Obtain the expression of electric field at any point by continuous distribution of charge on a  $(i)$ line $(ii)$ surface $(iii)$ volume.

$(1)$ Suppose, line is divided into smaller elements of $d l$ length and $\vec{r}$ is the position vector of any smaller element and its linear charge density is $\lambda$ and its charge is $\lambda d l$.

Suppose a point $\mathrm{P}$ (inside or outside) whose, position vector is $\overrightarrow{\mathrm{R}}$. $\mathrm{P}$ is at $r^{\prime}$ distance from $\Delta l$ element and unit vector is $\hat{r}^{\prime}$. Electric field at $P$ due to $\lambda \Delta l$

$\overrightarrow{\Delta \mathrm{E}}=\frac{k \lambda \Delta l}{\left(r^{\prime}\right)^{2}} \cdot \hat{r}^{\prime}$

Total electric field at $P$ by superposition principle,

$\overrightarrow{\mathrm{E}}=\sum_{\Delta l} \frac{k \lambda \Delta l}{\left(r^{\prime}\right)^{2}} \cdot \hat{r}^{\prime}$

By integration method,

$\overrightarrow{\mathrm{E}}=\int_{l} \frac{k \lambda d l}{\left(r^{\prime}\right)^{2}} \hat{r}^{\prime}$

$(2)$ Suppose, surface $\Delta S$ is divided into small elements and $\vec{r}$ is the position vector on anyone element.

$\sigma$ is the surface charge density hence, charge on $\Delta \mathrm{S}$ surface element $\Delta \mathrm{Q}=\sigma \Delta \mathrm{S} . \quad \therefore \quad \sigma=\frac{\Delta \mathrm{Q}}{\Delta \mathrm{S}}$

Suppose a point $\mathrm{P}$ (inside or outside) the surface whose position vector is $\overrightarrow{\mathrm{R}}$ and distance from $\Delta \mathrm{S}$ is $r^{\prime}$ and unit vector is $\hat{r}^{\prime}$.

Electric field at $P$ due to charge on $\sigma \Delta \mathrm{S}$,

$\overrightarrow{\Delta \mathrm{E}}=\frac{k \sigma \Delta \mathrm{S}}{\left(r^{\prime}\right)^{2}} \cdot \hat{r}^{\prime}$

Total electric field at$ P$ by superposition principle,

$\overrightarrow{\mathrm{E}}=\sum_{\mathrm{S}} \frac{k \sigma \Delta \mathrm{S}}{\left(r^{\prime}\right)^{2}} \hat{r}^{\prime}$

By integration method,

$\overrightarrow{\mathrm{E}}=\int_{\mathrm{S}} \frac{k \sigma \Delta \mathrm{S}}{\left(r^{\prime}\right)^{2}} \hat{r}^{\prime}$

$(3)$ Suppose a continuous charge distribution in space has a charge density $\rho$.

Choose any convenient origin $\mathrm{O}$ and let the position vector of any point in the charge distribution be $\vec{r}$.

Divide the charge distribution into small volume elements of size $\Delta \mathrm{V}$.

The charge in a volume element $\Delta \mathrm{V}$ is $\rho \Delta \mathrm{V}$.

Now, consider any general point $P$ (inside or outside the distribution with position vector $\vec{R}$.

Electric field due to the charge $\rho \Delta \mathrm{V}$ is given by Coulomb's law.

$\overrightarrow{\mathrm{E}}=\sum \frac{k \rho \Delta \mathrm{V}}{\left(r^{\prime}\right)^{2}} \cdot \hat{r}^{\prime}$

where $r^{\prime}$ is the distance between the charge element and $\mathrm{P}$ and $\hat{r}^{\prime}$ is a unit vector in the direction from the charge element to $\mathrm{P}$.

By the superposition principle, the total electric field,

$\overrightarrow{\mathrm{E}}=\int_{\mathrm{V}} \frac{k \rho \Delta \mathrm{V}}{\left(r^{\prime}\right)^{2}} \hat{r}^{\prime}=k \cong \frac{\rho \Delta \mathrm{V}}{r^{\prime 2}} \hat{r}^{\prime} $

$\mathrm{OR} \overrightarrow{\mathrm{E}}=k \int \frac{\rho \Delta \mathrm{V}}{r^{\prime 2}}$ $\hat{r}$

In short, using Coulomb's law and the superposition principle, electric field can be determined for any charge distribution, discrete or continuous or part discrete and part continuous.