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sounds like physics bud

but i will help you anyway. you just need to find the NET electric field.

if you want me to help. . .

Do you know the formula for electric field?

is it E=F(R)/Q ??

\[E= kq/r^2\]

okay. then?

|dw:1340433733455:dw|

does the question say the side length of the square is 52.5cm?

okay, i'm going to assume the "side length" is 52.5cm

okay. i think it is the length.

|dw:1340433904871:dw|

okay.

hmm i think thats it.

you need the length from the corner to the charges. do you know how to find that?

sorry i dont know.

can you do that?

52.5 * sqrt of 2 ?

yes = ?

74.24

but why sqrt of 2 ?

because a square is in the form of x^2+y^2=2.. you can look into it later, and just trust me.

okay sure.

its 37.12

okay so now we can use our formula.

okay.

E=(kq)/r^2

our k is 9x10^9 ?

close enough, in physics we typically use 8.99 x 10^9

oh but can we use 9x10^9 ?

we have to calculate each individual electric field. sure, it is close enough.

okay. then our Q is what? our R is 37.12 ?

our q is the charge, there are 4 different charges. +45uC and -27uC, -27uC, -27uC

so we will use the four different charges? one by one?

we will do two separate computations as three of them will be the same (-27uC)

so lets get started

okay sure lets get started

fill this in with what we know: E=(kq)/r^2

or just do it on you're calculator and tell me the two answers

E=(9x10^9) (+45uC)/37.12 ?

close... E=(9x10^9*45E-6)/(37.12)^2
dont forget to square the denominator

and also remember the units for the charge is in micro coulombs

therefore 45E-6

why is the exponent is -6 ?

i just told you, because the units are in uC

that little "u" mean E-6

oh yeah

okay, so what your answer?

then the C means coulombs?

correct

0.00079

you used this?
E=(9x10^9*45E-6)/(37.12)^2

yup i did

well i'm afraid that is wrong, you must have typed it in wrong

http://www.wolframalpha.com/input/?i=%28%289x10%5E9%29%2845E-6%29%29%2F%2837.12%29%5E2

let me check

E means exponent right?

its -27uC

right, -27uC still the same answer.

i think you should be able to do it from here.. i've been helping you for over an hour.

okay thank you so much.