91Ó°ÊÓ

The force experienced by such a unit positive charge put at a spot is the electric field intensity at that point. The intensity of an electric field is a vector quantity.

$E=\frac{kq}{R^2}$

where $R$denotes distance, $k$denotes Coulomb's constant, and $q$denotes charge

$dE=\frac{k\mathrm{Δ}q}{R^2}$

and

$x=R\mathrm{θ},\mathrm{Δ}x=R\mathrm{Δ}\mathrm{θ}$

(From figure)

$q=\mathrm{λ}x,\mathrm{Δ}q=\mathrm{λ}\left(\mathrm{Δ}x\right)$

(charge in terms of density)

$âŸ\P \mathrm{Δ}x=R\mathrm{Δ}\mathrm{θ}$and $\mathrm{Δ}q=\mathrm{λ}\left(R\mathrm{Δ}\mathrm{θ}\right)$

Fill in the blanks in the equation with these values.,

$dE=\frac{k\mathrm{λ}\mathrm{Δ}\mathrm{θ}}{R}$

As a result, the electric field's $x$component will be,

$$\begin{gathered} d{E}_x=dE\cos \mathrm{θ} \\ =\frac{k\mathrm{λ}\mathrm{Δ}\mathrm{θ}}{R}\cos \mathrm{θ} \end{gathered}$$

$→d{E}_x=\frac{k\mathrm{λ}}{R}\cos \mathrm{θ}\mathrm{Δ}\mathrm{θ}$

As a result, the electric field's $y$component will be,

$→d{E}_y=\frac{k\mathrm{λ}}{R}\sin \mathrm{θ}\mathrm{Δ}\mathrm{θ}$

$x$component equation,

$d{E}_x=\frac{k\mathrm{λ}}{R}\cos \mathrm{θ}\mathrm{Δ}\mathrm{θ}$

Integrate this equation,

$E_x=\frac{k\mathrm{λ}}{R}{∫}_0^{\mathrm{π}/2}\cos \mathrm{θ}\mathrm{Δ}\mathrm{θ}$ $\left(1\right)$

$y$component equation,

$d{E}_y=\frac{k\mathrm{λ}}{R}\sin \mathrm{θ}\mathrm{Δ}\mathrm{θ}$

Integration of this equation,

$E_y=\frac{k\mathrm{λ}}{R}{∫}_0^{\mathrm{π}/2}\sin \mathrm{θ}\mathrm{Δ}\mathrm{θ}$ $\left(2\right)$

Linear charge density of the rod,

$\mathrm{λ}=\frac{2Q}{\mathrm{π}R}$

substitute this value in equation $\left(1\right)$and $\left(2\right)$

$E_x=\frac{k2Q}{\mathrm{π}{R}^2}{∫}_0^{\mathrm{π}/2}\cos \mathrm{θ}\mathrm{Δ}\mathrm{θ}$

$E_y=\frac{k2Q}{\mathrm{π}{R}^2}{∫}_0^{\mathrm{π}/2}\sin \mathrm{θ}\mathrm{Δ}\mathrm{θ}$

Calculate the integral using both equations,

${∫}_0^{\mathrm{π}/2}\cos \mathrm{θ}d\mathrm{θ}=[\sin \mathrm{θ}{]}_0^{\mathrm{π}/2}=+1$

${∫}_0^{\mathrm{π}/2}\sin \mathrm{θ}d\mathrm{θ}=[-\cos \mathrm{θ}{]}_0^{\mathrm{π}/2}=+1$

the electric field can be written as:

$\stackrel{→}{E_{net}}=E_x\hat{i}+E_y\hat{j}$

Expand:

$E_x\hat{i}=\left[\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left(\frac{2Q}{\mathrm{Ï€}{R}^2}\right){∫}_0^{\mathrm{Ï€}/2}\cos \mathrm{θ}\mathrm{Δ}\mathrm{θ}\right]\hat{i}$

$E_y\hat{j}=E_x\hat{i}=\left[\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left(\frac{2Q}{\mathrm{Ï€}{R}^2}\right){∫}_0^{\mathrm{Ï€}/2}\sin \mathrm{θ}\mathrm{Δ}\mathrm{θ}\right]\hat{j}$

the sum of equations will be,

localid="1650218257527" $\stackrel{→}{E_{net}}=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left(\frac{2Q}{\mathrm{Ï€}{R}^2}\right)(\hat{i}+\hat{j})$