Two point charges (-q) and (+4q) are placed at separation 'r'. Where…

The correct answer is Option (1) → at separation 'r' from \( (-q) \) on the extreme side of \( -q \).

For the third charge \( Q \) to be in equilibrium, the forces of \( -q \) and \( +4q \) on it must be equal and opposite. Between two unlike charges both forces on \( Q \) point the same way, so \( Q \) must lie outside them, on the side of the smaller charge \( -q \).

Let \( Q \) be at distance \( x \) from \( -q \), on the side away from \( +4q \), so it is at \( r + x \) from \( +4q \). Using Coulomb's law:

    \( \displaystyle \frac{k\,q\,Q}{x^{2}} = \frac{k\,(4q)\,Q}{(r + x)^{2}} \)

    \( \displaystyle (r + x)^{2} = 4x^{2} \;\Rightarrow\; r + x = 2x \;\Rightarrow\; x = r \)

For the whole system to be in equilibrium, the force on \( -q \) must also vanish. The attraction by \( +4q \) (distance \( r \)) must be balanced by an equal attraction by \( Q \) (distance \( r \)):

    \( \displaystyle \frac{k\,q\,(4q)}{r^{2}} = \frac{k\,q\,Q}{r^{2}} \;\Rightarrow\; Q = +4q \)

With \( Q = +4q \) the forces on \( +4q \) also cancel: attraction \( 4kq^{2}/r^{2} \) by \( -q \) and repulsion \( k(4q)(4q)/(2r)^{2} = 4kq^{2}/r^{2} \) by \( Q \).