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[SOLVED] A few electricity problems
1. Two 1.0 g spheres are charged equally and placed 2.0 cm apart. When released, they begin to accelerate at 225m/s^2.
What is the magnitude on each sphere.
2. You have a lightweight spring whose unstretched length is 4.0 cm. You're curious to see if you can use this spring to measure charge. First, you attach one end of the spring to the ceiling and hang a 1.0 g mass from it. This stretches the spring to a length of 5.0 cm. You then attach two small plastic beads to the opposite ends of the spring, lay the spring on a frictionless table, and give each plastic bead the same charge. This stretches the spring to a length of 4.5 cm.
What is the magnitude of the charge (in nC) on each bead?
1. F=ma => (K)(q^2)/(r^2)
2. F = kx, (K)(q^2)/(r^2)
1. q = [tex]\sqrt{(r^2)(m)(a)/(r^2)}[/tex]
q = [tex]\sqrt{(0.0001)(225)(0.002^2)/(9x10^9)}[/tex] = 3 x 10 ^-9 C
but it says this is incorrect
2.
First: I calculated the spring constant using the information from the hanging mass part of the question.
F = kx
(0.001 kg)(9.81 m/s^2)= k (0.01 m)
0.981 kg/s^2 = k
Second: Using the spring constant from the previous part I calculated the force needed to push the spring apart when the beads are on either end.
F = kx
F = (0.981 kg/s^2)(0.005 m)
F = 0.004905 kgm/s^2
Third: Using the force from above, and using coulombs law calculate the charges. I know that q1 and q2 are equal, q1 =q2 = q
F = K((q1)(q2))/r^2
0.004905 kgm/s^2 = ((9×10^9 Nm^2/C^2)(q^2)/(0.045 m)^2
q = 3.3 x 10 ^ -8 and now I need to convert it to nC so (3.3 x 10 ^ -8) x (10x9) = 33 nC
Is this correct?
Homework Statement
1. Two 1.0 g spheres are charged equally and placed 2.0 cm apart. When released, they begin to accelerate at 225m/s^2.
What is the magnitude on each sphere.
2. You have a lightweight spring whose unstretched length is 4.0 cm. You're curious to see if you can use this spring to measure charge. First, you attach one end of the spring to the ceiling and hang a 1.0 g mass from it. This stretches the spring to a length of 5.0 cm. You then attach two small plastic beads to the opposite ends of the spring, lay the spring on a frictionless table, and give each plastic bead the same charge. This stretches the spring to a length of 4.5 cm.
What is the magnitude of the charge (in nC) on each bead?
Homework Equations
1. F=ma => (K)(q^2)/(r^2)
2. F = kx, (K)(q^2)/(r^2)
The Attempt at a Solution
1. q = [tex]\sqrt{(r^2)(m)(a)/(r^2)}[/tex]
q = [tex]\sqrt{(0.0001)(225)(0.002^2)/(9x10^9)}[/tex] = 3 x 10 ^-9 C
but it says this is incorrect
2.
First: I calculated the spring constant using the information from the hanging mass part of the question.
F = kx
(0.001 kg)(9.81 m/s^2)= k (0.01 m)
0.981 kg/s^2 = k
Second: Using the spring constant from the previous part I calculated the force needed to push the spring apart when the beads are on either end.
F = kx
F = (0.981 kg/s^2)(0.005 m)
F = 0.004905 kgm/s^2
Third: Using the force from above, and using coulombs law calculate the charges. I know that q1 and q2 are equal, q1 =q2 = q
F = K((q1)(q2))/r^2
0.004905 kgm/s^2 = ((9×10^9 Nm^2/C^2)(q^2)/(0.045 m)^2
q = 3.3 x 10 ^ -8 and now I need to convert it to nC so (3.3 x 10 ^ -8) x (10x9) = 33 nC
Is this correct?