Potential infinite hollow cylinder and infinite wire placed side by side
Answers
I am trying to find the electric field perpendicular to the surface of the hollow cylinder. I think the easiest way is Gauss' law which is;
ϕE=∫SEdA=Qϵ0
Thus when we apply the Gaussian surface(whom I chose as a cylinder) we should take the surface integral of it. However, I can not figure the rest out. What would be the final equation when we try to find E in means of q, r, h, π and ϵ0?
Moreover, would the electric field inside the cylinder be 0 because of the symmetry and since all electric field vectors that are exerted by each charge would cancel each other out?
I'm assuming here that the cylinder is "infinitely long", or at least very long so that h>>r. Otherwise, there are complicated non-integrable "end effects", but it doesn't look like you're interested in those.
As you said, you need Gauss's law. You're losing something important if you drop the vector notation, though, so I've added it back in.
ϕE=∫SE⃗ ⋅dA⃗ =Qϵ0
The critical part, which you've already done, is to choose a surface on which E is constant, so the integral is easy to evaluate.
You should pick a cylindrical surface of radius R and height L, centered on the axis of the charged cylinder. An illustration of that surface (in green) is shown here:enter image description here
With that choice of surface, simply by symmetry, the E field must have the same value, and the same direction relative to the normal vector dA⃗ , everywhere on the curved part of the surface!
On the flat ends of your cylindrical surface, the E field is not constant, but it is parallel to the surface vector dA⃗ , so the dot product E⃗ ⋅dA⃗ is zero on those surfaces, and we can ignore them in the integral.
Here I've split the integral into the two parts - the integral over the curved part of the surface, and the flat ends, and I've evaluated both parts of the integral.
∫curvedE⃗ ⋅dA⃗ +∫endsE⃗ ⋅dA⃗ =Qϵ0
∫curvedEdA+0=Qϵ0
E∫curveddA=Qϵ0
E Acurved=Qϵ0
E 2πRL=Qϵ0
Notice that on the curved part, since E⃗ and dA⃗ are in the same direction, their dot product is simply EdA, and since the magnitude E is the same everywhere, we can remove it from the integral as a constant. Then we're simply integrating dA, which simply gives us the area of that part of the surface.
The last job we have is to find how much charge (Q) is inside our surface. The charge density is σ=q2πrh
There are two cases. 1. If R>r, then the Q=2πrLσ=qLh. 2. If R<r, then no charge is inside our surface, so Q=0.
So, using our final version of Gauss's law:
E 2πRL=Qϵ0
E=Q2πϵ0RL
And our final answer is:
R<r;
E(R)=0
R>r;