A charge of magnitude q is distributed on semicircular thin non conductive wire of radius r such that half of the semi circle has a positive charge while other half has negative charge .The field at the centre of semi circle is

a)q/²₀r²

b)q/4²₀r²

c)q/2²₀r²

d)2q/²₀r²

Let, ABC be the semicircle, L is the centre of the semicircle & AL = radius of the semicircle = r

 

AB has a positive charge of q/2 and the length AB =  r / 2

So, linear charge density of AB =  = q /  r

BC has a negative charge of - q/2 and the length BC =  r / 2 

So, linear charge density of AB = -  = - q /  r

 

Consider two infinitesimally small segments of dx at a distance x from B on either side at M & N. These two pints M and N are exactly symmetrical.

 

Now, let the electric field at the centre L of the semicircle due to the segment dx be dE.  The electric fields make angles    with the line LC.  Refer to the fig.1

 

By symmetry, the sine components of the segments dx at M & N cancell each other as they are equal in magnitude and opposite in direction. The cosine components of the segments dx at M and N are along the same direction and gets added. Refer to the figure 2.

 

Charge on dx at M =  dx

Electric field at L due to the segment at M = dE =  dx / 4₀ r²  

 

Now, we know,   x = r   ( by geometry )

                  i.e dx = r d

 

Therefore, dE =  r d / 4₀ r²   

Now, the electric field at the centre L due to the segments at M and N

= 2dE cos =   r cos  d / 2₀ r²   

 

The net electric field due to the semicircle ABC at the centre L

 

= _[0]^{[/2 ]}  r cos  d / 2₀ r²        {when M is at A, =0, M is at B, =/2}

 

=  / 2₀ r   . sin   ] ^/2  ₀  =  / 2₀ r  = q / 2² ₀ r²

 

So, the correct option is (c)   q/2²₀r²

 

CHEERS !!