Question

In a Gauss's Law problem where we use a cylinder for the Gaussian surface, when we compute the surface we ignore the end caps of the cylinder and just use 2*pi*r*L as an answer. Explain why we drop out the geometry for the end caps.

Answer 1

Suppose you have an infinite rod having uniform charge density. Clearly, this rod has cylindrical symmetry. Also, due to symmetry, the electric field is in the radial direction only.

Gauss law says,over a closed surface, .d=q/₀, where is electric field vector,d is differential area element, q is the enclosed charge in the closed surface, ₀ is permittivity in free space.

Now,in the left hand side of the gauss law, there is dot product term between electric field and area element. At the end caps of the cylinder, electric field is along the surface of the end cap(remember, due to symmetry, the electric field is in the radial direction only.) while the area element is perpendicular to the surface.

So, the dot product in the left hand side of the gauss law(which is given by, E*dA*cos,where is the angle between electric field and area vectors) turns out to be zero for the end caps as =90 and cos(90)=0.

So,when we use gauss law, while calculating flux in the left hand side, we drop out the end caps of the cylinder.