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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?

Akshita Goyal , 9 Years ago
Grade 12
anser 4 Answers
Nicho priyatham

Last Activity: 9 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

Last Activity: 9 Years ago

can i know which book u r solving
 

Akshita Goyal

Last Activity: 9 Years ago

nicho priyatham im solving the modules from askittians only.

Umrao Krishna Singh

Last Activity: 9 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.

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