Two charges q (at A) and \( -3q \) (at B) are fixed on the x-axis, a distance d apart. At what distance from A, on the side of A away from B, should a third charge 2q be placed so that it experiences no net force?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \frac{d(\sqrt{3}+1)}{2} \)
On the side of A away from B, the charge \( q \) pushes \( 2q \) away from A, and \( -3q \) pulls it towards B. These two forces are opposite. So the net force is zero where their Coulomb forces are equal in magnitude.
Let the charge \( 2q \) be at a distance \( x \) from A, so its distance from B is \( x+d \). Equating the magnitudes of the two forces:
\( \displaystyle \frac{k\,q\,(2q)}{x^{2}}=\frac{k\,(3q)\,(2q)}{(x+d)^{2}} \)
\( \displaystyle \frac{(x+d)^{2}}{x^{2}}=3 \)
\( \Rightarrow\; x+d=\sqrt{3}\,x \)
\( \begin{aligned} x &= \frac{d}{\sqrt{3}-1} \\ &= \frac{d(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} \\ &= \frac{d(\sqrt{3}+1)}{2} \end{aligned} \)
Therefore, the third charge must be placed at \( d(\sqrt{3}+1)/2 \) from A, on the side away from B.