Three point charges ${{Q}_{1}}$,${{Q}_{2}}$,${{Q}_{3}}$are placed along a line at same distance apart. If ${{Q}_{1}}$and ${{Q}_{2}}$are opposite in nature and net force on ${{Q}_{3}}$is zero. Find magnitude of ${{Q}_{1}}$.
A.\[1\]
B.$\sqrt{2}$
C.$2{Q_2}$
D.$4{Q_2}$

Hint: When two objects have an electrical charge and exert an electric force on each other at a specific distance, the overall phenomena can be explained by Coulomb’s law. This law is also applicable for calculating net force since given three-point charges are equally spaced.



Formula used:
${{F}_{A,B}}=\dfrac{1}{4\pi {{\in }_{O}}}\dfrac{\left| {{q}_{A}}{{q}_{B}} \right|}{r_{AB}^{2}}$
Here ${{F}_{A,B}}=$ Electric force between two point charges ${{q}_{A}}$ and ${{q}_{B}}$
${{\in }_{O}}=$ constant = $8.85\times {{10}^{-12}}$${{C}^{2}}/N{{m}^{2}}$
${{r}_{AB}}=$ Distance between two-point charges ${{q}_{A}}$ and ${{q}_{B}}$

Complete answer:
First, we draw a possible diagram to explain the exerted force between the point charges.

According to the question, ${{Q}_{1}},{{Q}_{2}}$ ${{Q}_{3}}$point charges are placed along a straight line.
Let, ${{Q}_{1}}$ be a positive charge but ${{Q}_{2}}$ ${{Q}_{3}}$have the opposite sign. That means there exists an attractive force between ${{Q}_{2}}$and ${{Q}_{3}}$. But we have to keep the net force on ${{Q}_{3}}$is zero, that’s why the repulsive force acts between ${{Q}_{1}}$and ${{Q}_{3}}$.
Thus, the overall process reaches the equilibrium position as two forces that are attractive and repulsive must balance each other.
Applying Coulomb’s law between${{Q}_{1}}$ and ${{Q}_{3}}$ ,
${{F}_{1,3}}=\dfrac{1}{4\pi {{\in }_{0}}}\dfrac{{{Q}_{1}}{{Q}_{3}}}{r_{1,3}^{2}}$
 Or, ${{F}_{1,3}}=\dfrac{1}{4\pi
{{\in }_{0}}}\dfrac{{{Q}_{1}}{{Q}_{3}}}{{{(2r)}^{2}}}$ [Since ${{r}_{1,3}}=2r$] …….(i)
And electric force between ${{Q}_{2}}$ and ${{Q}_{3}}$ ,
${{F}_{2,3}}=\dfrac{1}{4\pi {{\in }_{0}}}\dfrac{{{Q}_{2}}{{Q}_{3}}}{r_{2,3}^{2}}$
Or, ${{F}_{2,3}}=\dfrac{1}{4\pi
{{\in }_{0}}}\dfrac{{{Q}_{2}}{{Q}_{3}}}{{{(r)}^{2}}}$ [Since ${{r}_{2,3}}=r$] ……….(ii)
Due to the equilibrium conditions, ${{F}_{1,3}}=-{{F}_{2,3}}$ [Since $(-)$sign indicates the attaractive forces between $2$and $3$]
 $\dfrac{1}{4\pi
{{\in }_{O}}}\dfrac{{{Q}_{1}}{{Q}_{3}}}{{{(2r)}^{2}}}=-\dfrac{1}{4\pi {{\in
}_{o}}}\dfrac{{{Q}_{2}}{{Q}_{3}}}{{{(r)}^{2}}}$
Or, $\dfrac{{{Q}_{1}}{{Q}_{3}}}{4{{r}^{2}}}=-\dfrac{{{Q}_{2}}{{Q}_{3}}}{{{r}^{2}}}$
Or,${{Q}_{1}}=-4{{Q}_{2}}$
Here also the minus (-) sign indicates the attractive forces between charge ${{Q}_{1}}$and ${{Q}_{2}}$.
Thus the charge of ${{Q}_{1}}$and ${{Q}_{2}}$ are opposite to each other
Therefore, (D) is the correct option.





Note:In this question three charges are placed on a line. So pay attention while assigning polarities to these three charges and also to displacement values between these charges. You should also understand the concept of repulsive and attentive electrostatic forces between charges.