Two free charges q and 4q are placed a distance d apart. A third char…

The correct answer is Option (1) → \( Q=-\displaystyle\frac{4q}{9},\ x=\displaystyle\frac{d}{3} \)

For the whole system to be in equilibrium, the net force on each of the three charges must be zero.

• The forces of \( q \) and \( 4q \) on \( Q \) must balance. Since \( q \) and \( 4q \) have the same sign, this happens between them, nearer the smaller charge \( q \).

• The charge \( 4q \) repels \( q \), so \( Q \) must attract \( q \). Hence \( Q \) has the opposite sign to \( q \).

For the charge \( Q \), equating the magnitudes of the forces due to \( q \) and \( 4q \) by Coulomb’s law:

    \( \displaystyle \frac{kq\,|Q|}{x^{2}}=\frac{k\,(4q)\,|Q|}{(d-x)^{2}} \)

    \( (d-x)^{2}=4x^{2} \)

    \( \Rightarrow\; d-x=2x \)

    \( \displaystyle x=\frac{d}{3} \)

For the charge \( q \), the attraction of \( Q \) at distance \( x \) must balance the repulsion of \( 4q \) at distance \( d \):

    \( \displaystyle \frac{kq\,|Q|}{x^{2}}=\frac{kq\,(4q)}{d^{2}} \)

    \( \begin{aligned} |Q| &= \frac{4q\,x^{2}}{d^{2}} \\ &= \frac{4q\,(d/3)^{2}}{d^{2}} \\ &= \frac{4q}{9} \end{aligned} \)

Since \( Q \) must attract \( q \), it is negative. Therefore, \( Q=-4q/9 \), placed at \( x=d/3 \) from \( q \).