Question

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?

• A. $-\left(\frac{4}{9}\right)q$
• B. $+\left(\frac{4}{9}\right)q$
• C. $-\left(\frac{16}{9}\right)q$
• D. $+\left(\frac{16}{9}\right)q$

Answer 1

Answer: (A)

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