Coulomb's Law Hexagon

1. Jun 4, 2004

sandplasma

I'm having a little trouble with this one..

Three pointlike charges Q are located on three successive vertices of a reguar hexagon with sides "l". Find the electrical force on another charge q located at the center of the hexagon. Assume all the charges are like charges. ( all positive )

I know I'm supposed to use Coulombs law as well as the principle of superposition but I'm having trouble with the direction.

Thanks in advance for the help!

2. Jun 4, 2004

Gza

The first step of any physics problem is defining a coordinate system. A "regular" coordinate system (no rotations or anything) with its origin at the center of the hexagon makes the most sense for this problem since it utilizes a lot of simplifying symmetry. Next find the contribution each individual charge on the vertices of the hexagon makes to the net force on the charge at the origin. This of course is done with as you mentioned, Coulomb's law. Sticking with the vectors throughout your solution makes it much easier to determine directions for forces.

Coulombs Law

$$\vec{F_{q2q1}} = \frac{k q_1q_2}{r^2} \hat{r_{q1q2}}$$

where $$\hat{r_{q1q2}}$$ represents a relative unit vector, or going by the subscripts, the position of charge 1 relative to charge 2. This vector simply "points" from charge 2 to charge 1. Charge 2 will represent the three charges on the vertices of the hexagon since we want to find their force which points at charge 1 at the origin; in other words the force of q2 on q1, as stated on the left side of coulombs law.

Have a look at the picture to get a feel for the setup. By the geometry of a hexagon, the distance from the vertex to the center is simply "l". (verify this) So the only important task left is finding the relative unit vectors for each charge on the hexagon in $$\hat{i},\hat{j}$$ components, sticking them into Coulombs law and adding the whole thing up, by principal of superposition.

The easiest technique i've found for determining relative vectors is through the use of something my mechanics teacher taught me a few years back, which he simply called the neumonic (i think that's how you spell it.) It looks like this:

$$\vec{r_{12}} = \vec{r_{1O}} + \vec{r_{O2}}$$
where O represents the origin of your coord sys.

This is easy for me to remember because the origin shows up on the "inside" of the subscripts of the two added vectors.

or flipping around the second vector being added(since it makes more sense in problem solving to find the position of an object relative to the origin, not vice versa)

$$\vec{r_{12}} = \vec{r_{1O}} - \vec{r_{2O}}$$

of course by dividing through by the magnitude (which is "l") you obtain the unit vector, which is what you need. I hope this helped you get started!

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