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A few electricity problems

by cse63146
Tags: electricity, solved
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cse63146
#1
Jan17-08, 11:20 AM
P: 452
1. The problem statement, all variables and given/known data

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?

2. Relevant equations

1. F=ma => (K)(q^2)/(r^2)

2. F = kx, (K)(q^2)/(r^2)

3. 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 = ((910^9 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?
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berkeman
#2
Jan17-08, 11:58 AM
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P: 40,687
Your equation for #1 looks wrong. Try working it out using more intermediate steps. Your starting equation is correct:

[tex]ma = K \frac{Q^2}{R^2} = \frac{Q^2}{4{\pi}{\epsilon}R^2}[/tex]
Shooting Star
#3
Jan17-08, 12:14 PM
HW Helper
P: 1,982
Quote Quote by cse63146 View Post

q = [tex]\sqrt{(0.0001)(225)(0.002^2)/(9x10^9)}[/tex] = 3 x 10 ^-9 C
Should be 0.02

Your methods look correct. Perhaps redo the arithmetic? Since they are mostlly using CGS, why don't you use that to make the numbers easier to handle, and convert afterward?

cse63146
#4
Jan19-08, 12:21 PM
P: 452
A few electricity problems

Quote Quote by berkeman View Post
Your equation for #1 looks wrong. Try working it out using more intermediate steps. Your starting equation is correct:

[tex]ma = K \frac{Q^2}{R^2} = \frac{Q^2}{4{\pi}{\epsilon}R^2}[/tex]
how do I obtain the value to {\epsilon}?

Quote Quote by Shooting star View Post
Should be 0.02

Your methods look correct. Perhaps redo the arithmetic? Since they are mostlly using CGS, why don't you use that to make the numbers easier to handle, and convert afterward?
I also made a mistake on converting from g to kg, so 1g should be 0.001kg

so it should be:

q = [tex]\sqrt{(0.001)(225)(0.02^2)/(9x10^9)}[/tex] = 1 x 10 ^-7 C
Shooting Star
#5
Jan20-08, 06:53 AM
HW Helper
P: 1,982
Correct. But I still recommend that if the values are given mostly in CGS, do the calculation in CGS and convert. Avoid so many decimals, whenever possible. You'll make mistakes in the exam.
cse63146
#6
Jan20-08, 08:32 AM
P: 452
It seems easier to me this way.
berkeman
#7
Jan20-08, 05:14 PM
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berkeman's Avatar
P: 40,687
Quote Quote by cse63146 View Post
how do I obtain the value to {\epsilon}?
You look it up. It's a physical constant.


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