Three charges q,q and -q are placed at the corners of an equilatereal triangle The ratio of the force on a +ve charge to the force on the negative charge will be equal to

Dear pallavi bhardwaj

let side of triangle is a

 F= 1/4∏ε q²/a²

 force on positiive charge = [F² +F² + 2FFcos120]^1/2

                                                = F

 force on negatieve  charge = [F² +F² + 2FFcos120]^1/2

                                                = F

so ratio =F/F =1:1

Please feel free to post as many doubts on our discussion forum as you can.
 If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly.
 We are all IITians and here to help you in your IIT JEE  & AIEEE preparation.

 All the best.
 
Regards,
Askiitians Experts
Badiuddin

Dear pallavi bhardwaj

let side of triangle is a

 F= 1/4∏ε q²/a²

 force on positiive charge = [F² +F² + 2FFcos120]^1/2

                                                = F

 force on negatieve  charge = [F² +F² + 2FFcos60]^1/2

                                                = F√3

so ratio =F/F√3 =1:√3

Please feel free to post as many doubts on our discussion forum as you can.
 If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly.
 We are all IITians and here to help you in your IIT JEE  & AIEEE preparation.

 All the best.
 
Regards,
Askiitians Experts
Badiuddin