Difference between revisions of "SolPr1"
From 105/106 Lecture Notes by OBM
(Created page with "==Question1== We consider the cylinder as two cylindrical capacitors in parallel. The two “negative plates” are the (connected) halves of the inner cylinder (half of which...") |
|||
Line 6: | Line 6: | ||
vapor). | vapor). | ||
− | <math>C=C_{\textrm{liq}}+C_{\textrm{V}}=\frac{2\pi\epsilon_0 K_{\textrm{liq}}h}{ln(R_a/R_b)}+\frac{2\pi\epsilon_0 K_{\textrm{V}}(l-h)}{ln(R_a/R_b)}=\frac{2\pi\epsilon_0 l}{ln(R_a/R_b)}\left[ | + | <math>C=C_{\textrm{liq}}+C_{\textrm{V}}=\frac{2\pi\epsilon_0 K_{\textrm{liq}}h}{\ln(R_a/R_b)}+\frac{2\pi\epsilon_0 K_{\textrm{V}}(l-h)}{\ln(R_a/R_b)}=\frac{2\pi\epsilon_0 l}{\ln(R_a/R_b)}\left[(K_{\textrm{liq}}-K_{\textrm{V}})\frac{h}{l}+K_{\textrm{V}}\right]=C</math> |
+ | |||
+ | <math>\frac{h}{l}=\frac{1}{(K_{\textrm{liq}}-K_{\textrm{V}})}\left[\frac{C\ln(R_a/R_b)}{2\pi\epsilon_0 l}-K_{\textrm{V}}\right]</math> |
Revision as of 01:35, 18 March 2019
Question1
We consider the cylinder as two cylindrical capacitors in parallel. The two “negative plates” are the (connected) halves of the inner cylinder (half of which is in contact with liquid, and half of which is in contact with vapor). The two “positive plates” are the (connected) halves of the outer cylinder (half of which is in contact with liquid, and half of which is in contact with vapor).