Answer :
Answer:
[tex]r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}[/tex]
Explanation:
We know that the electric force equation is:
[tex]F=k\frac{q_{1}*q_{2}}{r^{2}}[/tex]
- k is the electric constant [tex]9*10^{9} Nm^{2}/C^{2}[/tex]
- r is the distance between the particles
- q1 and q2 are the particle
Now, we have three particles, the first one at x=0, the second one at x=2a and the third in some place between these two particle.
1. Let's find the electric force between the first particle and the third particle.
[tex]F_{31}=k\frac{q_{3}*q_{1}}{r_{31}^{2}}[/tex]
[tex]F_{31}=k\frac{q*2q}{r_{31}^{2}}[/tex]
[tex]F_{31}=k\frac{2q^{2}}{r_{31}^{2}}[/tex]
r(31) is the distance between 3 and 1
2. Now, let's find the electric force between the third particle and the second particle.
[tex]F_{32}=k\frac{q_{3}*q_{2}}{x_{32}^{2}}[/tex]
[tex]F_{32}=k\frac{q*(-q)}{r_{32}^{2}}[/tex]
[tex]F_{32}=-k\frac{q^{2}}{r_{32}^{2}}[/tex]
r(32) is the distance between 3 and 2.
Now, [tex]r_{31}+r_{32}=2a[/tex] or [tex]r_{32}=2a-r_{31}[/tex]
The net force must be zero so:
[tex]F_{31}+F_{32}=0[\tex]
[tex]k\frac{2q^{2}}{r_{31}^{2}}-k\frac{q^{2}}{r_{32}^{2}}=0[\tex]
[tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{r_{32}^{2}})=0[\tex]
[tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}})=0[\tex]
It means that:
[tex]\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}}[/tex]
We just need to solve it for r(31)
[tex]r_{31}^{2}=2(2a-r_{31})^{2}[/tex]
[tex]r_{31}^{2}=2(2a-r_{31})^{2}[/tex]
[tex]r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}[/tex]
Therefore the distance from the origin will be:
[tex]r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}[/tex]
I hope it helps you!