Cylindrical capacitor

An air-filled capacitor is made from two concentric conducting cylinders. The inner cylinder has radius R1, the outer radius R2. Let the breakdown electric field in air be Eb.

i. What value of R1 allows for the maximum potential difference between conductors before breakdown in air?

ii. Suppose the outer cylinder has radius 1 cm and the breakdown electric field is 3 MV/m. Calculate the maximum potential difference just before breakdown.

Take a Gaussian cylindrical surface in the space between the 2 armatures. The total flux on the cylinder bases is zero (opposite electric fields). On the lateral surface of this cylinder one has:


$E(r)*2\pi r*L=\lambda L/\epsilon$   thus $E(r)=\lambda/2\pi \epsilon r$

The electric field is maximum Eb at the surface of the inner armature (smaller r). Thus

$E_b=\lambda/(2\pi\epsilon R_1 )$,   thus $\lambda=2\pi \epsilon R_1*E_b$

Capacitance of cylindrical capacitor (per unit length) is:

$C/L=(2\pi\epsilon)/(ln⁡(R_2/R_1 ))$

(This is found by integrating the above field between R1 and R2.)

For a given charge Q on the capacitor the potential difference is

$U=Q/C=(Q/L)/(C/L)=(\lambda/2\pi\epsilon)*ln⁡(R_2/R_1 )=R_1*E_b*ln⁡(R_2/R_1 )$

The condition of maximum for U is

$0= d U/(d R_1 )=E_b*ln⁡(R_2/R_1 )-R1*E_b*(1/R_1 )=E_b*[ln⁡(R_2/R_1 )-1]$

The value of R_1 for maximum voltage on capacitor is thus

$ln⁡(R2/R1)=1$  or $R_1=R_2/e$

For $R2 =1 cm$, $R1 = (1/e)=0.3679 cm$, $Eb =3 MV/m$ this implies

$U=R_1*E_b*ln⁡(R2/R1)=0.3679*10^{-2}*3*10^6*ln⁡(1)=1.104*10^4  V=11036 V$