Thus,
Φ = ∫E•dA = E∫dA = E•2πrl
The net charge enclosed by the surface is given by:
q_net = λl
So using gauss theorem, we have;
Φ = E•2πrl = q_net/εo = λl/εo
E•2πrl = λl/ε0
Making E the subject, we obtain ;
E = λ/2πrε0
