- #1

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## Homework Statement

See attachment

## Homework Equations

## 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.

[itex]E_{1} = \frac{1}{4\pi\epsilon_{0}} \frac{q_{1}}{r^{2}} \hat{i}[/itex]

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

[itex]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}[/itex]

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

[itex]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)[/itex]

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

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

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

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

For a final formula of:

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

**However, the value of the number within the arccos function is the**

*negative*of what it should be.This gives me the answer correctly:

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

Why?