Question

Three point charges $4q,Q$ and $q$ are placed in a straight line of length $L$ at points $0,L/2$ and $L$ respectively. The net force on charge $q$ is zero. The value of $Q$ is

• A. 4q
• B. -q
• C. 0.5q
• D. -2q
• E. 3q

Answer 1

Answer: (B)

$\because$ Net force on charge, $q=0$
$\Rightarrow f_1+f_2=0$...(i)
Here, $f_1$ is the force on charge $q$ due to charge $4q$
$f_2$ is the force on charge $q$ due to charge $Q$
$\therefore$ From eq. (i),
$\Rightarrow \frac{K(4q)q}{L^2}+\frac{KQq}{(L/2{)}^2}-0$
$\Rightarrow 4q^2+4Qq=0$
$\Rightarrow Q=-q$