Question

Two free charges $+q$ and $+4q$ are placed at a distance ${}^{\prime }{r}^{\prime }$ between them. A third charge is placed in between them and find the third charge when the system is in equilibrium.

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

Answer 1

The correct option is B $\frac{4q}{9}$

Let the third charge be $Q$ placed at distance $a$ from the charge $q$.

So distance of $Q$ from the $4q$ is $r-a$.

Electrostatic force on $Q$ from $q$ is: $kQq/a^2$

where $k=\frac{1}{4\pi {ϵ}_o}$

Similarly, Electrostatic force on $Q$ from $4q$ is:

$4kQq/(r-a{)}^2$

These forces are opposite in direction, balancing these forces on $Q$,

$kQq/a^2=4kQq/(r-a{)}^2$ $\Rightarrow (r-a{)}^2/a^2=4\Rightarrow a=r/3$

Electrostatic force on $q$ from $Q$ is:

$kQq/a^2$

Electrostatic force on $q$ from $4q$ is:

$4k{q}^2/(r{)}^2$

These forces are in same direction, To balance these forces on $q$,

$kQq/a^2+4k{q}^2/r^2=0$ $\Rightarrow Q=-4q(a/r{)}^2=4q(1/3{)}^2=-4q/9$