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

- anonymous

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- anonymous

|dw:1443486683805:dw|

- anonymous

\[E=\frac{ kQ }{ r^2 }\]
where
E is electric field
k is Coulomb constant
Q is charge

- 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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## More answers

- anonymous

Give me a minute.
Be back.

- anonymous

ok!

- anonymous

|dw:1443487357841:dw|

- 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

Understood that?

- anonymous

wait how is it known that P is positive?

- anonymous

P is an electric field.
The electric field relative to the charges.

- anonymous

Alright, about the electric field |dw:1443487887846:dw| is it something like that?

- anonymous

|dw:1443488158566:dw|

- anonymous

Yes

- anonymous

It moves in opposite direction from the charges.
So if there is angle it move diagonally

- 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

Opposite charge attract
Same charges repel.

- anonymous

Why not on that side? |dw:1443488417223:dw|

- anonymous

Oh yeah, I forgot.

- anonymous

But thats not important in this problem.
Lets label the electric field vectors now. E1, E2, E3
|dw:1443488495113:dw|

- anonymous

Alright.

- anonymous

First convert the cm to meters.
Then use
\[E _{1}=\frac{ kQ }{ r^2 }\]

- anonymous

Solve for electric field one first.

- anonymous

|dw:1443488673433:dw|

- 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

Yes, right units of meters and charges

- anonymous

Remember, that electric field move away from the positive charge?

- anonymous

Which charge is E1 referring to?

- anonymous

the 5nC right?

- anonymous

|dw:1443488995264:dw|

- anonymous

Yes.

- anonymous

And what is the distance from P to 5.0nC?

- anonymous

\[E=\frac{ (8.9*10^9)(5.0*10^{-9}) }{ r^2 }\] what is r? distance between P and 5nC?

- anonymous

or rather, the radius so 0.02m, correct?

- anonymous

\[E=\frac{ (8.9*10^9 )(5.0*10^{-9})}{(0.02)^2 }\]\[E=111250\] seems odd

- anonymous

Where did you get 0.02m?

- anonymous

|dw:1443489283742:dw|

- anonymous

|dw:1443489266073:dw|

- anonymous

isn't r = radius? 0.04m = diameter

- anonymous

|dw:1443489310614:dw|
r is just distance

- anonymous

oh... physics..

- anonymous

\[E=\frac{ (8.9*10^9)(5.0*10^{-9}) }{ (0.04)^2 }\]

- anonymous

No circles involved :)

- anonymous

Do the same for E2

- anonymous

\[E_1= 27812.5\]

- anonymous

Show me the formula.

- anonymous

\[E_2= 988.9\]

- anonymous

aww

- anonymous

\[E_2=\frac{ (8.9*10^9)(1.0*10^{-10}) }{ (0.03)^2 }\]

- anonymous

Correct.
Sorry I was lagging.

- anonymous

Now do E3, which is the hardest one.

- anonymous

|dw:1443489911056:dw|

- 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

Right :D?

- anonymous

Yes, nice job.

- 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

uh?

- anonymous

x= -251.7 ; y=-251.7

- anonymous

|dw:1443490394186:dw|

- anonymous

oh ok, this is to find it's magnitude?

- anonymous

or you can call it \[E _{3x}\]

- anonymous

Give me a second to write this.

- anonymous

|dw:1443490526909:dw|
|dw:1443490697051:dw|

- anonymous

oooooo I see

- anonymous

Add all the x and y component together
\[x^2+y^2=r^2\]\
R is the resultant field or magnitude.

- anonymous

Direction is \[\tan^{-1}( \frac{ y }{ x })\]

- anonymous

\[E_1= 27812.5\]\[E_2=988.9\]\[E_3=-356\]

- anonymous

\[x=-251.7 , y=-251.7\]

- anonymous

Wait so you said E_1 and E_2 are -x and -y?

- anonymous

E1x=-27812.5
E1y=0
E2x=0
E2y=-988.9
E3x=251.7
E3Y=251.7

- anonymous

ah ok I see

- 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

\[r=\sqrt{(251.7-27812.5)^2+(251.7-988.9)^2}\]

- anonymous

\[r=27570.65\]

- anonymous

Yes

- anonymous

Now find direction.

- anonymous

direction?

- anonymous

oh

- anonymous

Direction is the angle in which it moves.
Read what i wrote several post above.

- anonymous

well since it's magnitude its positive then the direction should be a positive type of slope right? |dw:1443491606328:dw|

- 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

I got to go.
Go find your direction.

- anonymous

y is the resultant y
and x is the resultant x
Same as the one you would use to find the magnitude.

- anonymous

|dw:1443491766486:dw|

- anonymous

I'll be back in like 75 minutes.
I can check your calculations when I get back on.

- anonymous

ok :[

- anonymous

What are you doing?
Use the formula.

- anonymous

Read every post I post.
I am heading off for sure now.

- anonymous

\[\tan^{-1}(\frac{ y }{ x }) = \tan^{-1}(\frac{ -737.2 }{ -27560.3 }) = 0.027\]

- anonymous

Your calculator should be radian mode and not degree.

- anonymous

its in radian

- 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

oh....

- 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

My calculator is in radians and I get 0.027, and in degrees 1.53

- 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

Redo your calculator for E2 and E3 since you wrote the C incorrectly.

- anonymous

calculations*

- anonymous

Alright but once I get the answer, how can I tell the direction by just the number?

- anonymous

1.53 means 1.53 degrees which is like this direction.
|dw:1443499160657:dw|

- anonymous

Direction is just angle.
No need to write northwest, southwest, etc.
But you can if you want.

- anonymous

oh ok, I got it now. Somehow you made me memorized all the steps on my head (which is amazing) thanks you.

- 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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