anonymous
  • anonymous
What are the magnitude and direction of the electric field produced at P by the three charges in the figure below?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1443486683805:dw|
anonymous
  • anonymous
\[E=\frac{ kQ }{ r^2 }\] where E is electric field k is Coulomb constant Q is charge
anonymous
  • anonymous
\[E=\frac{ (8.9*10^9)Q }{ r^2 }\] how exactly do you apply the charges? I actually have no much idea

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anonymous
  • anonymous
Give me a minute. Be back.
anonymous
  • anonymous
ok!
anonymous
  • anonymous
|dw:1443487357841:dw|
anonymous
  • anonymous
Had to switch computers earlier Electric fields goes away from positive charge in the opposite direction. (repel) Electric fields goes toward negative charge in the same direction. (attract) So your diagram of the electric field would be |dw:1443487695840:dw|
anonymous
  • anonymous
Understood that?
anonymous
  • anonymous
wait how is it known that P is positive?
anonymous
  • anonymous
P is an electric field. The electric field relative to the charges.
anonymous
  • anonymous
Alright, about the electric field |dw:1443487887846:dw| is it something like that?
anonymous
  • anonymous
|dw:1443488158566:dw|
anonymous
  • anonymous
Yes
anonymous
  • anonymous
It moves in opposite direction from the charges. So if there is angle it move diagonally
anonymous
  • anonymous
Dont worry about the movement of the charges. The question only concerns with the field not the charges. But the charges should be like this if you want to know |dw:1443488272987:dw|
anonymous
  • anonymous
Opposite charge attract Same charges repel.
anonymous
  • anonymous
Why not on that side? |dw:1443488417223:dw|
anonymous
  • anonymous
Oh yeah, I forgot.
anonymous
  • anonymous
But thats not important in this problem. Lets label the electric field vectors now. E1, E2, E3 |dw:1443488495113:dw|
anonymous
  • anonymous
Alright.
anonymous
  • anonymous
First convert the cm to meters. Then use \[E _{1}=\frac{ kQ }{ r^2 }\]
anonymous
  • anonymous
Solve for electric field one first.
anonymous
  • anonymous
|dw:1443488673433:dw|
anonymous
  • anonymous
\[E=\frac{ kQ }{ r^2 }\] \[E=\frac{ (8.9*10^9)Q }{r^2 }\] I am unsure what to plug for Q and r
anonymous
  • anonymous
Yes, right units of meters and charges
anonymous
  • anonymous
Remember, that electric field move away from the positive charge?
anonymous
  • anonymous
Which charge is E1 referring to?
anonymous
  • anonymous
the 5nC right?
anonymous
  • anonymous
|dw:1443488995264:dw|
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
And what is the distance from P to 5.0nC?
anonymous
  • anonymous
\[E=\frac{ (8.9*10^9)(5.0*10^{-9}) }{ r^2 }\] what is r? distance between P and 5nC?
anonymous
  • anonymous
or rather, the radius so 0.02m, correct?
anonymous
  • anonymous
\[E=\frac{ (8.9*10^9 )(5.0*10^{-9})}{(0.02)^2 }\]\[E=111250\] seems odd
anonymous
  • anonymous
Where did you get 0.02m?
anonymous
  • anonymous
|dw:1443489283742:dw|
anonymous
  • anonymous
|dw:1443489266073:dw|
anonymous
  • anonymous
isn't r = radius? 0.04m = diameter
anonymous
  • anonymous
|dw:1443489310614:dw| r is just distance
anonymous
  • anonymous
oh... physics..
anonymous
  • anonymous
\[E=\frac{ (8.9*10^9)(5.0*10^{-9}) }{ (0.04)^2 }\]
anonymous
  • anonymous
No circles involved :)
anonymous
  • anonymous
Do the same for E2
anonymous
  • anonymous
\[E_1= 27812.5\]
anonymous
  • anonymous
Show me the formula.
anonymous
  • anonymous
\[E_2= 988.9\]
anonymous
  • anonymous
aww
anonymous
  • anonymous
\[E_2=\frac{ (8.9*10^9)(1.0*10^{-10}) }{ (0.03)^2 }\]
anonymous
  • anonymous
Correct. Sorry I was lagging.
anonymous
  • anonymous
Now do E3, which is the hardest one.
anonymous
  • anonymous
|dw:1443489911056:dw|
anonymous
  • anonymous
|dw:1443489917026:dw|\[(0.04m)^2 + (0.03m)^2 = c^2\]\[c=0.05m\]\[E_3=\frac{ (8.9*10^9)(-1.0*10^{-10}) }{ (0.05)^2 }\]\[E_3=-356\]
anonymous
  • anonymous
Right :D?
anonymous
  • anonymous
Yes, nice job.
anonymous
  • anonymous
Since it is diagonal and makes an angle of 45 degrees |dw:1443490189435:dw| E3cos45 to get x component. E3sin45 to get y component
anonymous
  • anonymous
uh?
anonymous
  • anonymous
x= -251.7 ; y=-251.7
anonymous
  • anonymous
|dw:1443490394186:dw|
anonymous
  • anonymous
oh ok, this is to find it's magnitude?
anonymous
  • anonymous
or you can call it \[E _{3x}\]
anonymous
  • anonymous
Give me a second to write this.
anonymous
  • anonymous
|dw:1443490526909:dw| |dw:1443490697051:dw|
anonymous
  • anonymous
oooooo I see
anonymous
  • anonymous
Add all the x and y component together \[x^2+y^2=r^2\]\ R is the resultant field or magnitude.
anonymous
  • anonymous
Direction is \[\tan^{-1}( \frac{ y }{ x })\]
anonymous
  • anonymous
\[E_1= 27812.5\]\[E_2=988.9\]\[E_3=-356\]
anonymous
  • anonymous
\[x=-251.7 , y=-251.7\]
anonymous
  • anonymous
Wait so you said E_1 and E_2 are -x and -y?
anonymous
  • anonymous
E1x=-27812.5 E1y=0 E2x=0 E2y=-988.9 E3x=251.7 E3Y=251.7
anonymous
  • anonymous
ah ok I see
anonymous
  • anonymous
If it goes left or down it is negative, just like the coordinate system. Up or right is positive. Refer tot eh diagram for the direction of the field. brb
anonymous
  • anonymous
\[r=\sqrt{(251.7-27812.5)^2+(251.7-988.9)^2}\]
anonymous
  • anonymous
\[r=27570.65\]
anonymous
  • anonymous
Yes
anonymous
  • anonymous
Now find direction.
anonymous
  • anonymous
direction?
anonymous
  • anonymous
oh
anonymous
  • anonymous
Direction is the angle in which it moves. Read what i wrote several post above.
anonymous
  • anonymous
well since it's magnitude its positive then the direction should be a positive type of slope right? |dw:1443491606328:dw|
anonymous
  • anonymous
No, you find the direction to know whether its positive or negative. It looks more negative to me. Find the direction first It can have a magnitude of 100000 and moves down diagonally (negative slope)
anonymous
  • anonymous
I got to go. Go find your direction.
anonymous
  • anonymous
y is the resultant y and x is the resultant x Same as the one you would use to find the magnitude.
anonymous
  • anonymous
|dw:1443491766486:dw|
anonymous
  • anonymous
I'll be back in like 75 minutes. I can check your calculations when I get back on.
anonymous
  • anonymous
ok :[
anonymous
  • anonymous
What are you doing? Use the formula.
anonymous
  • anonymous
Read every post I post. I am heading off for sure now.
anonymous
  • anonymous
\[\tan^{-1}(\frac{ y }{ x }) = \tan^{-1}(\frac{ -737.2 }{ -27560.3 }) = 0.027\]
anonymous
  • anonymous
Your calculator should be radian mode and not degree.
anonymous
  • anonymous
its in radian
anonymous
  • anonymous
Sorry I did not catch the mistake you made earlier. 10nC is not 1.0 x 10^-10 C It is 1.0 x 10^-8 C
anonymous
  • anonymous
oh....
anonymous
  • anonymous
For the arctan it should not be in radians, not degree. When I used it on degree mode I got your answer 0.027
anonymous
  • anonymous
My calculator is in radians and I get 0.027, and in degrees 1.53
anonymous
  • anonymous
Strange. Well I am using the online right now. There might be a bug. But just use the one that doesnt have a 0.0XX something
anonymous
  • anonymous
Redo your calculator for E2 and E3 since you wrote the C incorrectly.
anonymous
  • anonymous
calculations*
anonymous
  • anonymous
Alright but once I get the answer, how can I tell the direction by just the number?
anonymous
  • anonymous
1.53 means 1.53 degrees which is like this direction. |dw:1443499160657:dw|
anonymous
  • anonymous
Direction is just angle. No need to write northwest, southwest, etc. But you can if you want.
anonymous
  • anonymous
oh ok, I got it now. Somehow you made me memorized all the steps on my head (which is amazing) thanks you.
anonymous
  • anonymous
Lol, haha. Yep, it is pretty easy to memorize if you do it. It is long but fun! Good luck with the rest!

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