# Homework Help: Coulomb's law for particles

1. Feb 9, 2006

### heelp

Three positive particles of equal charge,+11.0uc, are located at the corners of an equilateral triangle of side 15,0cm calculate the magnitude and direction of the net force on each particle.
Here is what I got so far f1= kq1q2/r^2 =(9.0*10^9N*m^2/c^2/)(11.0*10^-6)^2= 48.4N
F1 = COS30(48.4)= 41.9N
SIN30(48.4) =24.2N
I tried to find the square root of 41.9^2+24.2^2 and I didn't get my didn't my desired answer. How would I find the direction of each particle?

This one I have no idea on how to solve.Two small nonconducting spheres have a total charge of 90.0uc. (a) when placed 1.06m apart, the force each exerts on the other is 12N and is repulsive. What is the charge on each? (b) what if the force were attractive?

Here is another one

Two point charges have a total charge of 560uc. When placed 1.10m apart, the force exerts on the other is 22.8N and is replsive. What is the charge on each?

What is the best way to solve all Coulomb's law Problems? Thanks in advance for your help.

2. Feb 10, 2006

### Astronuc

Staff Emeritus
The Coulomb force law is as you wrote it.

For two charges, i.e. no influence from other charges,
$$F\,=\,k\,\frac{q_1\,q_2}{r^2}$$ where F is the magnitude of the force, $q_1$ and $q_2$ are the charges, and r is the distance separating the charges.

Now more correctly it should be written

$$\vec{F}\,=\,k\,\frac{q_1\,q_2}{r^2}\,\hat{r}$$

where $\vec{F}$ is a vector, and $\hat{r}$ is the unit vector along the line passing between the two charges. Remember mathematically, two points define a line.

Now, force is a vector, so for multiple charges, the Coulomb force must be added (or subtracted) as vectors. The resultant force is provided by the linear combination of each individual force vector.

So in the first problem, one must recognize that the triangle is an equilateral triangle, and the charges are at the three vertices. So for one charge, find the influence of the other two.

Draw an equilateral triangle, such that the base is horizontal and one vertex is pointing upward. Then resolve the force vectors of the two charges on the base into vertical and horizontal components. Due to symmetry, the horizontal components of two charges cancel (equal but opposite), while the vertical components are additive - (equal but in same direction).

In problem 2, apply Coulomb's force law, with a distance of 1.06 m. ONe can try the solution that the charges are equal, i.e. each charge of 45 uC, and if that is different from 12N, then the magnitude of charges is obviously different so then one must try q1 = 90 uC - x, and the other charges is q2 = x. Then one must solve a quadratic equation for x. If the force is repulsive, both charges have the same polarity (+ or -), but if the force is attractive, the charges have different polarity, i.e. one is + and the other -.

3. Feb 11, 2006

### heelp

Thanks for your help but I'm a little confuse on the first question. When I get the answer for the first equation do I multiply by cos30 or 60? I would I know what to do?

4. Feb 11, 2006

### Astronuc

Staff Emeritus
Taking an equilateral triangle with the base on the horizontal, the angle is 60° between the other legs (line of action) and the base. The vertical component of the force on the charge at the top vertex would require sin 60° or cos 30°.

Draw a vector diagram, resolve the force vectors into vertical and horizontal components, and note the angles.