a charge q is placed at the center of the line joining 2 equal charges Q. system of the 3 charges will be in equilibrium if q is equal to (a)-Q/2 (b) -Q/4 (c) Q /4 (d) Q/2

Hello ,

This question is quite simple and can be solved with the help of Columbs law easily .

See there are 3 charges in equilibrium , suppose you don't have the third central charge yet . You can imagine that two charges having equal charges repel each other but when a third charge inserted in system , the system comes to equilibrium this means the third charge must provide attraction force to overcome repulsion as two charges were +ve this charge should be -ve so the two answers having +ve charges got cancelled now you only have two option with -ve charges left . Now check which is correct for this apply Columnb's law and write equation for any charge .

Write for Q charge it faces two forces one repulsion from other Q charge and one attraction from q charge , now equation :-

KQQ/a^2+KQq/(a/2)^2=0

=> KQQ/a^2 = -kQq4/a^2

=> q = -Q/4 .

Hello!!!!

Hope you are doing great!!!!

Given that a charge q is placed at the center C of line joining two equal charges Q each,at points A and B.

Let AB=2a

If the net force on each charge will be 0(zero) ,then the system will be in equilibrium.

let the charge Q is present at A.

At equilibrium state ,the total force on this charge is given by

1/4π e0*qQ/a^2 + 1/4π e0*qQ/2a^2 =0

=1/4π e0*qQ/a^2 = - 1/4π e0*qQ/2a^2

on solving we get,

q=-Q/4

Hope it helps!!!!

Thank you!!!!