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

(N/A) $(1)$ Line charge distribution: Suppose a line is divided into smaller elements of length $dl$. Let $\vec{r}$ be the position vector of a small element with linear charge density $\lambda$,so its charge is $dq = \lambda dl$.
Consider a point $P$ with position vector $\vec{R}$. Let $r^{\prime}$ be the distance from the element $dl$ to $P$,and $\hat{r}^{\prime}$ be the unit vector pointing from the element to $P$. The electric field at $P$ due to the element is:
$\vec{dE} = \frac{k \lambda dl}{(r^{\prime})^{2}} \hat{r}^{\prime}$
By the superposition principle,the total electric field at $P$ is:
$\vec{E} = \int_{l} \frac{k \lambda dl}{(r^{\prime})^{2}} \hat{r}^{\prime}$
$(2)$ Surface charge distribution: Suppose a surface $S$ is divided into small elements $\Delta S$. Let $\sigma$ be the surface charge density,so the charge on an element is $dq = \sigma dS$.
The electric field at point $P$ due to the surface element is:
$\vec{dE} = \frac{k \sigma dS}{(r^{\prime})^{2}} \hat{r}^{\prime}$
By the superposition principle,the total electric field at $P$ is:
$\vec{E} = \int_{S} \frac{k \sigma dS}{(r^{\prime})^{2}} \hat{r}^{\prime}$
$(3)$ Volume charge distribution: Suppose a volume $V$ has a charge density $\rho$. The charge in a small volume element $dV$ is $dq = \rho dV$.
The electric field at point $P$ due to the volume element is:
$\vec{dE} = \frac{k \rho dV}{(r^{\prime})^{2}} \hat{r}^{\prime}$
By the superposition principle,the total electric field at $P$ is:
$\vec{E} = \int_{V} \frac{k \rho dV}{(r^{\prime})^{2}} \hat{r}^{\prime}$