# Homework Help: Electric Flux on one side

1. Jun 13, 2008

### K3nt70

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

A 1.3microC point charge is placed at the center of a cube with a volume of 9.8m^3. Calculate the electric flux through one side of the cube.

2. Relevant equations

E = $$\frac{k*q}{r^2}$$

$$\Phi$$ = E*A

3. The attempt at a solution

I cube root the volume of the cube. This gives me the length of any side since its a cube. So then, i calculate E where k = 9.0E9 q = 1.3E-6 and r = 1.07 (this is half of one length of one side). I then put E back into the electric flux equation where A is (2.14^2)*6 (the total surface area of the cube) i then take my answer and divide by 6 to get the flux for one side which comes out to 4.68E4 N*m^2/C which is incorrect. Im thinking ive made a mistake calculating the Electric field (the r value specifically, since it isnt really constant. ie if we wanted to know the flux in the corner of the cube, we would need to use trig to get the r value. But i cant see what else the r value would be.) A little direction would be great :D

my picture (not given):
http://img214.imageshack.us/img214/1397/chargeyp3.png [Broken]

Last edited by a moderator: May 3, 2017
2. Jun 13, 2008

### gendou2

My advice is to think in terms of electric field lines.
In order to avoid crazy integrals, simplify the problem to a sphere.
Solve for the flux on a sphere of arbitrary radius, then divide by six.
You see, the surface area of a sphere varies with r2, whereas the E-field varies as r-2, which cancel.

3. Jun 13, 2008

### K3nt70

Fantasic, i got it right. Thanks!