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Question Number 60219 by ANTARES VY last updated on 19/May/19

C=((2š›‘š›œš›œ_0 L)/(ln((R_2 /R_1 )))). prove.

$$\boldsymbol{\mathrm{C}}=\frac{\mathrm{2}\boldsymbol{\pi\varepsilon\varepsilon}_{\mathrm{0}} \boldsymbol{\mathrm{L}}}{\boldsymbol{\mathrm{ln}}\left(\frac{\boldsymbol{\mathrm{R}}_{\mathrm{2}} }{\boldsymbol{\mathrm{R}}_{\mathrm{1}} }\right)}. \\ $$$$\boldsymbol{\mathrm{prove}}. \\ $$

Commented by ANTARES VY last updated on 19/May/19

Answered by ajfour last updated on 19/May/19

C=(q/V) = (q/(āˆ«_R_1 ^( R_2 ) E.dr)) from Gaussā€² law E=((q/L)/(2Ļ€ĪµĪµ_0 r)) so C=(q/V)= (q/(āˆ«_R_1 ^( R_2 ) (((q/L))/(2Ļ€ĪµĪµ_0 r))dr)) ā‡’ C=((2Ļ€ĪµĪµ_0 L)/(āˆ«_R_1 ^( R_2 ) (dr/r))) =((2Ļ€ĪµĪµ_0 L)/(ln ((R_2 /R_1 )))) .

$$\mathrm{C}=\frac{\mathrm{q}}{\mathrm{V}}\:=\:\frac{\mathrm{q}}{\int_{\mathrm{R}_{\mathrm{1}} } ^{\:\:\mathrm{R}_{\mathrm{2}} } \mathrm{E}.\mathrm{dr}} \\ $$$$\mathrm{from}\:\mathrm{Gauss}'\:\mathrm{law}\:\:\mathrm{E}=\frac{\mathrm{q}/\mathrm{L}}{\mathrm{2}\pi\epsilon\epsilon_{\mathrm{0}} \mathrm{r}} \\ $$$$\mathrm{so}\:\:\:\:\mathrm{C}=\frac{\mathrm{q}}{\mathrm{V}}=\:\frac{\mathrm{q}}{\int_{\mathrm{R}_{\mathrm{1}} } ^{\:\:\mathrm{R}_{\mathrm{2}} } \frac{\left(\mathrm{q}/\mathrm{L}\right)}{\mathrm{2}\pi\epsilon\epsilon_{\mathrm{0}} \mathrm{r}}\mathrm{dr}} \\ $$$$\:\:\:\:\:\:\Rightarrow\:\:\mathrm{C}=\frac{\mathrm{2}\pi\epsilon\epsilon_{\mathrm{0}} \mathrm{L}}{\int_{\mathrm{R}_{\mathrm{1}} } ^{\:\:\mathrm{R}_{\mathrm{2}} } \:\frac{\mathrm{dr}}{\mathrm{r}}}\:=\frac{\mathrm{2}\pi\epsilon\epsilon_{\mathrm{0}} \mathrm{L}}{\mathrm{ln}\:\left(\frac{\mathrm{R}_{\mathrm{2}} }{\mathrm{R}_{\mathrm{1}} }\right)}\:. \\ $$

Commented by ANTARES VY last updated on 19/May/19

?

$$? \\ $$

Commented by ajfour last updated on 19/May/19

Calculate Electric field from Gaussā€² law, rest i have explained how to use that result.

$$\mathrm{Calculate}\:\mathrm{Electric}\:\mathrm{field}\:\mathrm{from} \\ $$$$\mathrm{Gauss}'\:\mathrm{law},\:\mathrm{rest}\:\mathrm{i}\:\mathrm{have}\:\mathrm{explained} \\ $$$$\mathrm{how}\:\mathrm{to}\:\mathrm{use}\:\mathrm{that}\:\mathrm{result}. \\ $$

Commented by ANTARES VY last updated on 19/May/19

E=((q/L)/(2š›‘š›œš›œ_0 r)) (q/L)=?. (1/(2š›‘š›œš›œ_0 r))=((2k)/(š›œr))=(2/C) C=((š›œR)/k).

$$\boldsymbol{\mathrm{E}}=\frac{\frac{\boldsymbol{\mathrm{q}}}{\boldsymbol{\mathrm{L}}}}{\mathrm{2}\boldsymbol{\pi\varepsilon\varepsilon}_{\mathrm{0}} \boldsymbol{\mathrm{r}}}\:\:\:\frac{\boldsymbol{\mathrm{q}}}{\boldsymbol{\mathrm{L}}}=?.\:\:\:\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{\pi\varepsilon\varepsilon}_{\mathrm{0}} \boldsymbol{\mathrm{r}}}=\frac{\mathrm{2}\boldsymbol{\mathrm{k}}}{\boldsymbol{\varepsilon\mathrm{r}}}=\frac{\mathrm{2}}{\boldsymbol{\mathrm{C}}}\:\: \\ $$$$\boldsymbol{\mathrm{C}}=\frac{\boldsymbol{\varepsilon\mathrm{R}}}{\boldsymbol{\mathrm{k}}}.\:\: \\ $$

Commented by ajfour last updated on 19/May/19

Commented by ajfour last updated on 19/May/19

According to Gaussā€² law flux of field through a Gaussian surface = (1/Īµ_0 )(charge within Gaussian surface) assuming charge q and āˆ’q on capacitor plates ĪµE(2Ļ€rL)=(1/Īµ_0 )(q) ā‡’ E=((q/L)/(2Ļ€ĪµĪµ_0 r)) .

$$\mathrm{According}\:\mathrm{to}\:\mathrm{Gauss}'\:\mathrm{law} \\ $$$$\:\:\:\:\mathrm{flux}\:\mathrm{of}\:\mathrm{field}\:\mathrm{through}\:\mathrm{a}\:\mathrm{Gaussian} \\ $$$$\mathrm{surface}\:=\:\frac{\mathrm{1}}{\epsilon_{\mathrm{0}} }\left(\mathrm{charge}\:\mathrm{within}\:\mathrm{Gaussian}\:\mathrm{surface}\right) \\ $$$$\mathrm{assuming}\:\mathrm{charge}\:\mathrm{q}\:\mathrm{and}\:āˆ’\mathrm{q}\:\mathrm{on}\:\mathrm{capacitor} \\ $$$$\mathrm{plates} \\ $$$$\:\:\epsilon\mathrm{E}\left(\mathrm{2}\pi\mathrm{rL}\right)=\frac{\mathrm{1}}{\epsilon_{\mathrm{0}} }\left(\mathrm{q}\right) \\ $$$$\Rightarrow\:\:\mathrm{E}=\frac{\mathrm{q}/\mathrm{L}}{\mathrm{2}\pi\epsilon\epsilon_{\mathrm{0}} \mathrm{r}}\:. \\ $$

Commented by ANTARES VY last updated on 19/May/19

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