Asked by someone
Both threads are now lengthened such that L=1.00m, while the charges q1 and q2 remain unchanged. What new angle will each thread make with the vertical? Hint: you must solve for the angle numerically
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Answered by
bobpursley
I assume these are hanging from a "ceiling" from the same point.
The force of the charge keeps them apart, and part of gravity tends them together, and those forces at equilibrium have to be equal.
Frepulsion=kqq/d^2
force gravity=mg sinTheta
mgsinTheta=kqq/d^2
Now, looking at the string arrangement, sinTheta=(d/2)/1m
mg (d/2)=kqq/d^2
d^3=2kqq/mg
because it is a small angle,
sinTheta=Theta(inRadians)
then Theta= (d/2)=cubrt(2kqq/mg)
The force of the charge keeps them apart, and part of gravity tends them together, and those forces at equilibrium have to be equal.
Frepulsion=kqq/d^2
force gravity=mg sinTheta
mgsinTheta=kqq/d^2
Now, looking at the string arrangement, sinTheta=(d/2)/1m
mg (d/2)=kqq/d^2
d^3=2kqq/mg
because it is a small angle,
sinTheta=Theta(inRadians)
then Theta= (d/2)=cubrt(2kqq/mg)
Answered by
someone
so to get the answer I do the inverse sine of d?
Answered by
bobpursley
No, you compute cubrt(2kqq/mg) and that is the angle (in Radians)
Answered by
someone
oh ok thanks
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