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

Answer 1

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)

Answer 2

so to get the answer I do the inverse sine of d?

Answer 3

No, you compute cubrt(2kqq/mg) and that is the angle (in Radians)

Answer 4

oh ok thanks