Question

what should be the position of charge 5 mc for it be in equilibrium on the line joining two charges -4mc and 16 mc separated by 9cm?​

Answer 1

Answer:

for neutral points the force on this charge will be zero.

and neutral point will lie on line joining of two charges but not between the charges.

so Suppose the distance of q=5µC from small charge is X.

force on it due to q1 F1=k*5*(-4) / X2 (force will apply toward system)

force due to q2 F2 = k * 5 * 16 / (9+x)2 (force will apply away from system)

for equilibrium F1=F2

after solved out you get answer X= 9 cm from -4 µC

1) when you solved it you found that equilibrium postion will not depend upon charge at that point( i.e for any value of charge position will remain same)

2) you can solved it by a shortcut

X= d / √n - 1 {here n= q2/q1 (large charge / small charge)} d= distance between charges