## Cylinderical capacitor

- A cylinderical capacitor is made up of a conducting cylinder or wire of radius a surrounded by another concentric cylinderical shell of radius b (b>a).

- Let L be the length of both the cylinders and charge on inner cylender is +Q and charge on outer cylinder is -Q.

- For calculate electric field between the conductors using Gauss's law consider a Gaussian surface of radius r and length L
^{1} as shown in figure 4.

- According to Gauss's law flux through this surface is $ \frac {q}{\epsilon _0}$ where q is net charge inside this surface.

- We know that electric flux is given by

$\phi =E.A$

$=EA cos \theta$

$=EA$

since electric field is constant in magnitude on the Gaussian surface and is perpendicular to this surface. Thus,

$\phi=E(2 \pi rL)$

since $\phi=\frac {q}{\epsilon _0}$

=> $ E(2 \pi rL)=\frac {\lambda L}{\epsilon _0}$

where $\lambda = Q/L$ = charge per unit length

So,
$E=\frac{\lambda }{2\pi \epsilon _{0}r}$

- If potential at inner cylinder is V
_{a} and V_{b} is potential of outer cylinder then potential difference between both the cylinders is

V=V_{a} and V_{b}=∫Edr

where limits of integration goes from a to b.

- Potential of inner conductor is greater then that of outer conductor because inner cylinder carries positive charge. Thus potential difference is
$$V=\frac {Qln(b/a)}{2\pi \epsilon _{0}L}$$

- Thus capacitance of cylinderical capacitor is

$C=\frac {Q}{V}

or,

$C=\frac{2\pi \epsilon _{0}L}{ln (b/a)}$ ------(5)

- From equation 5 it can easily be concluded that capacitance of a cylinderical capacitor depends on length of cylinders.

- More is the length of cylinders , more charge could be stored on the capacitor for a given potential difference.

**Question**

A cylindrical capacitor is constructed using two coaxial cylinders of the same length 10 cm of radii 5 mm and 10 mm.

(a) calculate the capacitance

(b) another capacitor of the same length is constructed with cylinders of radii 8 mm and 16 mm. Calculate the capacitance .

**Solution**

Capacitance is given by

$C=\frac{2\pi \epsilon _{0}L}{ln (b/a))}$

(a) Here L=10 cm = .1 m

b=10 mm=, a=5 mm

$C= \frac {2 \times 3.14 \times 8.88 \times 10^{-12} \times .10}{ln (10/5}$

$=8 pF$

(b) Now b=16 mm and a=8 mm

Since ln 16/8 = ln 10/5

C= 8pF

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