Question

Starting from Coulomb's law in integral form in terms of the charge density and a volume integration, calculate the electric field at the origin of coordinates for a uniform line charge density λ that extends from (x,y,z) = (R,0,0) to (0,R,0) along an arc of radius R. 4.

Answer 1

Let us consider a small element which make an angle d$\theta$ at the origin , therefore
charge of this small element (dq) = $\lambda$*(Rd$\theta$)
Now electric field due to this small element will be dE as shown in the figure
$dE = \frac{kdq}{R^{2}}$
Now we will take the component along the horizontal and vertical direction
dE_X = dECos$\theta$
In vertical direction
dE_Y = dE Sin$\theta$
now we will integrate it to find the net electric field in the X and Y direction
$E_{X} =\int_{0}^{90} dE_{X} = \int_{0}^{90}\frac{k\lambda (Rd\theta )*Cos\theta }{R^{2}}$
$E_{X} =\int_{0}^{90}\frac{k\lambda*Cos\theta (d\theta )}{R}$
$E_{X} =\left ( \frac{k\lambda }{R} \right )\left [ Sin\theta \right ]_{0}^{90}$
$E_{X} =\left ( \frac{k\lambda }{R} \right )$
Now similarly in the Y direction
$E_{Y} =\int_{0}^{90} dE_{Y} = \int_{0}^{90}\frac{k\lambda (Rd\theta )*Sin\theta }{R^{2}}$
$E_{Y} =\int_{0}^{90}\frac{k\lambda*Sin\theta (d\theta )}{R}$
$E_{X} =\left ( \frac{k\lambda }{R} \right )\left [ -Cos\theta \right ]_{0}^{90}$
$E_{X} =\left ( \frac{k\lambda }{R} \right )$
Now we know that net electric field will be
$E= \sqrt{E_{X}^{2}+ E_{Y}^{2}}$
$E= \frac{\sqrt{2}k\lambda }{R}$