Question

Two point-charges $+9e$ and $+e$ are is kept $16\text{cm}$ apart from each other. Where should a third charge $\text{q}$ be placed between them so that the system is in the equilibrium state?

• A. $24\text{cm from}+9e$
• B. $12\text{cm from}+9e$
• C. $24\text{cm from}+e$
• D. $12\text{cm from}+e$

Answer 1

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


Let the charge $\text{q}$ is placed at a distance $\text{x}$ from $+9e$ and inbetween charges $+9e$ and $+e$
Using Coulomb's law,

$\frac{K.9e.q}{{\text{x}}^2}=\frac{K.q.e}{(16-\text{x}{)}^2}\{F=\frac{K{q}_1{q}_2}{r^2}\}$

${\left(\frac{16-\text{x}}{\text{x}}\right)}^2=\frac{1}{9}$

$\frac{16}{\text{x}}-1=\pm \frac{1}{3}$

$\Rightarrow \frac{16}{\text{x}}=\frac{4}{3}\text{or}\frac{2}{3}$

$\Rightarrow \text{x}=\frac{3\times 16}{4}\text{or}\frac{3\times 16}{2}$

$\therefore \text{x}=12\text{cm}\text{or}24\text{cm}$

As the charge $q,$ can be equilibrium present only when its is kept inbetween charges $+9e$ and $+e$, correct answer is $\text{x}=12\text{cm}$