Answer to Question #88861 in Electricity and Magnetism for Shivam Nishad

Question

Two point charges +4e and +e are fixed at a
distance 'a'. A third charge q is placed on a
straight line joining these two charges so that
q is in equilibrium. Determine the position
of q.

Answer 1

Let q be r distant from e. 

Force on q due to e 

$$=k\frac{qe}{r^2}$$

 

Force on q due to 4e 

$$=k\frac{q4e}{(a-r)^2}$$

For it to be in equilibrium : 

The forces must be equal and opposite. 

$$k\frac{qe}{r^2}=k\frac{4eq}{(a-r)^2}$$

$$\implies \frac{1}{r^2}=\frac{4}{(a-r)^2}$$

$$\implies {a^2}+{r^2}-2{a}{r}=4{r^2}$$

$$\implies 3{r^2}+2{a}{r}-{a^2}=0$$

 

$$\implies 3{r^2}+3{a}{r}-{a}{r}-{a^2}=0$$

$$\implies 3r(r+a)-a(r+a)=0$$

 

$$\implies (3r-a)(r+a)=0$$

$\implies r=-a$ or $r = \frac{a}{3}$

Since distance is non-negative 

$r = \frac{a}{3}$ should be the answer .