Two identical spheres are each attached to silk threads of length L= 0.500 m and

Question

Two identical spheres are each attached to silk threads of length L= 0.500 m and hung from a common point. Each sphere has mass m= 8.00 g. The radius of each sphere is very small compared to the distance between the spheres, so they may be treated as point charges. One sphere is given positive charge q1, and the other a different positive charge q2 ( q2 < q1 ); this causes the spheres to separate so that when the spheres are in equilibrium, each thread makes an angle = 20.0 degrees with the vertical.

A small wire is now connected between the spheres, allowing charge to be transferred from one sphere to the other until the two spheres have equal charges; the wire is then removed. Each thread now makes an angle of 30.0 degrees with the vertical. Determine the original charges q1 and q2. (Hint: The total charge on the pair of spheres is conserved.)   q1= Coulombs, q2 = Coulombs

Explanation / Answer

we need to do the second half first:

weight of ball = mg = .008 kg * 9.8 = .0784 N

This is also the y component of the tension in the string.

tan(30) = x / .0784

x = .0453 N

This is the also the force of electricity on the sphere. Lets use Coloumbs law F = kq1q2/r2

But we need to find the distance separating, r. We know the length L, so sin30 = x / .5

x = .25 m. But we need to double this distance for the total distance between the spheres, so the total is .5 m. This is the value of r.

F = kq1q2/r2

.0453 N = (9e9)(q)(q)/(.5 m)2

q = 1.12e-6 C

Now, back to the first half of the question:

according to conservation of charge, q1 + q2 = total q. The total q is double the q we solved for ( we only solved for one ball, the other is identical ) So total is 2.24e-6, and

q1 + q2 = 2.24e-6 C.

With this knowledge, we repeat similar math as we did before, but now the angle is 20

tan(20) = x / .0784

x = .0285 N

We need to find the distance separating, r. We know the length L, so sin20 = x / .5

x = .171 m. But we need to double this distance for the total distance between the spheres, so the total is .342 m. This is the value of r.

F = kq1q2/r2

.0285 N = (9e9)(q1)(q2)/(.342 m)2

3.7e-13 C = q1q2

Lets rewrite this, so we can solve with the other equation 3.7e-13 C / q2 = q1

q1 + q2 = 2.24e-6 C.

3.7e-13 C / q2 + q2 = 2.24e-6 C

3.7e-13 C + q22 = ( 2.24e-6 C )(q2)

This is quadratic, but when we solve it, we get q2 = 2.06e-6 C

Therefore, to get q1:

q1 + q2 = 2.24e-6 C

q1 + 2.06e-6 C = 2.24e-6 C

q1 = .18e-6 C or 1.8 e-7 C

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