Sphere A and B each have 1/3 the original charge.
The electrostatic force separates them.
Electrostaticforce/mg= tanTheta
but sinTheta= d/2 / length where d is the sphere separation. Using the small angle approximation
Force/mg= d/2*4
K(1/3 q)^2/d^2 = d/8
solve for d. q is given as the initial charge.
When d is found, you can find theta.
Hey, ok so I am supposed to be helping my cousin this with this physics problem, and this is the only one that has ever stumped me! Neutral metal spheres A and B each of mass 0.2 kg hang from insulating wires that are 4.0 m long and are initially touching. An identical metal sphere C, with a charge of -6.0 x 10^-6C is brought into contact with both spheres simultaneously and then removed. Spheres A and B then repel. What is the angle between the wires? Note>> use small angle approximation Tan theta is approximately equal to sin theta. I think I the answer is 9.6 or 9.4 degrees, roughly, but please help me!! Thanks!
2 answers
Ok good, I did something similar to you! I had slightly different values then the ones I gave in this question for certain purposes, but; I did this: tan theta =f horizontal (which is Electric force)/Fg , but Fg =mg
mgtantheta= Fhorizontal (tan theta = 1/2r over 4)
mg1/2r/4= Kq^2/r^2 >>>I have written down what 1/3 of the original charge is... so I have accounted for it)
and then finally r^3=4Kq^2/mg and then I solved and wound up getting the angle to one side, but because there are two small triangles in the big one, added the two small angles together to get a value near 17-18 degrees... before I added, the angle was roughly 9 degrees...
Does all of this sound correct? Thanks sooo much!
mgtantheta= Fhorizontal (tan theta = 1/2r over 4)
mg1/2r/4= Kq^2/r^2 >>>I have written down what 1/3 of the original charge is... so I have accounted for it)
and then finally r^3=4Kq^2/mg and then I solved and wound up getting the angle to one side, but because there are two small triangles in the big one, added the two small angles together to get a value near 17-18 degrees... before I added, the angle was roughly 9 degrees...
Does all of this sound correct? Thanks sooo much!