 #1
 1,331
 45
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?
Attachments

10.2 KB Views: 278