Maximum Voltage Between Long Coaxial Cylinders

[cwill53]

Homework Statement

Derive an expression for the maximum voltage allowable between long coaxial cylinders of radii $R_1$, $R_2$ ($R_2> R_1$), if the field strength in the insulation between them is not to exceed $E_m$.

Answer: $E_mR_1ln\frac{R_2}{R_1}$

Relevant Equations

$C=\frac{Q}{V}$
$E=\frac{Q-q}{\varepsilon _0S}$

At first, I started with the result from an earlier problem regarding the capacitance of a cylindrical capacitor:

$$C=\frac{Q}{V}=\frac{2\pi \varepsilon _0\varepsilon _rl}{ln(R_1/R_2)}$$
$$\Rightarrow V=\frac{Qln(R_2/R_1)}{2\pi \varepsilon _0\varepsilon _rl }$$

Then I used the equation involving the electric field between the plates of a parallel capacitor filled with a dielectric material:
$$E_m=\frac{Q-q}{\varepsilon _0S}$$
$$C=\frac{Q}{V}=\frac{Q}{Q-q}\frac{\varepsilon _0}{d}$$
$$C=\frac{Q}{V}=\frac{Q}{Q-q}\frac{\varepsilon _0}{d}$$
where d would be the distance between the parallel plate capacitor's plates, but I'll take to be $R_2-R_1$ for the cylindrical plates.
$$\frac{Q}{V}=\frac{Q\varepsilon _0S}{(Q-q)d}\Rightarrow V=\frac{Q(Q-q)d}{Q\varepsilon _0S}=E_md$$
I'm not sure where to proceed from here.

 

[TSny]

Then I used the equation involving the electric field between the plates of a parallel capacitor filled with a dielectric material:
$$E_m=\frac{Q-q}{\varepsilon _0S}$$

You cannot assume that the parallel plate expression for E is applicable to the cylindrical plates. I assume that $S$ is the surface area of one of the plates and $q$ is the induced charge on the surface of the dielectric. Note that the plates of the cylindrical capacitor do not have the same area.

The electric field is not uniform between the plates of a cylindrical capacitor. You can use Gauss' law to find how the field varies between the plates. Then you can see where the field is a maximum. You might want to start with the case of no dielectric material between the plates. Adding the dielectric just changes the field in a simple manner.

 

[cwill53]

You cannot assume that the parallel plate expression for E is applicable to the cylindrical plates. I assume that $S$ is the surface area of one of the plates and $q$ is the induced charge on the surface of the dielectric. Note that the plates of the cylindrical capacitor do not have the same area.

The electric field is not uniform between the plates of a cylindrical capacitor. You can use Gauss' law to find how the field varies between the plates. Then you can see where the field is a maximum. You might want to start with the case of no dielectric material between the plates. Adding the dielectric just changes the field in a simple manner.

I decuded from this that E is maximum when R is minimum.
$$\Phi =EA=E2\pi Rl=\frac{Q}{\varepsilon}\Rightarrow E=\frac{Q}{2\pi \varepsilon Rl};\varepsilon =\varepsilon _r\varepsilon _0$$
$$V=\int -EdR=-\frac{Q}{2\pi \varepsilon l}\int_{R_2}^{R_1}\frac{dR}{R}=\frac{Q}{2\pi \varepsilon l}\int_{R_1}^{R_2}\frac{1}{R}dR$$
From this,
$$V=\frac{Q}{2\pi \varepsilon l}ln\left ( \frac{R_2}{R_1} \right )$$
The minimum that R can be is $R_1$, and this sets the maximum corresponding charge $Q_m$.
$$E_m=\frac{Q_m}{2\pi \varepsilon R_1l}\Rightarrow Q_m=E_m2\pi \varepsilon R_1l$$
$$V_m=\frac{Q_m}{2\pi \varepsilon l}ln\left ( \frac{R_2}{R_1} \right )=\frac{E_m2\pi \varepsilon R_1l}{2\pi \varepsilon l}ln\left ( \frac{R_2}{R_1} \right )$$
$$\Rightarrow V_m=E_mR_1ln\left ( \frac{R_2}{R_1} \right )$$

 

[TSny]

Looks very good