Question

A semi infinite line charge of linear charge density λ has the shape as shown in the figure. Portion ABC forms three-fourth of a circle of radius R while the straight portion from C to infinity is parallel to B9 BOA. The field at the centre of circle (O) is

Answer 1

Explanation:

Ex=2kD/r(sinƟ1+sinƟ2)

Ey=2kD/r(cosƟ2-cosƟ1)

So

For semi infinite wire

Ex=kd/r

Ey=-kd/r

And resultant will be

E=√2kD/r

(Shown in fig)

Now E due to an arc is given by

E=(2kD/r )sin(Ɵ/2)

Here Ɵ=270 degree

 

E=√2kD/r

At the center of circle,

The direction of the electric fields is opposite with equal magnitude

So resultant will be zero