Posted by Phy on .
Two conducting thin hollow cylinders are co-aligned. The inner cylinder has a radius R1 , the outer has a radius R2 . Calculate the electric potential difference V(R2)−V(R1) between the two cylinders. The inner cylinder has a surface charge density of σa=−σ , where σ>0 , and the outer surface has a surface charge density of σb=3σ ,
The cylinders are much much longer than R1 . Thus, you may ignore end effects and neglect the thickness of the cylinders.
a. What is the electric potential difference between the outer cylinder and the inner cylinder V(R2)−V(R1)?
Express your answer in terms of R1, R2,σ, and ϵ0
b. What is the magnitude of the electric field outside the cylinders, r>R2?
Express your answer in terms of r,R1, R2,σ, and ϵ0.
c. What is the electric potential difference between a point at a distance r=2*R2 from the symmetry axis and the outer cylinder V(2*R2)−V(R2)?
Express your answer in terms of R1, R2,σ, and ϵ0.
a. Gauss law applies. I would set V=0 on the inner cylinder, since V =0 is an arbitrary location.
Then, between the cylinders, you know E due to the enclosed charge (gauss law).
Potential then is INTEGRAL of E dr. That is straightforward.
b. E outside? You know the NET charge enclosed, find E for that for an infinitely long cylinder
c. Again, now set V=0 at R2, then compute V at any r as in a.
and the aswers are?
a little more explanation please