- #1
cwill53
- 220
- 40
- 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.
$$C=\frac{Q}{V}=\frac{2\pi \varepsilon _0\varepsilon _rl}{ln(R_1/R_2)}$$
$$\Rightarrow V=\frac{Qln(R_2/R_1)}{2\pi \varepsilon _0\varepsilon _rl }$$
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.