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Three point charges lie in a straight line along the y-axis. a charge of q1 = -9.10 µc is at y = 6.30 m, and a charge of q2 = -7.90 µc is at y = -4.50 m. the net electric force on the third point charge is zero. where is this charge located?

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Answer:

 Electric field E = kQ/r^2  

Distance between charges = 6.30 - (-4.40) = 10.70m  

Say the neutral point, P, is a distance d from q1. This means it is a distance (10.70 - d) from q2.  

Field from q1 at P = k(-9.50x^10^-6) / d^2  

Field from q2 at P = k(-8.40x^10^-6) / (10.70-d)^2  

These fields are in opposite directions and are equal magnitudes if the resultant field = 0  

k(-9.50x^10^-6) / d^2 = k(-8.40x^10^-6) / (10.70-d)^2  

9.50 / d^2 =8.40 / (10.70-d)^2  

d^2 / (10.70-d)^2 = 9.50/8.40 = 1.131  

d/(10.70-d) = sqrt(1.1331) = 1.063  

d = 1.063 ((10.70-d)  

= 10.63 - 1.063d  

2.063d = 10.63  

d = 5.15m  

The y coordinate where field is zero is 6.30 - 5.15 = 1.15m

Explanation:

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