Electrical Force and Coulomb's Law

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SUMMARY

The discussion centers on calculating the electrical force between an alpha particle and a gold nucleus using Coulomb's Law. The alpha particle has a charge of +2.0e, equating to 3.2e-19 C, while the gold nucleus has a charge of +79e, or 1.264e-17 C. The formula used is F = k (|q1 * q2| / r^2), where k is Coulomb's constant. The user successfully resolved their confusion after clarifying the charge values and applying Coulomb's Law correctly.

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Trista
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I have been working on a problem and can't seem to figure out what I'm doing wrong. Here goes:

An alpha particle (charge = +2.0e) is sent at high speed toward a gold nucleus (charge = +79e). What is the electrical force acting on the alpha particle when it is 2.0 X 10^-14 m from the gold nucleus?

I figured it would be an easy plug and chug, like this:

F=k (|e|^2 / r^2) which when I looked at it I found I'm confused. the problem said that the charge = +2.0e. Does that mean I multiply 2 * 1.6X10^-19? Then, there are 2 charges. In order to get the correct 'e' do I add them together and multiply by 1.6x10^-19? I understand they are both positive, and so they repel from each other, but the alpha particle is sent at high speed toward the gold nucleus.

I can tell I'm not going to enjoy this chapter...

thank you for your help!
 
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2e= 2*1.6e-19=3.2e-19 C
79e=79*1.6e-19=1.264e-17 C
Just use Coulomb`s law.
 
:rolleyes: Thanks. I must be brain fried. I must have tried this 10 times and it wouldn't come out. But after your post it worked just fine.
 

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