# 2 point charges q &4q are located at r1 & r2,respectively,on the x-y plane.The magnitude Q & location R of a third charge to be placed on the plane such that the total force on each charges vanishes are what?

Arun Kumar IIT Delhi
9 years ago
$\vec r_{1},\vec r_{2}\\ \vec r_{assumed}=\vec r_{1}-\vec r_{2}$
assume Q,q are positive.
$\vec F_{on\, 1st}=KQq(\vec r_{1}-\vec r_{3})/|\vec r_{1}-\vec r_{3}|^2+K4qq(\vec r_{1}-\vec r_{2})/|\vec r_{1}-\vec r_{2}|^2=0 \\\vec F_{on\, 2nd}=KQ4q(\vec r_{2}-\vec r_{3})/|\vec r_{2}-\vec r_{3}|^2+K4qq(\vec r_{2}-\vec r_{1})/|\vec r_{2}-\vec r_{1}|^2=0 \\\vec F_{on\, 3rd}=KQq(\vec r_{3}-\vec r_{1})/|\vec r_{3}-\vec r_{1}|^2+KQq(\vec r_{3}-\vec r_{2})/|\vec r_{3}-\vec r_{2}|^2=0 \\=>Qq(\vec r_{1}-\vec r_{3})/|\vec r_{1}-\vec r_{3}|^2=-4qq(\vec r_{1}-\vec r_{2})/|\vec r_{1}-\vec r_{2}|^2 \\=>Q(\vec r_{1}-\vec r_{3})/|\vec r_{1}-\vec r_{3}|^2=-4q(\vec r_{1}-\vec r_{2})/|\vec r_{1}-\vec r_{2}|^2$
See direction vectors are in same.
=> they are collinear.
The big part is done.
Now asssume
$\\\vec t_{3}=\vec r_{1}-\vec r_{3}=\lambda_1 \vec t \\\vec t_{2}=\vec r_{1}-\vec r_{2}=\lambda \vec t \\\vec t_{1}=\vec r_{2}-\vec r_{3}=\lambda_3 \vec t$
Its easier to solve now.
Thanks & Regards
Arun Kumar
IIT Delhi