Homework Help: Moving point charges - help debugging

1. Aug 25, 2012

1MileCrash

1. The problem statement, all variables and given/known data

See attachment

2. Relevant equations

3. The attempt at a solution

I was successful with finding a solution. But, only because I tried the practice version of that (where you can see the answers) and noticed the equation I derived gave angles that, when summed with the correct answer, yielded 180*.

Can someone tell me why?

START:

Since Bead 1 is stationary, we know its electric field will be in the direction of the positive x axis.

$E_{1} = \frac{1}{4\pi\epsilon_{0}} \frac{q_{1}}{r^{2}} \hat{i}$

While Bear 2 can be moved, thus its field will be a function of theta.

$E_{1} = \frac{1}{4\pi\epsilon_{0}} \frac{q_{2}}{r^{2}}cos(\theta) \hat{i} + \frac{1}{4\pi\epsilon_{0}} \frac{q_{2}}{r^{2}}sin(\theta) \hat{j}$

Thus, we add these two vector equations together for net E, and take the magnitude. After a bit of algebra we get:

$E^{2}(4\pi\epsilon_{0})^{2} = q^{2}_{1} + 2q_{1}q_{2}cos(\theta) + q^{2}_{2}cos^{2}(\theta) + q^{2}_{2}sin^{2}(\theta)$

By factoring out the q2 squared and noting the trigonometric identity,

$E^{2}(4\pi\epsilon_{0})^{2} = q^{2}_{1} + 2q_{1}q_{2}cos(\theta) + q^{2}_{2}$

$E^{2}(4\pi\epsilon_{0})^{2} - q^{2}_{1}- q^{2}_{2}= 2q_{1}q_{2}cos(\theta)$

$\frac{E^{2}(4\pi\epsilon_{0})^{2} - q^{2}_{1}- q^{2}_{2}}{2q_{1}q_{2}}= cos(\theta)$

For a final formula of:

$arccos(\frac{E^{2}(4\pi\epsilon_{0})^{2} - q^{2}_{1}- q^{2}_{2}}{2q_{1}q_{2}}) = \theta$

However, the value of the number within the arccos function is the negative of what it should be.

This gives me the answer correctly:

$arccos(- \frac{E^{2}(4\pi\epsilon_{0})^{2} - q^{2}_{1}- q^{2}_{2}}{2q_{1}q_{2}}) = \theta$

Why?

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2. Aug 25, 2012

TSny

Consider the direction of the electric field of bead 2 and decide on the signs of its x and y components.

3. Aug 25, 2012

1MileCrash

I think I understand you.

While cos(45) is positive, at an angle 45 B's electric field is pointing down and towards the left.

Right?

4. Aug 25, 2012

TSny

Yes

5. Aug 25, 2012

1MileCrash

Thank you very much.