# Homework Help: Gauss' Law -- Clarification about using the equation please

1. Mar 3, 2017

### grandpa2390

1. The problem statement, all variables and given/known data

Is the da on the left side the surface area of the the Gaussian surface?

and on the right side when I am integrating. I say that q = the charge density multiplied by something. Is that something surface area of the original shape?

if I am doing a line charge and draw a cylindrical gaussian surface. on the left, da= the surface area of the cylinder (the round side). and on the right, q = lambda*l (the surface area of the line charge)

if I am doing a gaussian cylinder around a cylinder then da is the surface area of the cylinder on the left, and on the right it is the surface of the original cylinder (integrated over the radius)?

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: Mar 3, 2017
2. Mar 3, 2017

### haruspex

It depends how the charge density is defined. For a line charge it will be charge per unit length, so multiply by (or, for non-uniform charge, integrate wrt) the length. For a surface charge, e.g. on a conductor, multiply by/integrate wrt area. For a spatial distribution of charge, volume.