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Net force acting on positive charge

  • Thread starter ardour
  • Start date

ardour

1. Homework Statement
upload_2017-10-3_18-55-36.png


2. Homework Equations
F= (k*q*Q)/r^2

3. The Attempt at a Solution
The answer key gives the answer as D. I thought it was C. If the net electric force acting on P is zero, doesn't that mean that the force between Q1 and P and the force between Q2 and P need to cancel each other out? If you added them together, one would have to be positive, and the other would have to be negative to cancel out. Also, why should Q1 have a magnitude greater than Q2? Since Q2 is closer to P, I would have thought that it would have had a greater charge than Q1.
 

Marc Rindermann

Gold Member
30
9
if one ##Q## was positive, the other negative, let's just say ##Q_1## positive and ##Q_2## negative, then the force of ##Q_1## on ##P## would push ##P## to the right and the force of ##Q_2## on ##P## would pull it to the right. So both forces would act in the same direction and the net force on ##P## cannot be zero. That means the charges cannot have opposite signs.

Electric force falls off with distance with ##\frac{1}{r^2}##. So if ##Q_1## is farther away from ##P## but still acts with an equal force on ##P## as ##Q_2## does then the magnitude of ##Q_1## must be greater than the magnitude of ##Q_2## to compensate for the greater distance.
 

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