Gauss' Law vector form problem

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.