Question

Two free positive charges $4q$ and $q$ are kept 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. $Q=-\frac{4}{9}q$ at $\frac{l}{3}$
• B. $Q=\frac{4}{9}q$ at $\frac{l}{3}$
• C. $Q=q$ at $\frac{l}{3}$
• D. $Q=-q$ at $\frac{l}{3}$

Answer 1

The correct option is A $Q=-\frac{4}{9}q$ at $\frac{l}{3}$

$4q$ and $q$ are given to be free positive charges placed at a distance $l$ apart. So, charge $Q$ needed to achieve equilibrium for the entire system, should be of negative sign. Let $Q$ is at a distance $x$ from $q$ charge, and therefore $(l-x)$ from $4q$ charge.

From Coulomb's Law, force between two charges is $\frac{k{q}_1{q}_2}{r^2}$

So, for equilibrium, force between $q$ and $Q$ should be equal to force between $4q$ and $Q$ i.e, $\frac{kqQ}{x^2}=\frac{k4qQ}{(l-x{)}^2}$ ...(i)

$(l-x{)}^2=4x^2$

Taking square root on both sides:

$l-x=2x$

$l=3x$

$x=\frac{l}{3}$

i.e, the charge $Q$ should be placed at a distance $\frac{l}{3}$ from charge $q$.

Now, applying the condition of equilibrium on $+q$ charge,

$\Rightarrow \frac{kQq}{{\left(\frac{l}{3}\right)}^2}=\frac{k\left(4q\right)q}{l^2}$

$\Rightarrow l^{2}Q=4q{\left(\frac{l}{3}\right)}^2$

$\Rightarrow Q=\frac{4q{l}^2}{9\times l^2}$

$\Rightarrow Q=\frac{4q}{9}$

Since $Q$ should be negative as explained above, so $Q=\frac{-4q}{9}$