Question

The distance between two point charges $4q$ and $-q$ is $r$. A third charge $Q$ is placed at their midpoint. The resultant force acting on $-q$ is zero then $Q=$ _________.

• A. $-q$
• B. $q$
• C. $-4q$
• D. $4q$

Answer 1

The correct option is B $q$

Let $"r"$be the distance from the charge $q$ where $Q$ is in equilibrium.

Total Force acting on Charge $q$ and $4q$:

$F=\frac{kqQ}{r²}+\frac{k4qQ}{(l-r)²}$

For $Q$ to be in equilibrium, $F$ should be equated to zero.

$\frac{kqQ}{r²}+\frac{k4qQ}{(l-r)²}$

$=0(l-r)²=4r²=l-r=2r=l=3r=r=\frac{1}{3}$

Taking the third charge to be-$Q$(say) and the non applying the condition of equilibrium on+q charge $\frac{kQ}{\left(\frac{l}{3}\right)²}$

$=\frac{k\left(4q\right)}{L²}\frac{9kQ}{L²}$

$=\frac{k\left(4q\right)}{L²}9Q$

$=4qQ$

$=\frac{4q}{9}$There fore a point charge-$\frac{4q}{9}$should be place data distance of $\frac{l}{3}$ right wards from the point charge $+q$ on the line joining the $2$charges.