Question

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?

Answer 1

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!