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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?

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Answer:

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

Step-by-step explanation:

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


E_(1)+E_(3)-E_(2)=0

We know that the electric field is:


E=k(q)/(r^(2))

Where:

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

So, we have:


k(q_(1))/(r_(1)^(2))+k(q_(3))/(r_(3)^(2))-k(q_(2))/(r_(2)^(2))=0

Let's solve it for r(3).


(3.01)/(0.0429^(2))+(9.03)/(r_(3)^(2))-(6.02)/(0.0429^(2))=0


r_(3)=0.0743\:

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

I hope it helps you!

User Brady Maf
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