anonymous
  • anonymous
Three 1.0 nC charges are placed as shown in the figure. Each of these charges creates an electric field vector E at a point 3.0 cm in front of the middle charge.What are the three fields vector E1, vector E2, and vector E3 created by the three charges? Write your answer for each as a vector in component form.
Physics
schrodinger
  • schrodinger
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anonymous
  • anonymous
I mainly need help on starting it.
TuringTest
  • TuringTest
Got a picture?
anonymous
  • anonymous
sorry yes there is just hold on

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anonymous
  • anonymous
|dw:1327883171291:dw|
anonymous
  • anonymous
q1,q2q3 all equal 1nC
TuringTest
  • TuringTest
so you need the field vector from each charge at that point on the right?
anonymous
  • anonymous
Yes
TuringTest
  • TuringTest
|dw:1327883385609:dw|The easiest one to find is the field due the the second charge\[\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\overrightarrow{r}=<0,\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}>\]You can just plug the numbers in, and you know the angle will be 0 because the vector points in the +x direction. Got it so far?
anonymous
  • anonymous
I do understand that so far but it says to put it in vector component form. How would i do that? Like (Ex, Ey). I know Ey should be zero but how would you find Ex?
TuringTest
  • TuringTest
this is component form:\[<0,\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}>\]^^^ ^^^ Ex=0 Ey=that^
anonymous
  • anonymous
ohhh ok
TuringTest
  • TuringTest
sorry Ex and Ey are backwards above...
TuringTest
  • TuringTest
\[<\frac{1}{4\pi\epsilon_0}\frac{q}{r^2},0>\] Ex^^ ^^Ey=0 I just noted that it says that the point is directly to the right of q2, so since the sine of 0=0 the y-component disappears.
anonymous
  • anonymous
now how do you find the components q1?
TuringTest
  • TuringTest
let's put a grid on our point so we can do a little trigonometry:|dw:1327884328041:dw|The next one I want to find is the vector due to q3. You will see why soon.
anonymous
  • anonymous
ok
TuringTest
  • TuringTest
notice all the geometry we can do here, and all the places this angle theta turns up|dw:1327884443364:dw|we can find r with the pythagorean theorem, and the angle with the inverse tangent.
TuringTest
  • TuringTest
\[r=\sqrt{3^2+1}\]\[\theta=\tan^{-1}(\frac13)\]so we have\[\overrightarrow{E_3}=\frac{1}{4\pi\epsilon_0}\frac{q_3}{r^2}\overrightarrow{r}=<\frac{1}{4\pi\epsilon_0}\frac{q_3}{r^2}\cos\theta,\frac{1}{4\pi\epsilon_0}\frac{q_3}{r^2}\sin\theta>\]so this can be solved quickly as well now
TuringTest
  • TuringTest
^I suppose you have to put the centimeters in terms of meters above actually
TuringTest
  • TuringTest
\[r=\sqrt{0.03^2+0.01^2}\]\[\theta=\tan^{-1}(\frac13)\]
anonymous
  • anonymous
Ok, i understand all of that so far.
TuringTest
  • TuringTest
well if you have the radius and angle you can solve that guy, right? now E1 will be simple, because by symmetry it will have the same vector as E3, only the y-component will be negative\[\overrightarrow{E_1}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\overrightarrow{r}=<\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\cos\theta,-\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\sin\theta>\]where theta is the angle we used earlier. Same r as well.
TuringTest
  • TuringTest
|dw:1327885322988:dw|
anonymous
  • anonymous
Thanks for helping this the set up. I'll let you know if it doesn't work on the homework. (it can be finicky sometimes)
TuringTest
  • TuringTest
please do, good luck!
anonymous
  • anonymous
For some reason it keeps saying I'm wrong for my q2 answer. I wrote that the answer was (1,0). I'm pretty sure it's right but just want to check with you (the answer is in kiloNewtons by the way).
TuringTest
  • TuringTest
let me bust out the calculator...
anonymous
  • anonymous
thanks for doing that :).
TuringTest
  • TuringTest
wait, the answer is in V/m or N/c or something, not just N that's force
anonymous
  • anonymous
kN/c
TuringTest
  • TuringTest
I got about 9.9kN/c
anonymous
  • anonymous
Yep that worked. thanks, i now see what i did wrong.
TuringTest
  • TuringTest
Have you tried the others yet? if not be careful of the units is all I can say...
anonymous
  • anonymous
yes, they all worked. Thank you very much
TuringTest
  • TuringTest
awesome! glad to help :)
anonymous
  • anonymous
Your saving the world by helping one physics student at a time. lol

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