Write an expressions for electric potential due to a continuous distribution of charges.

There are three ways for a continuous charge distribution (as seen in chapter $1$).

For linear charge distribution, electric potential at point P,

$\mathrm{V}_{\mathrm{L}}=k \int_{\mathrm{L}} \frac{\lambda d \mathrm{~L}}{\vec{r}-\vec{r}_{i} \mid}$

where $\lambda=$ linear charge density

$d \mathrm{~L}=$ small segment of length

$\vec{r}=$ position vector of point $\mathrm{P}$

$\vec{r}_{i}=$ position vector of $d \mathrm{~L}$ segment $i=1,2, \ldots, \mathrm{N}$

For surface charge distribution electric potential at point $P$,

$\mathrm{V}_{\mathrm{S}}=k \int_{\mathrm{S}} \underset{\left|\vec{r}-\vec{r}_{i}\right|}{\vec{\rightarrow} d \mathrm{~S}}$

where $\sigma=$ surface charge density

$d \mathrm{~S}=$ area of small surface element

$\vec{r}=$ position vector of point $\mathrm{P}$

$\overrightarrow{r_{i}}=$ position vector of $d \mathrm{~S}$ element $i=1,2, \ldots, \mathrm{N}$

For volume charge distribution electric potential at point $\mathrm{P}$

$\mathrm{V}_{\mathrm{V}}=k \int \frac{\rho d \mathrm{~V}}{\overrightarrow{\mathrm{V}}\left|\vec{r}-\vec{r}_{i}\right|}$ where $\rho=$ volume charge density $d \mathrm{~V}=$ volume of small volume element $\vec{r}=$ position vector of point $\mathrm{P}$ $\overrightarrow{r_{i}}=$ position vector of $d \mathrm{~V}$ element $i=1,2, \ldots, \mathrm{N}$

$d \mathrm{~V}=$ volume of small volume element

$\vec{r}=$ position vector of point $\mathrm{P}$

$\overrightarrow{r_{i}}=$ position vector of $d \mathrm{~V}$ element $i=1,2, \ldots, \mathrm{N}$