Question

Coulomb's Law: Help

Answer 1

@matt101 I know that you said I had to add the forces to get the net force, but how would I know which numbers to add?

Answer 2

Well if we're looking at particle A, We want to find the force of B on A and C on A. Use Coulomb's Law to find each individual force and then add up the forces to find the net force.

Answer 3

Well the force of B on A is the same as the force on A on B, just in the opposite direction.

You just need to find 2 forces. Coulomb's Law tells you the magnitude of the force between two particles.

Answer 4

Okay. im going to try to solve now

Answer 5

I solve for both and add those two forces?

Answer 6

Yes

Answer 7

Correct! So the magnitude of the net force is -0.0675 N. The negative sign in front indicates that the net force on particle A is ATTRACTIVE.

What about for particle B? Be careful with the directions of each individual force.

Answer 8

Not quite - this one's a bit tricky.

Particle B is in the middle and is negatively charged. It is attracted to the positively-charged particles on either side. These two attractive forces, however, are pointed in opposite directions, so we will need to subtract the forces ultimately.

You also need to remember that particle C is only 0.2 m away from particle B (0.8-0.6). So, to find the net force on particle C, use the following:

\[F_{net}=F_{A \space on \space B}-F_{C \space on \space B}\]\[F_{net}={kq_Aq_B \over r_{AB}^2}-{kq_Cq_B \over r_{CB}^2}\]\[F_{net}=kq_B \left( {q_A \over r_{AB}^2}-{q_C \over r_{CB}^2} \right)\]

Answer 9

Okay So now I am finding the net force of C and coming back to B?

Answer 10

No the equation I provided is still for finding the net force on B.

Answer 11

By that I just mean the distance from A to B. The question says it's 0.6 m. For the distance of B to C, it's 0.2 m, which you know because if B is 0.6 m from A and C is 0.8 m from A, then C is 0.2 m from B.

Answer 12

What's your final answer?

Answer 13

Right! The magnitude of the force is 3.06 N.

The direction is a bit trickier to understand. We know the force needs to be attractive no matter what, but the problem is whether this net attractive force is towards A or towards C. Just looking at the numbers, you should realize it's towards C, because C has a larger positive charge and is also much closer to B than A is.

But if we look at the fact that the sign of your answer (+3.06 N) is positive, that also tells us the direction. We had to subtract forces to find the net force. In doing so you automatically set the first force direction as your reference direction. The attractive force in THAT direction would have a negative sign, as we saw for part (a) of this question. However, our answer here has a positive sign - that means the force is in the direction OPPOSITE the reference direction, but it doesn't change the fact that the force is still attractive.

Does that make sense?

Answer 14

Yes it does a bit. It is very tricky!

Answer 15

Give it shot and I'll have a look

Answer 16

That's right! And once again we can see right away it's a net attractive force because of the negative, since both forces originated on the same side of particle C.

Well done!

Answer 17

@matt101

Answer 18

Thank you for all of your help. You are amazing!

Answer 19

Those would be my answers. But to be more clear I would say something like this:

a) The net force on particle A has a magnitude of 0.0675 N and is attractive towards the right.

b) The net force on particle B has a magnitude of 3.06 N and is attractive towards the right.

c) The net force on particle C has a magnitude of 3.1275 N and is attractive towards the left.

This way there's no ambiguity in terms of whether the negative sign is referring to the direction of the force or whether the force itself is attractive or repulsive.

Answer 20

Thank you so much for everything!