Question

A negative point charge $2q$ and a positive charge $q$ are fixed at a distance $l$ apart. Where should a positive test charge $Q$ be placed on the line connecting the charge for it to be in equilibrium? What is the nature of the equilibrium with respect to longitudinal motions?

Answer 1

Let the charge $Q$ be placed at distance $x$ from charge $2q$. Now for the charge to be in equilibrium the forces due to both charges must cancel each other or the forces must be equal and opposite. Now the forces due to each charges is given by

$F_{2q}=2kQq/x^2$ and

$F_q=kQq/(l-x{)}^2$

for equilibrium we must have

$F_{2q}=F_q$ or,

$2kQq/x^2=kQq/(l-x{)}^2$ or,

$\frac{x^2}{(l-x{)}^2}=2$ or,

1. $\frac{x}{l-x}=\sqrt{2}$ or, $x=\frac{\sqrt{2}l}{1+\sqrt{2}}$
2. $\frac{x}{l-x}=-\sqrt{2}$ or, $x=\frac{\sqrt{2}l}{\sqrt{2}-1}$

Since second solution of $x$ lies outside the $l$ we will neglect it. Hence the answer is-

$$x=\frac{\sqrt{2}l}{\sqrt{2}+1}$$

Now if the charge is displaced longitudinally towards charge $q$ then $x$ will increase hence $F_{2q}$ will decrease and $F_q$ will increase. Hence the charge will be forced to move towards charge $q$.

Similarly if it is moved towards charge $2q$, $F_{2q}$ will increase and $F_q$ will decrease thus the charge $Q$ will be moved towards charge $2q$.

Hence the charge $Q$ is in unstable equilibrium with respect to longitudinal motions.