FIGURE shows a thin rod of length $L$with total charge $Q$. Find an expression for the electric field$\overrightarrow{E}$ at point $P$. Give your answer in component form.

A fundamental feature of matter held by some constituent particles that affects how the particles react to an external magnetic field Positive and negative electric charge exists in separate natural units and cannot be manufactured or destroyed.

At point, take a small amount of energy due to the axis charge electrical field.

$E_y=\frac{1}{4\pi {ϵ}_o}\times \left(\frac{Q}{L}dy\right)\times \frac{y}{{\left(n^2+y^2\right)}^{3/2}}$

Applying Integration to the equation,

$E_y=\frac{1}{4\pi {ϵ}_o}\times \frac{Q}{L}{\int }_0^L\frac{ydy}{{\left(n^2+y^2\right)}^{3/2}}$

Substituting the limits;

$E_y=\frac{Q}{4\pi {ϵ}_{o}L}\left[\frac{-1}{\sqrt{n^2+y^2}}\right]$

$E_y=\frac{Q}{4\pi {ϵ}_{o}L}\left[\frac{-1}{\sqrt{n^2+L^2}}+\frac{1}{n}\right]$

$E_y=-\frac{Q}{4\pi {ϵ}_{o}2n}\left[1-\frac{n}{\sqrt{2L^2+L^2}}\right]\hat{j}$

Applying the equation,

$\left(\cos \theta =\frac{n}{\sqrt{n^2+y^2}}\right)$

$E_n=\frac{1}{4\pi {ϵ}_o}\times \frac{Q}{2}\times \frac{dy\times n}{{\left(n^2+y^2\right)}^{3/2}}$

Applying Integration,

$E_n=\frac{1}{4\pi {ϵ}_o}\times \frac{Q}{L}n{\int }_0^L\frac{dy}{{\left(n^2+y^2\right)}^{3/2}}$

Substituting the limits value,

$E_n=\frac{Q}{4\pi {ϵ}_o}\times \frac{1}{n\sqrt{n^2+L^2}}\hat{i}$

$E={\overrightarrow{E}}_n+{\overrightarrow{E}}_y$

Applying the vectors value,we get

$E=\frac{Q}{4\pi {ϵ}_{o}n}\times \frac{1}{\sqrt{n^2+L^2}}i-\frac{Q}{4\pi {ϵ}_{o}Ln}\left(1-\frac{n}{\sqrt{n^2-L^2}}\right)\hat{j}$