# Homework Help: Coulombs law problem

1. Jan 20, 2005

### radtad

Two charges, q1 and q2, are held a fixed distance d apart. a. Find the strength of the electric force that acts on q1. Assume that q1=q2=21.3 micro COulombs and d=1.52 m. b. A third charge q3=21.3 micro COulombs is brought in and placed as shown. Find the stregnth of the electric force on q1 now.
My best attempt at showing the diagrams:
a.
q1----------q2

b.
q1----------q2
__-----------
___--------
____-----
_____q3
b is an equilateral triangle if you cant figure out by the diagram with all sides a distance d. edit: i see the my attemtp at diagrams isnt working too well buts the general idea

I got part a with Fe=1.77 N
I don't know what to do for part b other than its some application of Fe=kq1q2/r^2 . Thanks for the help

Last edited: Jan 20, 2005
2. Jan 20, 2005

### MathStudent

Part a looks alright
For part B
Hint:
The net force acting on q1 is the vector sum of the individual forces caused by the other two charges.

3. Jan 20, 2005

### radtad

thanks i think i got it
3.07 N sound about right

4. Jan 20, 2005

### MathStudent

yep i got same ting

5. Jan 20, 2005

### DB

I have a simple question: how do you convert Micro Coulombs to coulombs? (I couldn't find it on the net)
Once I do that I use

$$\frac{\mid q_1q_2\mid}{4(1.52^2)*8.854*10^{-12}*\pi}$$

right?

P.S you did say that q1*q2 was 21.3 mC right?

6. Jan 20, 2005

### MathStudent

micro => 10^-6 (has the greek symbo "mu" )
=> 1 micro Coulomb = 10^-6 Coulombs

Last edited: Jan 20, 2005
7. Jan 20, 2005

### DB

Thanks, so 21.3 mC = 2.13e-5 C
But what have I done wrong here?

$$F=\frac{\mid 2.13*10^{-5}\mid}{4(1.52^2)*8.854*10^{-12}*\pi}$$

$$F=\frac{\mid 2.13*10^{-5}\mid}{2.57061216 * 10^{-10}}$$

It doesnt come to 1.77 N lol

8. Jan 20, 2005

### MathStudent

look at the equation of coulombs law that you posted... it says q1*q2 but in this case q1=q2... so it should be (2.13E-5)^2
you forgot to include the second charge

9. Jan 20, 2005

### DB

Awsome, my carelessness once again...

$$F=\frac{\mid 4.5369*10^{-10}\mid}{2.57061216 * 10^{-10}}\approx1.77 N$$

Thanks

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