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.(adsbygoogle = window.adsbygoogle || []).push({});

The charges and coordinates of two charged particles held fized in the xy plane are [tex]q_1[/tex] = +3.0 * 10^(-6) C and (0.035, 0.005) and [tex]q_2[/tex] = -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 [tex]q_2[/tex].

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:

[tex] F = \frac{q_1 q_2 }{4 \pi e d^2} [/tex]

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 [tex]q_3[/tex] = +4.0 * 10^(-6) C such that the net electrostatic force on [tex]q_2[/tex] is zero?

My work:

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

[tex] d_2 = 0.075 m [/tex]

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

[tex]m = \frac{y_2 - y_1}{x_2 - x_1} = -0.18[/tex]

Next I used the equation of a line:

[tex] y - 0.005 = -0.18(x - 0.035) [/tex]

[tex] y = -0.18(x) + 0.011 [/tex]

And then the distance formula again:

[tex] d_2^2 = 0.0056 = (-0.02 - x)^2 + (0.015 - y)^2 [/tex]

[tex] 0.0056 = (-0.02 - x)^2 + (0.015 - 0.011 + 0.18x)^2 [/tex]

[tex] 0.005184 = 1.0324 x^2 + 0.04144 x [/tex]

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~

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# Homework Help: Electric charge problems

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