Question

Charges $4Q,q$ and $Q$ are placed along x-axis at position $x=o,x=l/2$ and $x=l,$ respectively. Find the value of $q$, so that force on charge $Q$ is zero.

• A. $Q$
• B. $\frac{Q}{2}$
• C. $-\frac{Q}{2}$
• D. $-Q$

Answer 1

Answer: (D)

From Coulombs law, the force acting between two charges $\left(q_1,q_2\right)$
separated at a distance $r$ is given by
$F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$
Total force acting on charge $Q$ is
$F=\frac{1}{4\pi {\epsilon }_0}\frac{qQ}{(l/2{)}^2}+\frac{1}{4\pi {\epsilon }_0}\frac{4Q\cdot Q}{(l{)}^2}$
According to question, $F=0$
$\therefore \frac{1}{4\pi {\epsilon }_0}\frac{qQ}{(l/2{)}^2}+\frac{1}{4\pi {\epsilon }_0}\cdot \frac{4Q^2}{(l{)}^2}=0$
$\Rightarrow \frac{4q}{l^2}+\frac{4Q}{l^2}=0$