Introductry Physics- Coloumbs Law problem

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SUMMARY

The discussion centers on solving a Coulomb's Law problem involving a small spherical insulator with a mass of 8.00x10^-2 kg and a charge of 6.0x10^-5 C, positioned 0.150 m away from a charge of -9.0x10^-5 C. Participants clarified the approach to finding the angle of the wire and the tension, emphasizing the importance of vector addition of the forces involved: the gravitational force (mg) and the electrostatic force (Fe). A common mistake identified was the incorrect conversion of microcoulombs to coulombs, which led to erroneous angle calculations.

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  • Coulomb's Law and its application in electrostatics
  • Basic principles of vector addition in physics
  • Trigonometry for solving right triangles
  • Understanding of mass, charge, and gravitational force
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moephysics
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Homework Statement


A small spherical insulator of mass 8.00x10^-2 Kg and charge 6.0x10^-5C is hung by a thin wire of negligible mass. A charge of -9.0x10^-5 is held 0.150 m away from the sphere and directly to the right of it, so the wire makes an angle with the vertical (see the drawing). Find (a) the angle and (b) the tension in the wire.

Homework Equations


Fe=Kq1q2\r^2

The Attempt at a Solution


Well as far as the tension is concerned I have no problem, I have mg and Fe and from adding them vectorialy I can figure out the tension. However when I tried to figure out the angle I tried a trigonometrical approach where I extended the line and made a big triangle and a small triangle so I can try to figure out the length L of the thin wire and one of the sides of the triangle so I can calculate the angle. However when I tried that approach I seem to be going through trouble with having less information than needed, so if someone can help give me a hint or something that could push me further, that's all and thank you (and also I have attached a picture with this thread if anyone wants the problem shown in a diagram)
 

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moephysics said:
Well as far as the tension is concerned I have no problem, I have mg and Fe and from adding them vectorialy I can figure out the tension.
Good.
However when I tried to figure out the angle I tried a trigonometrical approach where I extended the line and made a big triangle and a small triangle so I can try to figure out the length L of the thin wire and one of the sides of the triangle so I can calculate the angle.
Not sure what you're doing here, since the length of the wire is not needed or relevant. The tension will be parallel to the wire, but will have nothing to do with its length.

The two vectors you must add are the Fe (to the right) and mg (downward). Hint: They form the sides of the right triangle that you want.
 
ooh I see where I made my mistake, I actually tried out what you said before I asked you, however I was getting this weird answer of 90 degrees, which you would get if you tried solving this problem right now, kind of embarrassing but turns out I converted from micro coulombs to coulombs wrongly :shy:, but any how thanks you very much for the help. :smile:
 

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