Question

Two free positive charges $4q$ and $q$ are at a distance $l$ apart. What charge $Q$ is needed to achieve equilibrium for the entire system and where should it be placed from charge $q$?

• A. $Q=\frac{4}{9}q\left(negativeat\frac{l}{3}\right)$
• B. $Q=\frac{4}{9}q\left(positiveat\frac{l}{3}\right)$
• C. $Q=q\left(positiveat\frac{l}{3}\right)$
• D. $Q=q\left(negativeat\frac{l}{3}\right)$

Answer 1

The correct option is A $Q=\frac{4}{9}q\left(negativeat\frac{l}{3}\right)$

Let $r$ be the distance from charge q where Q is in equilibrium.
Total force on Q due to q and 4q is $F=k\frac{qQ}{r^2}+k\frac{4qQ}{(l-r{)}^2}$

For equilibrium of Q , F should be zero. so $(l-r{)}^2=4r^2\Rightarrow l-r=2r$
$\therefore r=\frac{l}{3}$

Now to find out the the value of charge, lets equate the equilibrium of 4q charge:

$\therefore k4qQ\frac{1}{(l-r{)}^2}+4kqq\frac{1}{l^2}=0$

$Q=-\frac{4q}{9}$
Ans:(A)