Question

A long cylinder of radius a has a charge of Q coul/m. Find the potential difference between two points at distances r1 and r2 from the axis of the cylinder.

Answer 1

Given:

A long cylinder of radius a has a charge of Q coulomb.

To find:

Potential difference between r1 and r2.

Calculation:

We shall consider a cylindrical Gaussian surface of radius r and height h :

$\therefore \displaystyle \: \int E.ds = \dfrac{Q}{\epsilon_{0}}$

$= > E \times (2\pi rh) = \dfrac{Q}{\epsilon_{0}}$

$= > E = \dfrac{Q}{2\pi\epsilon_{0}rh}$

Now , we know that :

$\therefore \: dV = E \: \times dr$

$= > dV = \dfrac{Q}{2\pi\epsilon_{0}rh} \times dr$

$= > dV = \dfrac{Q}{2\pi\epsilon_{0}h} \times \dfrac{ dr}{r}$

$\displaystyle = > \int dV = \dfrac{Q}{2\pi\epsilon_{0}h} \times \int \: \dfrac{ dr}{r}$

$\displaystyle = > \int_{0}^{V } dV = \dfrac{Q}{2\pi\epsilon_{0}h} \times \int_{r1}^{r2} \: \dfrac{ dr}{r}$

$\displaystyle = > V = \dfrac{Q}{2\pi\epsilon_{0}h} \times \int_{r1}^{r2} \: \dfrac{ dr}{r}$

$= > V = \dfrac{Q}{2\pi\epsilon_{0}h} \times ln \bigg(\dfrac{r2}{r1} \bigg)$

So , potential difference will be :

$\boxed{ \bold{ \blue{ V = \dfrac{Q}{2\pi\epsilon_{0}h} \times ln \bigg(\dfrac{r2}{r1} \bigg) }}}$