Electrostatic force using vectors

Homework Statement

Charge A is +2C and is located at <4,0,0>
Charge B is +12.5C and is located at <0,-3,0>

What is the the Force(AonB)
What is the magnitude of F(AonB)

Homework Equations

F = (k)(Qa)(Qb)/(r^2) (r^)

For reference the answers are

< -0.8k, 0.6k, 0 > N (F vector)

and

1k N (magnitude)

The Attempt at a Solution

So I am confused as to how to solve this problem. I don't know if maybe it is worded strangely or what but I am not getting those answers.

I did a lot of these problems but not with vectors. How do you apply the answer you get into a vector. The equation is given as r(hat) after the regular F equation. Does this mean I distribute my answer into i+j+k? I am assuming the answer to that is no because that makes no sense mathematically.

This is not a homework problem it is a practice problem. I am trying to learn physics and would really appreciate if someone could explain how you apply your F into vector notation and how this particular problem can be solved so I can learn. I am sure it is something simple that I am just not getting and my teacher didn't explain properly (in class we never did any with vectors).

Answers and Replies

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ehild
Homework Helper

Homework Statement

Charge A is +2C and is located at <4,0,0>
Charge B is +12.5C and is located at <0,-3,0>

What is the the Force(AonB)
What is the magnitude of F(AonB)

Homework Equations

F = (k)(Qa)(Qb)/(r^2) (r^)

For reference the answers are

< -0.8k, 0.6k, 0 > N (F vector)

and

1k N (magnitude)

The Attempt at a Solution

So I am confused as to how to solve this problem. I don't know if maybe it is worded strangely or what but I am not getting those answers.

I did a lot of these problems but not with vectors. How do you apply the answer you get into a vector. The equation is given as r(hat) after the regular F equation. Does this mean I distribute my answer into i+j+k? I am assuming the answer to that is no because that makes no sense mathematically.
Yes, you need to write up the force in x,y,z components F=Fxi+Fyj+Fzk, or <Fx,Fy,Fz>

This is not a homework problem it is a practice problem. I am trying to learn physics and would really appreciate if someone could explain how you apply your F into vector notation and how this particular problem can be solved so I can learn. I am sure it is something simple that I am just not getting and my teacher didn't explain properly (in class we never did any with vectors).
The vector "r^" is the unit vector pointing from A to B when you want the force exerted by charge at A to the charge at B. The vector pointing to B from A is the difference of the position vectors r=rB-rA=<0,-3,0>-<4,0,0> (or rB-rA=-3j-4i). You need to divide this vector with its magnitude to get the unit vector r^. "r" means the magnitude of the vector r.

ehild

Ah that makes a lot more sense now. So the reason I am finding the unit vector is I need a vector that is pointing in the same direction as the Force caused by these two charges.

Whenever I see r hat from now on that is referring to the unit vector, correct?

Problem worked out perfectly after that. I really appreciate the help.

Also just for clarification. If the question had asked for the F (BonA) or F_BA it then the difference would be the unit vector being composed from A - B or <4,-3,0> and of course the magnitude would be the same (5).

So the F_BA would have been <4/5 ke, -3/5 ke, 0>

Could you explain why that is the case? As in why would two positive charges cause one to go in the positive X (AonB) and the other in the negative X (BonA) direction? And why are would both have the same Y component? I just want to get a better understanding of what I am doing. I hate to memorize formulas and plug things in I would rather understand the concept.

Once again thank you very much for the explanation I truly appreciate it.

Last edited:
jtbell
Mentor
If the question had asked for the F (BonA) or F_BA it then the difference would be the unit vector being composed from A - B
Correct.

or <4,-3,0>
Check your arithmetic.

As a reality check, remember Newton's Third Law of Motion. The force that B exerts on A must be equal in magnitude and opposite in direction to the force that A exerts on B.

Ah you're right. Thank you it all makes sense now! I really appreciate both of your help.