# Using Gauss' law for spherical charge distribution

## Homework Statement

The Earth is constantly being bombarded by cosmic rays, which consist mostly of protons. Assume that these protons are incident on the Earth’s atmosphere from all directions at a rate of 1366. protons per square meter per second. Assuming that the depth of Earth’s atmosphere is 110.0 km, what is the total charge incident on the atmosphere in 701.0 s? Assume that the radius of the surface of the Earth is 6378. km.

## The Attempt at a Solution

This is easily solved using some basic algebra.... by doing the following..

$Q = {[(4\pi(110+6378)^2)*10^6]}*1366* 701*(1.6*10^{-19})$

But, considering this is a spherical charge distribution (right?) we should be able to exploit symmetry and Gauss' Law as another way to solve the problem? I tried to start from the equations below as practice, but got stuck and it quickly turned into a mess. Could someone point me in the right direction.

We know that $E(x, y, z) = \int {p(x', y', z') \hat r dx' dy' dz'}/{r^2}$
$dQ = pdV$
$dE = \int \hat (r /r^2) dQ$

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## Answers and Replies

Doc Al
Mentor
But, considering this is a spherical charge distribution (right?) we should be able to exploit symmetry and Gauss' Law as another way to solve the problem?
Well, there is an element of symmetry, but I don't see how Gauss' law will help with this problem. For one thing, you don't have the electric field due to the protons so you cannot use Gauss' law to calculate their number.

Your initial solution is the way to go.