Question

Three charges +$Q_1$,$Q_2$ and q are placed on a straight line such that $q$ some where in between +$Q_1$ And +$Q_2$. If this system of charges is in equilibrium, what should be the magnitude and sign of charge q.

• A. $\frac{Q_1{Q}_2}{{\left(\sqrt{Q_1}+\sqrt{Q_2}\right)}^2},$ Positive
• B. $\frac{Q_1+Q_2}{2},$ Positive
• C. $\frac{Q_1{Q}_2}{{\left(\sqrt{Q_1}+\sqrt{Q_2}\right)}^2},$ Negative
• D. $\frac{\sqrt{Q_1{Q}_2}}{2},$ Negative

Answer 1

The correct option is A $\frac{Q_1{Q}_2}{{\left(\sqrt{Q_1}+\sqrt{Q_2}\right)}^2},$ Positive

The system is an equilibrium which suggest that force on $q$ due $+Q_A=$ force on $q$ due to $+Q_B$.

Also, at point C, Electrostatic field provided by $Q_1$ & $Q_2$ will be zero.

$\Rightarrow {\overrightarrow{E}}_{CA}={\overrightarrow{E}}_{CB}$

$\Rightarrow \frac{K{Q}_1}{y^2}=\frac{{KQ}_2}{{(x-y)}^2}$

$\Rightarrow {\left(\frac{x-y}{y}\right)}^2=\frac{Q_2}{Q_1}$

$\frac{x}{y}-1=\sqrt{\frac{Q_2}{Q_1}}$

$\frac{x}{y}=\sqrt{\frac{Q_1}{Q_2}}+1=\sqrt{\frac{Q_1+Q_2}{Q_2}}$

$y=\sqrt[x]{x\frac{Q_2}{Q_1+Q_2}}$

Now, given in the question that

${\overrightarrow{F}}_{CA}={\overrightarrow{F}}_{AB}$

$\Rightarrow \frac{K{Q}_{1}q}{y^2}=\frac{K{Q}_1{Q}_2}{x^2}$

$q=\frac{y^2}{x^2}\left\{Q_2\right\}$

Substituting value of $y$ into the above equ.

$q=\frac{Q_1{Q}_2}{{(\sqrt{Q_1}+\sqrt{Q_2})}^2}$ {negative}

$\therefore$ Option (C) is the correct option.