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Two 2.0g spheres are attached to each end of a silk thread 1.20m long. The spheres are given identical charges and the midpoint...Asked by Farah
Two 2.0g spheres are attached to each end of a silk thread 1.20m long. The spheres are given identical charges and the midpoint of the thread is then suspended from a point on the ceiling. The spheres come to rest in equilibrium, with their centers 15cm apart. What is the magnitude of the charge on each sphere?
Ans: I transformed the words into diagrams, and using vector addition and cosine rule, i found the angle, though i found an angle of 166deg, then i found the force on the string which then i multiplied to get the electrical force and i used then coloumbs law to get q, however i get the answer of 5.529*10^-8 however the answer is incorrect. Where am i going wrong?
Ans: I transformed the words into diagrams, and using vector addition and cosine rule, i found the angle, though i found an angle of 166deg, then i found the force on the string which then i multiplied to get the electrical force and i used then coloumbs law to get q, however i get the answer of 5.529*10^-8 however the answer is incorrect. Where am i going wrong?
Answers
Answered by
bobpursley
Look at one half. Triangle is hypotenuse .6m, bottom .075m
So SinTheta= .075/.6
Now, force downward on each mass is mg, force horizontal is kQQ/.15^2
so TanTheta=kqq/(.15^2 * mg)
or Tan (arcsin.075/.6)=kqq/.15^2
solve for q
check my thinking.
So SinTheta= .075/.6
Now, force downward on each mass is mg, force horizontal is kQQ/.15^2
so TanTheta=kqq/(.15^2 * mg)
or Tan (arcsin.075/.6)=kqq/.15^2
solve for q
check my thinking.
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