Another Cylindrical Capacitor (Electric Field and Potential)

Cylindrical Capacitor (Electric Field and Potential)

You are given a long cylindrical capacitor made from two aligned cylindrical conductors of radius $a$ (the inner massive conductor) and $b$ (the outer tube) with length $l$.

The capacitor is long, that is $l>>a$, or l is infinite. On the inner conductor there is a surface charge  $sigma$.

(a) Draw the capacitor with all the charges (initial and induced) on it, the electric field, and the Cartesian system.

(b) What is the electric field in between the conductors? Compute it by using Gauss law of your chosen surfaces.

(c) What is the potential difference V between the two conductors (it has to be positive)?

(d) What is the capacitance per unit length of this capacitor, that is $C/l$?

The figure is above. On the exterior surface of the inner cylinder we have surface charge density

$σ_1=+ Q/S(a)$    where $S(a)=2πa*L$
On the interior surface of the outer shell we have the surface charge density

$σ2=-Q/S(b)$   where $S(b)=2πb*L$

The total charge in absolute value, Q is the same on both inner and outer conductors.

Consider a Gaussian cylindrical surface of radius a< R< b and length L. (see the figure). On the bases of the cylinder electric field E is opposite and therefore total flux is zero( the two bases have the same surface).

On the lateral surface of the cylinder Gauss law is

$∅=E(R)*S(R)=Q(inside)/ϵ_0$

$Q(inside)=σ1*2πaL$ and $S(R)=2πRL$

$E(R)*R=σ1*a/ϵ0$

$E(R)=σ1/ϵ0*a/R$

Potential difference between two conductors is

$V(b)-V(a)=int_a^b (E(R)*d R)=(σ1*a)/ϵ0*ln⁡(b/a)$

Capacitance per unit length is

$C/L=(Q/L)/V(ab) =(2πa*σ1)/((σ1*a)/ϵ0*ln⁡(b/a))=2πϵ0*[1/ln⁡(b/a)]$