Why Is the Direction of Electric Force Important in Calculations?

[flyingpig]

Homework Statement

http://www.youtube.com/watch?v=5AiWe9-szPc&feature=related

Go to frame 7:50

The man explains why it shouldn't be -q

unfortunately I am too dumb to fully understand.

Let the charge on the most left side be Q1, the middle charge be Q2, and the chargeon the right be Q3

The question wanted to ask what the force on Q2 be, the total force.

Now he explained something about "it's already taken care of"


F₂ = F₂₃ + F₁₃

Is the man trying to say that F₂₃ is negative because it is pushing left?

I pulled that formula out of my book. The way he did it is F₃₂ because it is to the right

 

[0ddbio]

It appears that he is trying to bypass direction vectors to make it easier, but sometimes this just makes things more confusing.

I only watched about 10 seconds of the video starting at 7:50, but it appears that he is just using coulombs law like so:
$$F = \frac{k q_{1}q_{2}}{r^{2}}$$
right? Where $k = \frac{1}{4\pi\epsilon_{0}}$ in SI units.

The thing is that it should more generally be:
$$\vec{F} = \frac{k q_{1}q_{2}}{r^{2}}\hat{r}$$

notice that now F is a vector (the arrow on top) and that there is an $\hat{r}$ unit vector. A unit vector is a vector that has length 1 and basically is just used to show direction.
In the coulomb force the unit vector points from source to test charge since the source is on the right and the test charge (the charge that is measuring the force) is on the left then the unit vector is -1. (in 2-dimensions it is (-1,0) ).

When you do it this way it is less confusing because you just plug everything in and you don't have to care about which way the force should go like he says in the video. One charge is positive one is negative and you do plug them in as such, but the unit vector effectively multiplies this by a -1 so the force ends up pointing in the positive direction like it should.
If you were using the other force to measure the charge then the particle on the left is the source and the one on the right is the [/b]test charge[/b] so then the unit vector becomes (1, 0) instead. Then you plug in one positive charge and one negative and the force is negative as it should be.

The way he does it though is often how people start learning it though I don't see why.. unit vectors are not that confusing.. any way, the method in the video effectively does away with vectors and replaces the direction vector with your "intuition".

 

[flyingpig]

Sorry, but that explanation is just as confusing! Ahhh lol

 

[0ddbio]

Another way to put it, is basically you just ignore the signs on everything. Just stick in the values and get a number positive or negative. (go ahead and just take the absolute value to make it easier, so you should always have a positive magnitude for the force)
Then once you have that value you decide whether the force should be positive or negative based on the problem. That's it. Of course this isn't the most formal way to do it, but for simple 1-dimensional problems it works just fine.

That is basically the way he does it in the video. For example if right is the positive direction in your coordinate axis and you have a positive and negative charge like so:
(-) (+)
then the force (magnitude) is the same on both.. the only difference is direction. Since they are opposite charges they attract so the force on the positive charge is in this direction:
<----------
So, for the positive charge you might choose the force to be negative simply because the direction toward the negative charge is to the left which is the negative direction.
The negative charge will experience a positive force.
---------->

Similarly in this case:
(+) (-)
now the positive charge will experience the positive force and the negative charge will experience the negative force.

The way I explained it earlier may be more confusing but it gets rid of this "guess work". You don't choose based on your intuition like this, you simply follow the math and it will tell you the directions automatically which I find better.. But if you think this way is easier then by all means use it.

Keep in mind that the "positive" and "negative" meaning is arbitrary. Positive and negative forces only show direction in your particular coordinate system... If you do not specify what your system is then the direction is meaningless.
For example say you have a force in this direction:
--------->
most people would say this is positive, but is it?
what if my coordinate axis is such that the x-axis decreases as you go to the right and increases toward the left. Then the above force could be called "negative". So it can be positive or negative just depending on which direction you want to be positive.

The only important thing is that whatever direction you want to be positive must be consistent throughout your problem.

 

[rwisz]

.


I would like to clarify and provide a response to the content in the video. The man in the video is correct in stating that the force on Q2 is the sum of the forces from Q1 and Q3, which can be written as F2 = F23 + F13. However, the direction of the forces should be taken into account when calculating the total force. In this case, the direction of F23 is to the left, so it should be written as F2 = -F23 + F13. This is because forces in opposite directions cancel each other out, and the overall force on Q2 would be to the right. This is what the man means when he says "it's already taken care of" - the direction of the forces have been accounted for in the calculation. It is important to pay attention to the direction of forces when solving problems involving electric forces. I hope this clarification helps in your understanding.