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

$$\mathbf{C}=\frac{2\mathit{\pi }\mathit{\epsilon }{\mathit{\epsilon }}_0\mathbf{L}}{\mathbf{ln}\left(\frac{{\mathbf{R}}_2}{{\mathbf{R}}_1}\right)}.$$$$\mathbf{prove}.$$

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

Answered by ajfour last updated on 19/May/19

$$\mathrm{C}=\frac{\mathrm{q}}{\mathrm{V}}=\frac{\mathrm{q}}{{\int }_{{\mathrm{R}}_1}^{{\mathrm{R}}_2}\mathrm{E}.\mathrm{dr}}$$$$\mathrm{from}{\mathrm{Gauss}}^{\prime }\mathrm{law}\mathrm{E}=\frac{\mathrm{q}/\mathrm{L}}{2\pi ϵ{ϵ}_0\mathrm{r}}$$$$\mathrm{so}\mathrm{C}=\frac{\mathrm{q}}{\mathrm{V}}=\frac{\mathrm{q}}{{\int }_{{\mathrm{R}}_1}^{{\mathrm{R}}_2}\frac{\left(\mathrm{q}/\mathrm{L}\right)}{2\pi ϵ{ϵ}_0\mathrm{r}}\mathrm{dr}}$$$$\Rightarrow \mathrm{C}=\frac{2\pi ϵ{ϵ}_0\mathrm{L}}{{\int }_{{\mathrm{R}}_1}^{{\mathrm{R}}_2}\frac{\mathrm{dr}}{\mathrm{r}}}=\frac{2\pi ϵ{ϵ}_0\mathrm{L}}{\ln \left(\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}\right)}.$$

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

$$?$$

Commented by ajfour last updated on 19/May/19

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

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

$$\mathbf{E}=\frac{\frac{\mathbf{q}}{\mathbf{L}}}{2\mathit{\pi }\mathit{\epsilon }{\mathit{\epsilon }}_0\mathbf{r}}\frac{\mathbf{q}}{\mathbf{L}}=?.\frac{1}{2\mathit{\pi }\mathit{\epsilon }{\mathit{\epsilon }}_0\mathbf{r}}=\frac{2\mathbf{k}}{\mathit{\epsilon }\mathbf{r}}=\frac{2}{\mathbf{C}}$$$$\mathbf{C}=\frac{\mathit{\epsilon }\mathbf{R}}{\mathbf{k}}.$$

Commented by ajfour last updated on 19/May/19

Commented by ajfour last updated on 19/May/19

$$\mathrm{According}\mathrm{to}{\mathrm{Gauss}}^{\prime }\mathrm{law}$$$$\mathrm{flux}\mathrm{of}\mathrm{field}\mathrm{through}\mathrm{a}\mathrm{Gaussian}$$$$\mathrm{surface}=\frac{1}{{ϵ}_0}\left(\mathrm{charge}\mathrm{within}\mathrm{Gaussian}\mathrm{surface}\right)$$$$\mathrm{assuming}\mathrm{charge}\mathrm{q}\mathrm{and}-\mathrm{q}\mathrm{on}\mathrm{capacitor}$$$$\mathrm{plates}$$$$ϵ\mathrm{E}\left(2\pi \mathrm{rL}\right)=\frac{1}{{ϵ}_0}\left(\mathrm{q}\right)$$$$\Rightarrow \mathrm{E}=\frac{\mathrm{q}/\mathrm{L}}{2\pi ϵ{ϵ}_0\mathrm{r}}.$$

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

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