Question

What is the capacitance of a cylindrical capacitor of length $L$ and inner and outer radius $a$ and $b$ respectively (in farad)?

• A. $\frac{2\pi \epsilon L}{2.3{\log }_{10}\left(b/a\right)}$
• B. $\frac{2\pi \epsilon L}{2.3{\log }_{10}\left(a/b\right)}$
• C. $\frac{2\pi \epsilon L}{4.6{\log }_{10}\left(b/a\right)}$
• D. $\frac{2\pi \epsilon L}{1.3{\log }_{10}\left(b/a\right)}$

Answer 1

The correct option is A $\frac{2\pi \epsilon L}{2.3{\log }_{10}\left(b/a\right)}$

If V is the potential difference between inner and outer shells of radii $a,b$ respectively. Hence, capacitance of cylindrical capacitor will be

$C=q/V$

Let us draw a gaussian cylindrical surface around inner shell, of radius $r$, then by Gauss's theorem , total electric flux through gaussian surface,

$\varphi =q/{\epsilon }_0$ .........eq1 ($q=$ charge on inner shell)

Now, if magnitude of electric field at gaussian surface is $E$, then

$\varphi =E\times 2\pi rL$ ............eq2

equating eq1 and eq2 ,

$E\times 2\pi rL=q/{\epsilon }_0$

or $E=\frac{q}{2\pi {\epsilon }_{0}rL}$ ................eq3

therefore, potential difference between inner and outer shell ,

$V={\int }_a^{b}Edr={\int }_a^b\frac{q}{2\pi {\epsilon }_{0}rL}dr$

or $V=\frac{q}{2\pi {\epsilon }_{0}L}{\int }_a^{b}dr/r$

or $V=\frac{q}{2\pi {\epsilon }_{0}L}[{\log }_{e}r{]}_a^b$

or $V=\frac{q}{2\pi {\epsilon }_{0}L}{\log }_e\left(b/a\right)$

Therefore capacitance of capacitor is,

$C=\frac{q}{\frac{q}{2\pi {\epsilon }_{0}L}{\log }_e\left(b/a\right)}=\frac{2\pi {\epsilon }_{0}L}{{\log }_e\left(b/a\right)}$

or $C=\frac{2\pi {\epsilon }_{0}L}{2.3{\log }_{10}\left(b/a\right)}$