not in syllbus of iit jee
to solve such questions we genrally use some other complex integratios
so better drop this question 
approve if use ful
						

							
can i know which book u r solving
 
						

							
Actually it is an easy one. I’m not going to solve the entire problem for you but I will give you some hints and definitely it may come in JEE, so neglect the above answer.  In a distribution system as provided in the question you will have to perform the ritual of integration to find the centre of charge (just as you find center of mass.). You are lucky that the question is about a ring, we already know what is the position of center of mass of ring w.r.t. center, position of center of charge will be same. According to question, The distribution contains +q only i.e. at the position of center of charge (mass) you will denote +Q. (+Q= ∑+q) , Now you have -Q at centre and +Q at the position of center of charge (mass), 
=>                                  p= 2.Q.d                   where, d= distance between -Q and + Q
 
 
 
B.T.W., in case you don’t remember center of charge (mass) of a ring, it is = 
where  = angle substended by arc on center of ring.
 
Hope this helps.