Electric field from two point charges at a given location

[fball558]

Electric field?

Homework Statement

A point charge of -4 uC is located at x = 2 m, y = -2 m (called A) . A second point charge of 12 uC is located at x = 1 m, y = 2 m. (called B)

(a) Find the magnitude and direction of the electric field at x = -1 m, y = 0. (called C)

(b) find the angle. I know i can find angle using arctan after i find the x and y components above. so this I am not too worried about.

Homework Equations

Enet = Enetx + Enety

kq / r^3 * R where R is the vector (equation given by TA)

The Attempt at a Solution

i started by finding the two vectors
rAC = -3,2 magnitued is sqrt(13)
rBC = -2,-2 magnitued is sqrt(8)
then i jumped into Enet = EAC + EBC
did kq / r^3 * R for point AC
k(-4e-6)/((sqrt(13))^3) * <-3,2> this gives an x and a y
then the same for BC
k(12e-6)/((sqrt(8))^3) * <-2,-2> this gives an x and a y

i added x components togeather then the y components and said that was by
Enet vector. then just found the magnitued.
my Enet vector was some crazy numbers of
<-7241.832,-18323.896>
that is not right lol
and this is where I am stuck
any help would be great!

 

[tms]

Check your arithmetic again, especially for the $$\vec j$$ component.

 

[Accipiter]

Yeah, I did the problem, and your equations are correct (but I got a different answer!). You probably mistyped on your calculator, or added the components incorrectly.

 

[fball558]

ok... here is what i did.
first i started with k(-4e-6)/((sqrt(13))^3) * <-3,2> equation
k = 9e9
so 9e9*-4e-6 = -36000
(sqrt(13)) ^ 3 = 46.8722
so -36000/ 46.8722 = -768.04643
then take that number and * by <-3,2> to get
<2304.1393,-1536.09285> <x,y> respectivly
do the same thing for k(12e-6)/((sqrt(8))^3) * <-2,-2>
9e9 * 12e-6 = 108000
(sqrt(8))^3 = 22.6274
so 108000/22.6274 = 4772.970773
then * <-2,-2> to get
<-9545.94155,-9545.94155> <x,y> respectivly
at the two x and the two y's and i get the same answer as posted??

can you guys see where i went wrong?
might be using the wrong k value?
let me know
thanks

well... i do get a different J component and it is -11082.0344 not what i had in the original post but my I component i get the same.
is this right??

 

[fball558]

never mind...
i got it

thanks everyone for the help!
just a stupid plug-in error lol

 

[LeNhatLinh]

1.a Cos a = 4/*65, E1 = 12/13k, E2 = 4/5k, use cos function theorem to find electric field!