# Gauss' Law

1. Feb 1, 2016

### squeak

1. The problem statement, all variables and given/known data
(a) A spherical insulating shell of radius R = 3.00 m has its centre at the origin and carries a surface charge density σ = 3.00 nC/m2. Use Gauss’s law to find the electric field on the x axis at (i) x = 2.00m and (ii) x = 4.00 m. Give you answers in the vector form.

(b) A point charge q = 250 nC is added to the y axis at y = 2.00 m. Determine the new values of electric field at positions (i) and (ii). Give you answers in the vector form.

2. Relevant equations
∫|E|dA = Q/ε0
σ = Q/4πr2

3. The attempt at a solution
I think I've done the first one my sing a simple gaussian surface arriving at 0 electric field at x = 2.00 as inside the hollow sphere and E = σR20r2 = 190.6 Nc-1 i

However for part b i get confused as I'm not sure as to wether the charge not being in the centre affects it due to the distribution changing.
Currently I'm thinking that when x = 2.0m you create a gaussian surface where r=R and the charge enclosed is only that of q. For x = 4 could you do the same as before except the charge enclosed is now σrπR2 + q. That is what i would do if the additional charge was at the origin but as it is not i don't know how to take this into account.
Thanks

2. Feb 1, 2016

### Incand

The first part seems correct to me.
For the second part you're can only use Gauss's law to determine the $E-$field when you know the direction of the $E$ field. This is only possible when you have certain perfect symmetries (spherical, infinite plane, infinite cylinder symmetry) which isn't true unless the charge is at the centre. However luckily the $E$ fields obey the superposition principle. So The total field at a point $r$ is simply $E(r) = E_1(r) + E_2(r)$ if you have fields from two charge distributions.

3. Feb 1, 2016

### squeak

So could i just find the electric field of the point charge using kq/r2 and add it to the electric field found by using Gauss law?

4. Feb 1, 2016

### Incand

Yes but remember the fields also have a direction each so you would have to do vector addition.

5. Feb 1, 2016

### squeak

I often get the vector part wrong - so for when x = 4m would E due to q = kq/r2 where r = √(22+42) but that would be in-between the x/y directions. Would I then resolve this is the x direction by taking Esin(θ) where θ=arctan(4/2). And to this term i finally add the field gained in part 1).

6. Feb 1, 2016

### Incand

Involving the trigonometric functions works but is unnecessarily complicated imo. The direction of $r$ is simply $\hat r$ which just is the normed position vector (which you already know in Cartesian coordinates). Drawing a figure often helps when figuring out the direction as well if you are uncertain.

7. Feb 1, 2016

### squeak

Thanks! I'll try and do it that way as i think thats the way we're supposed to! Thank you so much for all of your help.