Question

Two charges $4q$ and $q$ are placed $3x$ apart in air $A$ charge $Q$ is placed at a point on the line joining so that the systein of these charges remains in equilibrium. Then, the distance of $Q$ from $4q$ and the value of $Q$ are

• A. $2x,-\frac{2q}{3}$
• B. $2x,-\frac{4q}{3}$
• C. $x,\frac{2q}{3}$
• D. $x,\frac{4q}{9}$

Answer 1

Answer: (B)

Given situation is shown in the figure.
To be all charges in equilibrium charge $Q$ should be negative and placed between $q$ and $4q$.
At equilibrium, force between $4q$ and $Q$ = force between $Q$ and $q$.
$\therefore k\frac{4q.Q}{r^2}=k\frac{q.Q}{{\left(3x-r\right)}^2}$
or, $\frac{4}{r^2}=\frac{1}{{\left(3x-r\right)}^2}$
or, $\frac{2}{r}=\frac{1}{3x-r}$
or, $6x-2r=r$
or, $3r=6x$
or, $r=2x.$
Again, force between $4q$ and $Q$ = force between $q$ and $4q$
$k\frac{4q.Q}{{\left(2x\right)}^2}=k\frac{4q.q}{{\left(3x\right)}^2}$
or, $\frac{Q}{4}=\frac{q}{9}or,Q=\frac{4q}{9}$
So charge $Q=-\frac{4q}{9}r=2x$