Electric Field Strength Calculation for a Point 0.01m to the Left of a Charge

[WardenOfTheMint]

Please help...! I tried!

I even draw this picture. Please help!
http://img504.imageshack.us/img504/5137/myillustrationql5.jpg

Homework Statement

Determine the electric field strength at a point 0.01m to the left of the middle charge.

Homework Equations

Electric Field
E = Ke (q / (r^2) )

Electric Force
F = Ke (q1q2 / r^2)

Constant
Ke = 9.00e9 Nm^2/C^2

The Attempt at a Solution

http://img504.imageshack.us/img504/5137/myillustrationql5.jpg

Let 7e-6C be q1, 1.0e-6 q2, and -3e-6C q3.
So if you start at the middle charge and go 0.01m to the left, I imagine a point P, and the distance between the middle charge and point P should be 0.01m I think...

So from there...

E q1 on P = 9e9 Nm^2/C^2 ( 7e-6C / 0.03m^2 ) = 7e7 N / C.

E q2 on P = 9e9 Nm^2/C^2 ( 1e-6C / 0.01m^2 ) = 9e7 N / C.

E q3 on P = 9e9 Nm^2/C^2 ( -3e-6C / 0.03m^2 ) = -3e7 N / C. 0.03m because 0.01m+0.02m I think.

Enet = 7e7 + 9e7 - 3e7 = 1.3e8 N/C.

But that's wrong!

Then you have to find the magnitude (in Newtons) if a charge of -3e-6C is placed at this imaginary point.

 

[learningphysics]

Your signs are wrong... take right as positive, left as negative... what is the direction of the field due to the first charge... what is the direction of the field due to the second charge... and the third...

 

[WardenOfTheMint]

first thank learnphysc, you are always around!

i'm not getting it. E q1 on p should be positive shouldn't it? q1 is acting on point P forcing it to move right?

 

[learningphysics]

first thank learnphysc, you are always around!

i'm not getting it. E q1 on p should be positive shouldn't it? q1 is acting on point P forcing it to move right?

no prob. Yes, but what about q2 and q3?

 

[WardenOfTheMint]

oh yea. so E q2 on P should be negative because it's being to the left. ok i got it. does the negative charge push it also push it to the left?

 

[WardenOfTheMint]

so maybe the answer should be 7e7 N / C + ( -9e7 N / C) + ( -3e7 N / C) = -5e7N/C
i don't know the answer so i can only hope it's right

 

[learningphysics]

oh yea. so E q2 on P should be negative because it's being to the left. ok i got it. does the negative charge push it also push it to the left?

no the negative charge creates a field to the right... so it's positive...

If you're not sure of the direction of the field use a positive test charge in the place where you're trying to get the field... so the direction of the force acting on a positive test charge, is the direction of the field...

So the negative charge would attract a positive test charge towards the right... so it creates a positive field at that point...

 

[learningphysics]

so maybe the answer should be 7e7 N / C + ( -9e7 N / C) + ( -3e7 N / C) = -5e7N/C

should be +3e7 N/C for the negative charge.

 

[WardenOfTheMint]

so in electric field questions a positive charge will always "push" and the negative will do the "pulling'?

 

[WardenOfTheMint]

thanks! i got a confirmation it's right 7e7 N / C + ( -9e7 N / C) + ( 3e7 N / C) IS RIGHT! woohoo

 

[learningphysics]

so in electric field questions a positive charge will always "push" and the negative will do the "pulling'?

Yes, in electric field questions it is like that since the test charge is positive. but not in electric force though.

 

[learningphysics]

thanks! i got a confirmation it's right 7e7 N / C + ( -9e7 N / C) + ( 3e7 N / C) IS RIGHT! woohoo

Cool!

 

[BlackWyvern]

E = kq/d^2

Just add up all of the three vectors you get.