Answer :
Answer:
The third charge placed is 0.80 m.
Explanation:
Given that,
Distance = 0.57 m
First charge = q
Third charge = 2q
We need to calculate the electrostatic force on charge q₁ due to q₂
Using formula of electrostatic force
[tex]F_{21}=\dfrac{kq_{1}q_{2}}{r_{1}^2}[/tex]
When placed another charge q₃ at certain distance from origin, then the net force on charge q₁ due to both charges is
[tex]F_{net}=F_{21}+F_{31}[/tex]
The net electrostatic force on the charge at the origin doubles.
[tex]2F_{21}=F_{21}+F_{31}[/tex]
[tex]F_{31}=F_{21}[/tex]
[tex]\dfrac{kq_{3}q_{1}}{r_{2}^2}=\dfrac{kq_{2}q_{1}}{r_{1}^2}[/tex]
[tex]\dfrac{q_{3}}{r_{2}^2}=\dfrac{q_{2}}{r_{1}^2}[/tex]
[tex]r_{2}^2=\dfrac{q_{3}}{q_{2}\timesr_{1}^2}[/tex]
[tex]r_{2}=\sqrt{\dfrac{q_{3}}{q_{2}}}r_{1}[/tex]
Put the value into the formula
[tex]r_{2}=\sqrt{\dfrac{2q}{q}}\times0.57[/tex]
[tex]r_{2}=\sqrt{2}\times0.57[/tex]
[tex]r_{2}=0.80\ m[/tex]
Hence, The third charge placed is 0.80 m from origin in x-axis.
