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

`E_x = (2kQ)/(\pi R^2)`

`E_y = (2kQ)/(\pi R^2)`

Step-by-step explanation:

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

`dE = (kdq)/(R^2)`

here we know that

`dq = (Q)/((\pi)/(2)R) Rd\theta`

`dq = (2Q d\theta)/(\pi)`

Now we will have two components of electric field given as

`E_x = \int dE cos\theta`

`E_x = \int (kdq)/(R^2) cos\theta`

`E_x = \int (k (2Qd\theta) cos\theta)/(\pi R^2)`

`E_x = (2kQ)/(\pi R^2) \int_0^(90) cos\theta d\theta`

`E_x = (2kQ)/(\pi R^2) (sin 90 - sin 0)`

`E_x = (2kQ)/(\pi R^2)`

similarly in Y direction we have

`E_y = \int dE sin\theta`

`E_y = \int (kdq)/(R^2) sin\theta`

`E_y = \int (k (2Qd\theta) sin\theta)/(\pi R^2)`

`E_y = (2kQ)/(\pi R^2) \int_0^(90) sin\theta d\theta`

`E_y = (2kQ)/(\pi R^2) (-cos 90 + cos 0)`

`E_y = (2kQ)/(\pi R^2)`