By Sunil Bhardwaj

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CAPACITANCE OF CYLINDRICAL CAPACITOR

Consider a cylindrical capacitor of length $$L$$. It is constructed with two coaxial cylinders such that inner cylinder has radius $$a$$ and charge $$+Q$$ while outer cylinder has radius $$b$$ and charge $$-Q$$ as shown in fig.

The electric field lines come out from positively charged cylinder and enter into negatively charged cylinder.

Now imagine a Gaussian surface in the form of a cylinder of radius r encloses the positively charged inner cylinder of capacitor. The Gaussian cylinde three surfaces.

The flux through top-capped Gaussian cylinder having area element $$da$$ is $$\phi_1\ =\oint{\vec{E}.\vec{da}}$$ $$\ \phi_1\ =\ \int{E\ da\ \cos{90}}$$ $$\phi_1\ =\ 0$$ The flux through bottom capped Gaussian cylinder having area element $$da$$ is $$\phi_2\ =\oint{\vec{E}.\vec{da}}$$ $$\ \phi_2\ =\ \int{E\ da\ \cos{90}}$$ $$\phi_2\ =\ 0$$ The flux through curved surface of Gaussian cylinder having area element $$da$$ is $$\phi_3\ =\oint{\vec{E}.\vec{da}}$$ $$\ \phi_3\ =\ \int{E\ da\ \cos{0}}$$ $$\phi_3\ =\ E\left(2\pi r\ L\right)$$ Where $$2\pi r$$ is circumference of Gaussian cylinder and $$L$$ is its length. Net flux through whole Gaussian cylinder is $$\phi_c\ =\ \phi_1+\phi_2+\phi_3$$ $$\phi_c\ =\ 0+0+E\left(2\pi r\ L\right)$$ $$\phi_c\ =\ E\left(2\pi r\ L\right)$$

. . . . . . . . . . . . . (1)

From Gauss's law $$\phi_c\ =\ \frac{Q}{\varepsilon_0}$$

. . . . . . . . . . . . . (2)

Comparing eq (1) and eq (2), we get $$E\left(2\pi r\ L\right)\ =\ \frac{Q}{\varepsilon_0}$$ $$E\ =\ \frac{Q}{\left(2\pi r\ L\right)\varepsilon_0}$$ The potential related to this electric field of cylindrical capacitor is $$V\ =\ \int_{a}^{b}{E\ ds}$$ $$V\ =\ \int_{a}^{b}{\frac{Q}{\left(2\pi r\ L\right)\varepsilon_0}\ dr}$$ $$V\ =\ \frac{Q}{\left(2\pi\ L\right)\varepsilon_0}\int_{a}^{b}{\frac{1}{r}\ dr}$$ $$V\ =\ \frac{Q}{\left(2\pi\ L\right)\varepsilon_0}\left[\ln{r}\right]_a^b$$ $$V\ =\ \frac{Q}{\left(2\pi\ L\right)\varepsilon_0}\left(\ln{b}-\ \ln{a}\right)$$ $$V\ =\ \frac{Q}{\left(2\pi\ L\right)\varepsilon_0}\ln{\left(\frac{b}{a}\right)}$$ $$\frac{\left(2\pi\ L\right)\varepsilon_0}{\ln{\left(\frac{b}{a}\right)}}\ =\ \frac{Q}{V}$$ The capacitance of a capacitor is related with voltage as $$C=\ \frac{Q}{V}$$ Comparing eq(1) and eq(2) $$C=\ \frac{\left(2\pi\ L\right)\varepsilon_0}{\ln{\left(\frac{b}{a}\right)}}$$ This is the capacitance of a cylindrical capacitor which depends upon length radius "$$a$$" of inner cylinder and radius "$$b$$" of outer cylinder.

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