Three charges +4q, Q and q are placed along x-axis at points, \( x = 0,\ l/2 \) and \( l \) respectively. The value of Q to make the net force on q to be zero :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( -q \)
By Coulomb's law, the force between two charges is \( kq_1q_2/r^{2} \). The force on \( q \) is zero when the forces from \( +4q \) and \( Q \) are equal and opposite.
\( \text{distance of } q \text{ from } +4q = l \)
\( \text{distance of } q \text{ from } Q = l/2 \)
The charge \( +4q \) repels \( q \), so \( Q \) must attract it, which means \( Q \) is negative. Setting the sum of the forces on \( q \) to zero:
\( \displaystyle \frac{k(4q)\,q}{l^{2}} + \frac{k\,Q\,q}{(l/2)^{2}} = 0 \)
\( \displaystyle \frac{4q}{l^{2}} + \frac{4Q}{l^{2}} = 0 \)
\( \displaystyle Q = -q \)