Electric charge problems

1. Jan 17, 2005

Silimay

I need a bit of help with two problems. I've just started using latex, so bear with me if the equations are a bit funky.

The charges and coordinates of two charged particles held fized in the xy plane are $$q_1$$ = +3.0 * 10^(-6) C and (0.035, 0.005) and $$q_2$$ = -4.0 * 10^(-6) C and (-0.02, 0.015). All coordinates are given in meters. A.) Find the magnitude and direction of the electrostatic force on $$q_2$$.

I used the distance formula to find the distance between the two coordinates and came up with d=0.065 m. Then I used Coulomb's law:

$$F = \frac{q_1 q_2 }{4 \pi e d^2}$$

I calculated this and got F = 25 N. I knew that the force was attractive, since the two particles are opposite in sign.

B. Where could you locate a third charge $$q_3$$ = +4.0 * 10^(-6) C such that the net electrostatic force on $$q_2$$ is zero?

My work:

$$\frac{(4 * 10^-6 C)^2}{4 \pi e d_2^2} = 25 N$$

$$d_2 = 0.075 m$$

I wasn't quite sure about how to proceed. At any rate I found the slope:

$$m = \frac{y_2 - y_1}{x_2 - x_1} = -0.18$$

Next I used the equation of a line:

$$y - 0.005 = -0.18(x - 0.035)$$

$$y = -0.18(x) + 0.011$$

And then the distance formula again:

$$d_2^2 = 0.0056 = (-0.02 - x)^2 + (0.015 - y)^2$$

$$0.0056 = (-0.02 - x)^2 + (0.015 - 0.011 + 0.18x)^2$$

$$0.005184 = 1.0324 x^2 + 0.04144 x$$

I tried using that one formula at this point (I don't remember what it's called...the one to solve equations involving Ax^2 + Bx + C = 0) and came up with complex numbers for answers! What did I do wrong? I know I probably messed up the actual numbers some...I only was writing 2 significant figures in the beginning in my calcs, and I was using 4 by the end

~Silimay~

Last edited by a moderator: Jan 17, 2005
2. Jan 17, 2005

stunner5000pt

its simpler than you think it is

first of all why did you use the slope of a line???!!

secondlythe charget o be placed in positive yes? So dont you think there would be a point in between(or to the side) of the charges that the negative charge would attract with the ssame for the postiive repels with??

then you'll have a diagram like this-

|-------------x--------------------|---------x-0.065----------------|
the above system is like so becuase the x must go away when you add x and x-0.065.
-4microC ----------------------New charge------------------+3microC
to calculate the point where the force is zero

Force (from 4 microC on new charge) = Force (from +3microC on new charge) because push and pull is equal here thus no NET force.
now that you found the distance draw a trangle and try and see how this distnace you computed above. What does the distance represent?

your triangle should be right angle triangle (for simplciity) Try and figure out how you plug the co-ordinates in for the vertices of the right angle triangle and THEN use the distance formula to compute the point (assume it is x,y and then solve)

3. Jan 17, 2005

vincentchan

the above post is absolutely definitely positively deadly unquestoinably don'tASKmeWAYly wrong, ignore it.. silimay, your method is right.. check your algebra and arithimetics...

4. Jan 17, 2005

vincentchan

i don't see any reason you will come up with a complex number.. it is real.. rearrange your equation like this

$$0= 1.0324 x^2 + 0.04144 x -0.005184$$

and do it again,

5. Jan 17, 2005

stunner5000pt

i'm going to ask anyway

WHYDOESNTITWORK!

6. Jan 17, 2005

vincentchan

read the question part B again.. it was asking you how could make the electric force on q2 equal to zero.. what you was doing is making the force on q3 equal to zero.. I am pretty sure you know how to do this problem.. you just didn't read it carefully....