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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

Give me a minute.
Be back.

ok!

|dw:1443487357841: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

Opposite charge attract
Same charges repel.

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

Oh yeah, I forgot.

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|

Right :D?

Yes, nice job.

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

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

\[r=27570.65\]

Yes

Now find direction.

direction?

oh

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

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....

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

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.