Charged Aluminum Spheres: Calculating Electrons and Attractive Force

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SUMMARY

The discussion revolves around calculating the number of electrons in two aluminum spheres, each with a mass of 0.025 kg, separated by 80 centimeters. The correct number of electrons per sphere is determined to be approximately 7.25 x 1024. To achieve an attractive force of 1.00 x 104 N between the spheres, participants clarify that the charge difference must be calculated correctly, emphasizing the need to divide the charge by the elementary charge (1.6 x 10-19 C) to find the number of electrons transferred. Misunderstandings regarding the calculations and the role of protons and electrons are addressed throughout the conversation.

PREREQUISITES
  • Understanding of Coulomb's Law (F=K*q1*q2/(r^2))
  • Knowledge of elementary charge (e = 1.6 x 10-19 C)
  • Familiarity with the concept of point charges
  • Basic knowledge of atomic structure, specifically regarding protons and electrons in aluminum
NEXT STEPS
  • Learn how to apply Coulomb's Law in various scenarios
  • Study the concept of electric charge and its quantization
  • Explore the relationship between mass, moles, and atomic structure in chemistry
  • Investigate the implications of charge transfer in electrostatics
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding electrostatic forces and charge calculations, particularly in the context of aluminum and similar materials.

Gaupp
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I'm having difficulty with a problem on MasteringPhysics (such wonderful software...) and as a last resort I'm posting on here. This is, I'm sure, a really simple problem but I'm getting no kind of feedback from MP and there isn't an example problem like this in the book.


Homework Statement



Two aluminum spheres of mass .025 kg are separated by 80 centimeters.

A) How many electrons does each sphere contain?

B)How many electrons would have to be removed from one sphere and added to the other to cause an attractive force between the spheres of magnitude 1.00 x 10^4 (roughly one ton)? Assume that the spheres may be treated as point charges.

C)What fraction of all the electrons in one of the spheres does this represent?

Homework Equations



F=K*q1*q2/(r^2)

q=e*(#protons-#electrons)

The Attempt at a Solution



A) I found Part A to be 7.25 x 10^24 electrons.

B) This is where I'm stuck.

If you were to remove electrons from one sphere and put them on the other I understand that their charges are to be equal but opposite, as in q1 = -q2. So using Coloumb's Law (F=K*q1*q2/(r^2)) I've set the Force to 1*10^4, divided that by K=9*10^9.

10000/(9*10^9) = q1*q2/(.8^2)

Then multiplying that by .8^2, I have just the charges on the other side of the equation. Since the charges in the formula are absolute value I can set q1=q2 and have q1^2. Taking the square root of the entire thing I have:

q=8.4327*10^-4.

So now I can use the formula q=e(#protons-#electrons). So:

8.4237*10^-4 = 1.6*10^-19(7.25*10^24-#electrons). Solving from this I get 7.249*10^24 electrons as my final answer.

However, MP says I'm wrong but there isn't any feedback as to where I went wrong, and it seems straightforward enough to me that no matter how I rework it I get the same thing every time.

Can anyone help me out here?

C) Can't do this one until B is done.

Any help is greatly appreciated.
 
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how you did 'a'?
I am getting something different [5.58E20]
 
rootX said:
how you did 'a'?
I am getting something different [5.58E20]

Code:
.025 kg x [U]1000g[/U] x [U]1mol[/U] x [U]6.023*10^23 atoms in a mol[/U] x [U]13 electrons[/U] = #electrons
           1kg  26.982g               1 mol             1 atom Al
 
Last edited:
oops, I forgot units...
but do you get 7. something when you do this cal. on calculator?

my cal's giving me 5.something

and you getting
A) I found Part A to be 7.25 x 10^24 electrons.
 
rootX said:
oops, I forgot units...
but do you get 7. something when you do this cal. on calculator?

my cal's giving me 5.something

and you getting



How much does each ball weigh in your problem? It could be different, especially if you're doing this in MP.
 
rootX said:
oops, I forgot units...
but do you get 7. something when you do this cal. on calculator?

my cal's giving me 5.something

and you getting

multiply by the atomic number of aluminimum (ie number of protons in an atom which equals the number of electrons in an atom)
 
I'm confused by what you did in part B... the charge you calculated looks correct... you just need to divide by 1.6*10^-19 to get the number of electrons.
 
learningphysics said:
multiply by the atomic number of aluminimum (ie number of protons in an atom which equals the number of electrons in an atom)

You have to? I think maybe not. (We never did that)

oo..O you don't know that part A is correct try without multiplying by #13, hopefully, you may get the right answer
 
Last edited:
rootX said:
You have to? I think maybe not. (We never did that)

oo..O you don't know that part A is correct try without multiplying by #13, hopefully, you may get the right answer

Gaupp did... he didn't write it out..

Remember you need the number of electrons... not the number of atoms...

So number of atoms * (number of electrons/atom) gives the answer... I get the same answer as Gaupp.
 
  • #10
learningphysics said:
Gaupp did... he didn't write it out..

Remember you need the number of electrons... not the number of atoms...

So number of atoms * (number of electrons/atom) gives the answer... I get the same answer as the Gaupp.

ooo.. yea you are right.. that gives the # of atoms, not the # of electrons
 
  • #11
learningphysics said:
I'm confused by what you did in part B... the charge you calculated looks correct... you just need to divide by 1.6*10^-19 to get the number of electrons.

q=e(#protons-#electrons).

Ok I see now. The difference between the #protons and the #electrons would be the number of electrons lost or gained. I kept solving for just the number of electrons, by moving around the #protons in the equation. DOH!

Thanks for the help there learningphysics.
 

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