Question

Two point charges $+9q$ and $+q$ are kept $16\text{cm}$ apart. Where should a third charge $q_0$ be placed between them so that the system remains in equilibrium?

• A. $6\text{cm}$ from $+9q$
• B. $12\text{cm}$ from $+9q$
• C. $6\text{cm}$ from $+q$
• D. $12\text{cm}$ from $+q$

Answer 1

The correct option is B $12\text{cm}$ from $+9q$


Since the given first two charges $+9q$ and $+q$ are of same polarity, so the third charge which is to be placed in between must have opposite polarity to establish equilibrium in the system.


For equilibrium, net force on every charge must be zero.

The forces acting on $+9q$ charge is shown below,


Where,
$F_q$ is the force due to charge $+q$ and $F_{-q_0}$ is the force due to charge $-q_0$.

Thus, from above diagram,

$\left|F_q\right|=\left|F_{-q_0}\right|$

$\Rightarrow \frac{k\left(9q\right)\left(q\right)}{{16}^2}=\frac{k\left(9q\right)\left(q_0\right)}{r^2}.......\left(1\right)$

Similarly, forces acting on charge $+q$,


$\left|F_{9q}\right|=\left|F_{q_0}\right|$

$\Rightarrow \frac{k\left(9q\right)\left(q\right)}{{16}^2}=\frac{k\left(q_0\right)\left(q\right)}{(16-r{)}^2}.........\left(2\right)$

Dividing eq.(1) by eq.(2), we get

$\Rightarrow 1=9\frac{(16-r{)}^2}{r^2}$

$\Rightarrow \frac{r^2}{(16-r{)}^2}=9$

$\Rightarrow \frac{r}{16-r}=3$

$\Rightarrow r=48-3r$

$\Rightarrow r=12\text{cm}$

So, distance of $q_0$ from $+9q$ is $12\text{cm}$.

Alternative:

Using the formula,
$x=\frac{d}{\sqrt{\frac{q_1}{q_2}}+1}$

$x=\frac{16}{\sqrt{\frac{9q}{q}}+1}=4\text{cm}$ from smaller charge($q$).

Hence, option $\left(b\right)$ is the correct answer.