# Draw a free body diagram for the electric force

1. Aug 28, 2016

### Cglez1280

1. The problem statement, all variables and given/known data
A positive charge Q1=7.4uC is located at a point X1=-2 m, a negative charge Q2=-9.7uC is located at a point X2=3m and a positive Q3=2.1uC is located at a point X3=9
A. Draw a free body diagram for the electric force acting on Q1, Q2, and Q3.
B. Find the magnitude of the force between Q1 and Q2
So like I have no idea what to do I'm completely lost this is ap physics summer work and I've never taken a physics class and this is just plain confusing

Not sure what would help.
Q1=-7.5 uC
Q2=-9.7 uC
Q3=2.1 uC
X1=-2 m
X2=3 m
X3=9 m
2. Relevant equations
I don't know

3. The attempt at a solution
I'm sorry but all I have is numbers that i don't know what to do with

2. Aug 28, 2016

### Lucas SV

The total force acting on a body is the sum of all the forces acting on the body due to all other particles. If you know Coulomb's law this will help drawing the free body diagrams. Make sure you know the definition of a free body diagram. If you can add vectors, you can sum forces, since forces are represented by vectors.

Last edited: Aug 28, 2016
3. Aug 28, 2016

### Cglez1280

Ok sorry to add that I already drew the free body diagrams but after that I'm confused.

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4. Aug 28, 2016

### Lucas SV

Next step is applying Coulomb's law to find the forces between particles. You can start for example, by computing $F_{12}$. You need to know the charges $Q_1$ and $Q_2$, as well as the distance between these two charges.

Last edited: Aug 28, 2016
5. Aug 28, 2016

### Cglez1280

Ok so how would I do the distance. Since X1 is at -2m and X2 is at 3m would it be 5? Sorry for the dumb questions I'm just trying to make sure I can do the next few parts without help

6. Aug 28, 2016

### Lucas SV

Correct! Actually more precisely you should put 5m. Always make sure you put units in physical calculations.

7. Aug 28, 2016

### Cglez1280

Ok I've been looking at Coloumb's law and it says k would be 9*10^9 N, is that right in this case because every time I do the math I get some other number from the answer I'm supposed to get.

8. Aug 28, 2016

### Lucas SV

Actually the unit is N m^2 / C^2.

Are the answers you got the same as the solution up to a power of 10?

9. Aug 28, 2016

### Cglez1280

So the picture shows my work and the answer I'm supposed to get is 2.58*10^-2. I'm not sure what I did wrong

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10. Aug 28, 2016

### Lucas SV

You made the mistake I thought you did. The charges are measured in micro coulombs, $\mu C$. You forgot about the micro. Watch out those units.

Last edited: Aug 28, 2016
11. Aug 28, 2016

### Cglez1280

Hmm ok I see. But would the negative sign be dropped in my work because that seems to be an issue also. Another question is what would I do when I'm asked for the magnitude and direction of the net electric force on charge Q1?

12. Aug 28, 2016

### Lucas SV

The negative sign is dropped precisely because question B asks for the magnitude of the force. The magnitude of a force is always non-negative, so you take the absolute value.
I already explained magnitude. The direction can be seen from your free body diagram and can be found by the rule: opposite charges attract and like charges repel.

13. Aug 28, 2016

### Cglez1280

Oh my god just when I get magnitude the last part completely lost me.

14. Aug 28, 2016

### Lucas SV

If the force is attractive, the force vector of charge 1 will be pointing towards charge 2. If the force is repulsive, the force vector of charge 1 will be pointing away from charge 2. The rule I just mentioned tells you if the force is attractive or repulsive.

Do you know what I meant by 'like' charges and 'opposite' charges?

15. Aug 28, 2016

### Cglez1280

So basically my answer for Q1 will be positive and Q2 and Q3 will be negative?
Is there a certain equation I should I use to be able to get my answer?

16. Aug 28, 2016

### Lucas SV

The force on Q1 due to Q2 is positive because they attract and the coordinate of Q2 is higher than the coordinate Q1. This is in fact true regardless of how you choose to orient your x axis. If you choose to orient it to the right, Q2 will be to the right of Q1, and so will be the force on Q1 due to Q2. So the fact that the x axis and the direction of the force have the same orientation makes the force positive. If instead you orient x to the left, Q2 will be to the left of Q1 (by definition of the position of Q1 and Q2).
It is the Coulomb law. The information about the direction of the force is contained in the law. Here is the equation for the coulomb force on $q_1$ due to $q_2$, if $q_1$ is located at $x_1$ and $q_2$ is located at $x_2$:
$$F=k q_1 q_2 \frac{x_1-x_2}{|x_1-x_2|^3},$$ where $k$ is the Coulomb's constant. It is easy to check that the magnitude of this equation yields what you probably used to compute the magnitude, while $F$ has the same direction obtained from the rules I gave.

The full law is actually written in terms of vectors in three dimensions. Note however that in your problem, the charges were collinear, which simplified the problem a lot, because you can neglect the extra dimensions and work in one dimension, by making sure that the x axis intersects all three charges.

Last edited: Aug 28, 2016