Question

Consider a rod of length ℓ that carries a uniform charge distribution. Is Gauss’s law appropriate for calculating the electric field at a radial distance r from the midpoint of the rod’s length?

Hint: For each case in which your answer is yes, describe the shape of the most useful Gaussian surface. For “no”, explain why.

• (a) When r ≪ ℓ
• (b) When r = ℓ
• (c) When r ≫ ℓ

Answer & Detailed Explanation

(a) When r ≪ ℓ:

In this case, the observation point is very close to the midpoint of a long rod. The central portion of the rod appears nearly infinite. The electric field is approximately radially outward (in the plane perpendicular to the rod) and nearly constant in magnitude on a cylindrical surface that is coaxial with the rod.

Therefore, we can use a cylindrical Gaussian surface where the field is perpendicular and constant on the curved side. The flux through the end caps is negligible. Hence, Gauss’s law is useful here.

✔ Yes – Use a cylindrical surface coaxial with the rod.

(b) When r = ℓ:

Here, the observation point is at a distance comparable to the rod’s length. The rod does not appear infinite anymore. The symmetry required for a Gaussian surface is lost, and the electric field varies in direction and magnitude across any closed surface.

While Gauss’s law remains true universally, it’s not practically useful for direct calculation in this case.

✖ No – The finite size of the rod destroys symmetry needed for an effective Gaussian surface.

(c) When r ≫ ℓ:

In this scenario, the observation point is far away from the rod. The entire rod behaves like a point charge located at its center with total charge Q. The electric field becomes spherically symmetric, depending only on the distance r.

Therefore, a spherical Gaussian surface centered at the rod’s midpoint can be used. The electric field is uniform on the surface and directed radially outward.

✔ Yes – Use a spherical surface centered at the midpoint of the rod.

Final Summary:

• (a) Yes – Use a cylindrical Gaussian surface coaxial with the rod.
• (b) No – Symmetry is not sufficient to use Gauss’s law effectively.
• (c) Yes – Use a spherical Gaussian surface; rod approximates a point charge.