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.

Explanation

For the system to be in equilibrium, the 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 (3),
`1/(4 pi epsilon_(0)) (qQ)/(x^2)=1/(4 pi epsilon_(0))(Q4q)/((l-x)^(2))`
or `[(l-x)/(x)]^(2) =4 or (l-x)/(x)=2 or x=l/3`
For equilibrium of `(1) 1/(4 pi epsilon_(0))(q4q)/(l^2)=1/(4 pi epsilon_(0))(Qq)/(x^2)`
or `Q=4q(x/l)^(2)=4q(1/3)^(2)=(4q)/(9)`
or `Q=-(4q)/(9)`.