Point charges q= 50 μC and 92=-25C areplaced

Meet Malaviya

Meet Malaviya

Answered question

2022-05-26

Point charges q= 50 μC and 92=-25C are

placed 1.0 m apart. What is the force on a third charge 93= 20 μC placed midway between 1 and 4₂?

Answer & Explanation

karton

karton

Expert2022-07-07Added 613 answers

Though the question doesn't state where the charges are in relation to the observer, for the sake of definiteness let's say q1 is on the left and q2 is on the right. By the Principle of Superposition of Forces, we know that we can consider each force on q3 individually and then just add them together (keeping in mind direction).
From Coulomb's law, we know that the magnitude of the force on q3 due to q1 is given by

F=k|qq|r2
In this case, r=0.5 m since q3 is halfway between the two charges. So then

 

F=8.988×109((50×10-6)(20×10-6)0.52

F=35.95 N


Now let's think about the direction of the force. q1 is a positive charge and so is q3. So those two particles will want to repel each other. Since q1 is on the left, q3 will feel a force in the positive direction.


Now let's consider the force due to charge q2. Again, we use Coulomb's law to get the magnitude.

 

F=8.988×109((25×10-6)(20×10-6)0.52

F=17.98 N


And this time the charges have opposite sign and so attract each other. Since q2 is on the right of q3, the force due to it will also point in the positive direction.


So the total force is just the sum of the two since they point in the same direction. That is, the force has a magnitude of 35.95+17.98=53.93 newtons and points in the direction of q2 from q3 (remember we said it was to the right, but the question didn't so all we can say is it's in the q2 direction).

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