Two particles (free to move) with charges `+q` and `+4q` are a distance L apart. A third charge is placed so that the entire system is in equilibrium.
(a) Find the location, magnitude and sign of the third charge.
(b) Show that the equilibrium is unstable.

Options

A. left of A at a distance `1//3` from A
B. right of A at a distance `1//3` fromB
C. between A and B at a distance `21//3` from A
D. between A and B at a distance `l//3` from A

Answer

D

Explanation

For system to be in equilibrium, net force on each charge should be zero, hence the third charge should be negative and it should be placed near q between (1) and (2). For equilibrium of (3), we have
`(1)/(4 pi epsilon_(0))(qQ)/(x)=(1)/(4 pi epsilon_(0))(Q4q)/((1-x)^(2)) or x=l/3`
For equilibrium of (1) we have
`(1)/(4 pi epsilon_(0))(q4q)/(l^(2))=(1)/(4 pi epsilon_(0))(Qq)/(x^(2))`
`Q=4q((x)/(l))^(2) = 4q((1)/(3))^(2)=(4q)/(9)`
Q should be negative of q, otherwise the resultant force on q cannot be zero.