What are the magnitude and direction of the electric field produced at P by the three charges in the figure below?

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What are the magnitude and direction of the electric field produced at P by the three charges in the figure below?

Physics
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|dw:1443486683805:dw|
\[E=\frac{ kQ }{ r^2 }\] where E is electric field k is Coulomb constant Q is charge
\[E=\frac{ (8.9*10^9)Q }{ r^2 }\] how exactly do you apply the charges? I actually have no much idea

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Other answers:

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