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|dw:1360530853167:dw|

\[\vec{F}_{net}=(\vec F_{41}-\vec F_{31}\cos(45))\hat i +(\vec F_{21}-\vec F_{31}\sin(45)) \hat j\]

|dw:1360531227825:dw|

\[q_1=3\times10^{-9}\]
\[q_2=9\times10^{-9}\]
\[q_3=-9\times10^{-9}\]
\[q_4=9\times10^{-9}\]

|dw:1360531777377:dw|

I get \[(72974.5 N)\hat i\]

which is wrong...

.000392 should be the answer

|dw:1360532089439:dw|

8 orders of magnitude is quite an error

it sure is ...haha

the charges were given in nano coulombs. I converted them to coulombs.
\[q_1=+3nC=3\times 10^{-9}\]

k=8.99E9

First, what units is the answer given in ?

Newtons \hat{i}

maybe I read the question wrong

your approach is ok.
I would write down the components of each vector

the k 9e9 times 9e-9 gives 81. that still leaves a 3e-9
are you double counting the k ?

|dw:1360532909843:dw|
oh my gosh ...yes I think soooo

I'll do it on wolfram... one sec

silly question...
\[F_{31}=F_{31x}+F_{31y}\]
is this true?

or is it
\[F_{31}=\sqrt{F_{31x}^2+F_{31y}^2}\]

why is
"the force from the bottom left is < -9*3/8 * 1/sqrt(2) i"
Why did you divide by 8

your 2.8 was just an approximation

Yep that's correct...I'm still trying to make sense of it...one second...

what happened to k

Oh I see

if we use 8.99e9 for k, we match the book's answer 0.000392

and \( 2 \sqrt{2} \) cm squared is written as 8e-4 m^2

finally!!!1

Good.