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lorasvet [3.4K]
4 years ago
14

A particle of charge +2q is placed at the origin and particle of charge -q is placed on the x-axis at x = 2a. Where on the x-axi

s can a third positive charge be placed so that the net electric force on it is zero? a is the unit of distance.
Physics
1 answer:
zzz [600]4 years ago
4 0

Answer:

r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}  

Explanation:

We know that the electric force equation is:

F=k\frac{q_{1}*q_{2}}{r^{2}}

  • k is the electric constant 9*10^{9} Nm^{2}/C^{2}
  • 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.

F_{31}=k\frac{q_{3}*q_{1}}{r_{31}^{2}}

F_{31}=k\frac{q*2q}{r_{31}^{2}}

F_{31}=k\frac{2q^{2}}{r_{31}^{2}}

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.

F_{32}=k\frac{q_{3}*q_{2}}{x_{32}^{2}}

F_{32}=k\frac{q*(-q)}{r_{32}^{2}}

F_{32}=-k\frac{q^{2}}{r_{32}^{2}}

r(32) is the distance between 3 and 2.

Now, r_{31}+r_{32}=2a or r_{32}=2a-r_{31}

The net force must be zero so:

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}}

We just need to solve it for r(31)

r_{31}^{2}=2(2a-r_{31})^{2}

r_{31}^{2}=2(2a-r_{31})^{2}

r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}  

Therefore the distance from the origin will be:

r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}  

I hope it helps you!        

                 

 

     

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