What is the mass of the charged balls in a magnetic field with given parameters?

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SUMMARY

The discussion focuses on calculating the mass of two charged balls in a uniform electric field of 100,000 N/C. The balls have charges of 100 nC and -100 nC, and the setup involves a string length of 50 cm and an angle of 10 degrees with the vertical. The initial calculation yielded a mass of 7.5 grams, but the correct answer is 4.1 grams, highlighting the importance of recognizing the forces at play, including the attractive forces between the charges and their respective potentials.

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  • Understanding of electric fields, specifically uniform electric fields.
  • Familiarity with Coulomb's law and the equation F = k|q1||q2|/r^2.
  • Knowledge of force balance equations in physics, particularly F = Eq and F = mg.
  • Basic trigonometry to resolve forces into components.
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  • Study Coulomb's law in detail, focusing on the interaction between different charges.
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2 identical small balls hang in a 100, 000 N/C electric field as shown. They have charges of 100 and -100 nC respectively. The string length is 50cm and the angle each makes with the vertical is 10 degrees. Find the mass of the balls.

http://img514.imageshack.us/img514/633/phees.th.jpg


Homework Equations



F = Eq , F = k|q1||q2|/r^2, F = mg
k = 9 x 10^9 Nm^2/C^2

The Attempt at a Solution



For the left ball :

Fx = Tsina - Eq - kq^2/(sina^2) = 0
T = (Eq + kq^2/sina^2)/ sina

Fy = Tcosa - mg = 0
m = Tcosa / g

m =(cosa/g) * (Eq + kq^2/sina^2) / sina

but when i evaluate this with a = 10 , g = 9.8, E = 100 000, k = 9x10^9, q = 1 x 10^7
i get m = 7.5 grams

the answer in the back of the book is 4.1 grams

where am i going wrong?
 
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First of all, its not a magnetic field. Looks like a uniform E-field of 100,000 V/m.

And it makes a difference.

Left to themselves the + and - charges on the balls would be attractive, not repulsive. Hence the m*g AND the attractive forces between the balls are being balanced by the attraction of the balls toward their respective potentials.
 
ohhhhhhhh that's right , thank you!
 

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