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A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each a distance 0.0429 m from the origin. Where should a third particle with charge 9.03 µC be placed so that the magnitude of the electric field at the origin is zero?

Respuesta :

whitneytr12

Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

Explanation:

Let's assume that the third charge is on the negative x-axis. So we have:

[tex]E_{1}+E_{3}-E_{2}=0[/tex]

We know that the electric field is:

[tex]E=k\frac{q}{r^{2}}[/tex]

Where:

  • k is the Coulomb constant
  • q is the charge
  • r is the distance from the charge to the point

So, we have:

[tex]k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0[/tex]

Let's solve it for r(3).

[tex]\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0[/tex]

[tex]r_{3}=0.0743\: [/tex]  

Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

 

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