Coulombs law problems

1. Aug 24, 2008

roy_ament

I'm new in this forum, so i don't know if this is in the correct place..
1. The problem statement, all variables and given/known data
I got 3 charges
q1= 2.0x10^-6C
q2=8.0x10^-6C
Q=4.0x10^-6C
They're placed in the corners of a triangle like in this draw

I have to know the force that q1 and q2 act on Q, the magnitude, and the direction

2. Relevant equations

F=(kq1q2)/r2
for the angle which is 38.65 i do the tan-1 of .40/.50

3. The attempt at a solution
Fq1Q=( (9x109 Nm2/c2) (2.0x10-6 C) (4.0x10-6C))/(.50m)2

Fq1Q= .288N

So know i split in the components of X and Y right?
x= .288(cos38.65) Y= .288 (sin38.65)
x=.22N Y= .1798

Fq2Q=( (9x109 Nm2/c2) (8.0x10-6 C) (4.0x10-6C))/(.50m)2

Fq2Q= 1.152N

x=1.152 (cos 38.65) Y=1.152 (sin38.65)
x=.8999n y=.7119n
Now for the magnitude

x= .22+.899=1.119N
y=-.1798+.7119= .5321N

sqrt((1.119)2(.5321)2)= 1.311N
as for the direction
tan-1 = (.5321/1.119)= 25.43°

so.. am I right?
all sugetionts are really apreciated, tips and stuff.. hope anyone can help me

sry if my descriptions are wrong, I'm from mexico and my english is not perfect:shy:
if someone also could give me other exercises for practice... it will be perfect!

Last edited: Aug 24, 2008
2. Aug 24, 2008

LowlyPion

Not quite finished. Without doing all the math, it looks like you understand the steps involved. I will presume that the magnitude and angle are correct. But what direction is the angle off of?

3. Aug 24, 2008

LowlyPion

As to more problems you can look through this thread. At the bottom of this thread are some links to Coulombs Law problems. You can also search the thread for other topics of interest.

Good Luck.

4. Aug 24, 2008

roy_ament

thanks for your help!, im looking the other threads right know, trying to solve them

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