# dipole moment of a system of charge +q distributed uniformly on an arc of radius R subtending an angle pi/2 at its centre where  another charge -q is placed is?

Nicho priyatham
625 Points
8 years ago
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
Nicho priyatham
625 Points
8 years ago
can i know which book u r solving

Akshita Goyal
17 Points
8 years ago
nicho priyatham im solving the modules from askittians only.
Umrao Krishna Singh
14 Points
8 years ago
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 = $(2.R.Sin(\Theta /2))/\Theta$
where $\Theta$ = angle substended by arc on center of ring.

Hope this helps.