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Mindscrape
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Two 3.0 g spheres on 1.0-m-long threads repel each other after being equally charged. What is the charge of q?
I don't know why I can't seem to get this right, but I am missing something somewhere. So, I figured I could find use the tension in the 1m string to find the horizontal component of the force, which would then equal the force of the repulsion between the charges. Which means that [tex]T_x = m g tan \theta[/tex]. Also using trig to find the x leg yields .324m. If [tex]F = \frac{kq^2}{r^2}[/tex] then q should be [tex]q = \sqrt{\frac{mgtan\theta 2x^2}{k}}[/tex], and plugging in the numbers would be [tex]q = \sqrt{\frac{.003kg*9.8\frac{m}{s^2}*tan(20)*.684^2m}{(9*10^9\frac{Nm^2}{C^2})}}[/tex]. But that gives that q=6.17*10^13C, which would be ripping atoms in air apart. Anyone chatch were I went wrong?
I don't know why I can't seem to get this right, but I am missing something somewhere. So, I figured I could find use the tension in the 1m string to find the horizontal component of the force, which would then equal the force of the repulsion between the charges. Which means that [tex]T_x = m g tan \theta[/tex]. Also using trig to find the x leg yields .324m. If [tex]F = \frac{kq^2}{r^2}[/tex] then q should be [tex]q = \sqrt{\frac{mgtan\theta 2x^2}{k}}[/tex], and plugging in the numbers would be [tex]q = \sqrt{\frac{.003kg*9.8\frac{m}{s^2}*tan(20)*.684^2m}{(9*10^9\frac{Nm^2}{C^2})}}[/tex]. But that gives that q=6.17*10^13C, which would be ripping atoms in air apart. Anyone chatch were I went wrong?
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