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Q.
Two charges +4q and +q are placed at a distance x apart and free to move. What charge must be placed in between two charges so the system will be in equilibrium?
Solution:
q and 4q are also in equilibrium so force on 4q will also be zero
$\frac{k\times (4q) (+q_o)}{r^2}+\frac{k(4q)q}{x^2}=0$
$\frac {q_o}{r^2}+\frac{q}{x^2}=0$
$q_o=-q\frac{r^2}{x^2}=\frac{-q\times 4x^2}{9x^2}$
$q_o=\frac{-4}{9}q$ negative in nature
