Question

Two charges q1 and q2 are placed on the x-axis, with q1 at x=a and q2 at x=3a

Answer 1

The force experienced by Q due to q1 is, F1= kQq1/(a+a/2)2

The force experienced by Q due to q2 is, F2= kQq2/(a/2)2

For the net force on Q to be zero,

[in this case q1 and q2 must be of same nature, we consider both to be positive]

F1 - F2 = 0

=> kQq1/(a+a/2)2 - kQq2/(a/2)2 = 0

=> q1/(3a/2)2 = q2/(a/2)2

=> q1 = 9q2

Similarly when Q is placed at 3a/2,

[in this case q1 and q2 must be of opposite nature, we consider q1 as positive and q2as negative]

The force experienced by Q due to q1 is, F1= kQq1/(a+3a/2)2

The force experienced by Q due to q2 is, F2= -kQq2/(a/2)2

For the net force on Q to be zero,

F1 - F2 = 0

=> kQq1/(a+3a/2)2 + kQq2/(a/2)2 = 0

=> q1/(5a/2)2 = -q2/(a/2)2

=> q1 = -25q2