Finding the Force of Nucleus on a Single Electron of Plutonium

AI Thread Summary
The discussion centers on calculating the force exerted by a Plutonium nucleus on a single electron located 0.563x10^-12 m away. The user initially miscalculated by considering the total charge of all 94 electrons instead of just one. The correct approach involves using Coulomb's law with the charge of one electron and the charge of the nucleus, which is equivalent to the number of protons. The participants clarify the formula needed and suggest using LaTeX for better formatting in future posts. Understanding these concepts is crucial for accurately solving similar physics problems.
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Homework Statement



A particular nucleus of the element Plutonium contains 94 protons and 150 neutrons. What is the magnitude of the force from the nucleus on a single electron that is a distance of 0.563x10-12 m from the Plutonium nucleus?

So basically this was a homework question that I got wrong, and I'm not entirely sure what the correct answer is. I'm reworking it to study for an exam, but I want to make sure this is right so I don't keep lousing it up!

Homework Equations



e=1.60x10-19 C
and F=kQ/r2
where k is Coulomb's constant 8.99x109 Nm2/C2

The Attempt at a Solution



Okay so if there are 94 protons, there should be 94 electrons.
(94)(e)=(94)(1.60x10-19 C)=1.5x10-17 C

Where I'm getting confused is what charge I should use to find the force? Because 1.5x10-17 C is the charge of all 94 electrons and I need the magnitude of the force of the nucleus on just one electron. I'm not quite grasping this concept.

Anyway, if I blindly plug in numbers without really understanding completely what I'm doing here, I get:

F=(k(1.5x10-17 C))/(0.563x10-12 m)2
so F=4.27x1017 N

Also, how do I type fractions on this forum? I found the button but I'm not sure where to enter my numbers so it works! Thanks!
 
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anomalocaris said:
and F=kQ/r2
You've left something out of that equation.
Okay so if there are 94 protons, there should be 94 electrons.
The problem asks for the force of the nucleus on a single electron. You can ignore the other 93 electrons.
 
Okay, so should I have used F=\frac{(k(q1)(q2))}(r<sup>2</sup>){}? And then I would use e as the charge of one electron (my q2) and the charge of all 94 electrons as my q1? Or am I completely off here? Thank you!
 
Oh, sorry, the fraction thing still is not working for me. I meant F=(k(q1q2))/(r^2)
 
anomalocaris said:
Okay, so should I have used F=\frac{(k(q1)(q2))}(r<sup>2</sup>){}? And then I would use e as the charge of one electron (my q2) and the charge of all 94 electrons as my q1? Or am I completely off here? Thank you!
That's the way to do it.
 
Oh! Yay! Thank you so much,tms! I really appreciate it!
 
anomalocaris said:
Oh, sorry, the fraction thing still is not working for me. I meant F=(k(q1q2))/(r^2)
You need to put the denominator in the second set of curly braces: \frac{(k(q1)(q2))}{(r2)}
F=\frac{(k(q1)(q2))}{(r2)}
You should also use subscripts and superscripts, and you can get rid of the extraneous parentheses: F=k\frac{q_1q_2}{r^2}
F=k\frac{q_1q_2}{r^2}
If you are planning to go further in math or science, it is worth learning LaTeX. You can google for some brief introductions and tutorials.
 
I tried doing that, but it didn't look right when I previewed it. My browser can be kind of wonky, though, so I'll keep trying. And I've never heard of LaTeX--sounds really cool. I'll definitely google that! Thanks again for all your help!
 
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