Two identical spheres are attached to strings of length L = 0.500 m and hung as
ID: 2269688 • Letter: T
Question
Two identical spheres are attached to strings of length L = 0.500 m and hung as shown. The mass of each sphere is m = 8.00 g. They may be treated as point charges. The spheres are given charges q1 and q2 as shown. This causes the spheres to separate until they are in equilibrium. The strings make an angle ? = 20.0 degrees as shown.
(a) Draw a Free Body Diagram for each sphere when in equilibrium, and label all the forces that act on each sphere.
(b) Find the magnitude of the electrostatic force that acts on each sphere, and find the tension in each string.
(c) Based on the given information, what can be said about the magnitudes and the signs of q1 and q2? Explain your answers.
(d) A conductive wire is now connected between the spheres which allows charge to flow from one sphere to the other until the charge on each sphere is the same; the wire is then removed. Each string now makes an angle of 30.0 degrees with the vertical. Determine the original charges of q1 and q2.
Two identical spheres are attached to strings of length L = 0.500 m and hung as shown. The mass of each sphere is m = 8.00 g. They may be treated as point charges. The spheres are given charges q1 and q2 as shown. This causes the spheres to separate until they are in equilibrium. The strings make an angle ? = 20.0 degrees as shown. Draw a Free Body Diagram for each sphere when in equilibrium, and label all the forces that act on each sphere. Find the magnitude of the electrostatic force that acts on each sphere, and find the tension in each string. Based on the given information, what can be said about the magnitudes and the signs of q1 and q2? Explain your answers. A conductive wire is now connected between the spheres which allows charge to flow from one sphere to the other until the charge on each sphere is the same; the wire is then removed. Each string now makes an angle of 30.0 degrees with the vertical. Determine the original charges of q1 and q2.Explanation / Answer
Step One Part B
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Calculate the vertical force represented by F = mg
m = 8.00 grams
g = 9.81 kg m/s^2
m = 8.00 grams kg/1000 grams
m = 0.008 kg Note: there are 1000 grams in 1 kg.
F = mg
F = 0.008 * 9.81
F = 0.07846 N
Step Two
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Draw a diagram of the forces (Can you do this).
Draw a vertical line. Label it Fg = 0.07846.
Draw a horizontal line, not too long. Label it Fe ???
Draw a line oppose the right angle. Label it T or Tension.
The angle 16.8 degrees is between the hypotenuse and the vertical tension.
Step Three
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Find the tension.
T = ??
Cos(16.5) = 0.9588
Fg = 0.07846 N
Cos(16.5) = adj/hyp
Cos(16.5) = Fg/hyp
hyp = T = 0.07846/0.9588
T = .08183
Step 4
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Find Fe
Sin(16.5) = opp/T
Fe = T*sin(16.5)
Fe = 0.08183 * 0.2840
Fe = 0.02324 N
Step Five Part C
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You know what Fe is from step 4. Solve for q1 * q2. You won't know what either of the is, but you will know what the product is.
Fe = k* q1*q2/r^2
But before we can do this, we need to know what r is equal to.
Sin(16.5) = R/Thread Length
R = ??
Thread Length = 0.5 meters
R = (Thread Length) * Sin(16.5)
R = .1420 m
But R is half the distance between the two spheres. Refer to diagram again. The right angle triangle on the horizontal only represents 1/2 the distance between the two masses.
r = 2*R = 2*0.1420 = 0.2840
Fe = k q1*q2/r^2
Fe = 0.02342 N
k = 9* 10^9
r = 0.2880 m
x = q1*q2 = ???
0.02342 = 9 * 10^9 * x /(0.2880)^2
0.02342 * (0.2880)^2 / 9 * 10^9 = x
2.1584 * 10^-13 = x
2.1584 * 10^-13 = q1*q2
Step 6 Part D
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Find out q1 = q2 = q
Fe = Fg*tan(34.3) How did this happen? Draw your diagram again to see how this happens.
Fe = 0.07847 * tan(34.3) Fg did not change.
Fe = 05369 N
Next
Just repeat the steps above to find r
r = 0.5635 m
Fe = kq^2/r^2
r= .5635
k = 9 * 10^9
Fe = 0.05369 N
q^2 = 0.5635 m^2 How come it's q^2?
0.05396 * 0.5636^2 / 9 * 10^9 = q^2
1.904 * 10^-12 = q^2
q = 1.379 ^ 10^ - 6 c
Step 7
=====
Now the tricky part.
We left step 5
2.1584 * 10^-13 = q1*q2
You have to realize from step 6 that a charge has been transferred from the higher to lower charge so both wind up with the same charge.
Let the amount transferred = x
q1 (the larger) = 1.379 ^ 10^ - 6 c + x
q2 (the smaller) = 1.379 ^ 10^ - 6 c - x
Multiply them together
q1*q2 = 1.904 * 10^-12 - x^2 But q1*q2 has a value
2.1584 * 10^-13 = 1.904 * 10^-12 - x^2 Now solve for x. I'll leave it for you, but I'll tell you that x = 1.299 * 10^6
Now go back and substitute this into q1 and q2
q1 = 1.379 ^ 10^ - 6 c - 1.299 * 10^-6
q1 = 8* 10^-7 c
q2 = 2.678 * 10^-6 c
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