# Coulomb's Law: forces between charges

1. Sep 11, 2004

### ACLerok

apparently i got the correct answer for this problem to be 8.62*10^(-5) N but when I try doing it again, I get a different answer.

A particle of charge 4.96nC is placed at the origin of an xy-coordinate system, and a second particle of charge 2.02nC is placed on the positive x-axis at 3.96cm. A third particle, of charge 6.01nC is now placed at the point 3.96cm, 3.01cm.

Find the x-component of the total force exerted on the third charge by the other two. Use episilon_0=8.85×10-12C/N*m^2 for the permittivity of free space.

I first converted each charge and distance from nC to C and cm to m. Then I used the equation F=(1/4piEpisilon_0)*(|product of charges|/distance between charges^2) to determine the electric force converted on one charge by another charge. I then multiplied this F by cos(45) to find the total x-component. I do this but i'm not getting the same answer i got before. Can someone help me out?

PS: how would I find the y component of the force exerted on charge 3 by charge 2? Would it just be the Force of 2 on 3 multiplied sin(90)?

2. Sep 11, 2004

### maverick280857

The first charge can be taken at the origin of a cartesian coordinate system. Next, you must compute the position vector of the second charge with respect to the first and in this case, with respect to the origin. Use the vector form of Coulomb's law:

$$\vec{F} = \frac{1}{4\pi \epsilon_{0}}\frac{Q_{1}Q_{2}}{r^{3}}\vec{r}$$

As far as resolving the forces is concerned, this one step will take care of the problem correctly if you compute the radius vector correctly.

Lets first talk of two charges, one at the origin and the other at (3.96, 3.01).

If you wish to use the scalar force equation (as you have done in the method indicated in your post), you must multiply F (magnitude) by the cosine of the angle made by the line joining the charges with the x-axis. Since charge 2 is at (3.96, 3.01) and charge 1 is at (0, 0), it is easiest to find first the tangent of that angle, which is 3.96/3.01. You can then use

$$\cos\theta = \frac{1}{\sqrt{1 + \tan^{2}\theta}}$$

(If you have a calculator, you can find the cosine directly. Alternatively, find the length of the hypotenuse of the triangle, i.e. the quantity $$\sqrt{3.96^2 + 3.01^2}$$ and then find the cosine. Your choice.)

In your problem, you can use this method to find the forces on charge 3 due to charge 1 and on charge 3 due to charge 2. But to find the net force and its direction, you should ideally resort to the vector method so you will get the direction automatically through the unit vectors. Otherwise, you will have to go through a tedious task of first computing scalars, then resolving them and adding them up with proper signs to account for directional differences and then finding the net force by squaring the x and y components of the net force and taking its square root.

Hope that helps.

Cheers
Vivek

(PS--Don't substitute for numbers until the algebra is complete...as a general rule but if you do see factors/numbers/quantities cancelling, go ahead and do that.)

3. Sep 12, 2004

### ACLerok

ok but how would i find the y component of force on charge 3 by charge 2? Multiply the Force on 3by2 by sin of what angle?

4. Sep 13, 2004

### maverick280857

You can do this the following way:

1. You need force vector expressions for forces on charge 3 due to charge 1 and 2. Now you can either set up an expression for the electric field due to superposition of the electric fields due to charge 1 and 2, at a point (x, y) and then plug in the coordinates of charge 3.

2. Alternatively, you can find the position vectors of charge 2 and charge 1 referred to charge 3. Suppose $$\vec{r_{1}}$$, $$\vec{r_{2}}$$ and $$\vec{r_{3}}$$ are the position vectors of charges 1, 2 and 3 respectively (in your case, one of the charges is at the origin). Then you need to find $$\vec{r_{13}}$$ and $$\vec_{r_{23}}$$ and find their unit vectors.

Follow either of these two ideas and let me know if things work out...I'll elaborate (if required) after you post your reply.

Cheers
Vivek

EDIT: Draw a diagram on a piece of paper showing the positions of the charges on a cartesian coordinate system (x-y plane). Try and work out the angle yourself first.