Question

A quarter circle of radius a is centered about the origin in the first quadrant and carries a uniform charge of −Q. Find the x- and y-components of the net electric field at the origin.

Answer 1

Answer:

$E_x = \frac{2kQ}{\pi R^2}$

$E_y = \frac{2kQ}{\pi R^2}$

Explanation:

Electric field due to small part of the circle is given as

$dE = \frac{kdq}{R^2}$

here we know that

$dq = \frac{Q}{\frac{\pi}{2}R} Rd\theta$

$dq = \frac{2Q d\theta}{\pi}$

Now we will have two components of electric field given as

$E_x = \int dE cos\theta$

$E_x = \int \frac{kdq}{R^2} cos\theta$

$E_x = \int \frac{k (2Qd\theta) cos\theta}{\pi R^2}$

$E_x = \frac{2kQ}{\pi R^2} \int_0^{90} cos\theta d\theta$

$E_x = \frac{2kQ}{\pi R^2} (sin 90 - sin 0)$

$E_x = \frac{2kQ}{\pi R^2}$

similarly in Y direction we have

$E_y = \int dE sin\theta$

$E_y = \int \frac{kdq}{R^2} sin\theta$

$E_y = \int \frac{k (2Qd\theta) sin\theta}{\pi R^2}$

$E_y = \frac{2kQ}{\pi R^2} \int_0^{90} sin\theta d\theta$

$E_y = \frac{2kQ}{\pi R^2} (-cos 90 + cos 0)$

$E_y = \frac{2kQ}{\pi R^2}$