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Order of magnitude problem involving the force of repulsion between two persons

  • Thread starter StrawHat
  • Start date
  • #1
33
0

Homework Statement



ruZ5R.png


Homework Equations



[itex]\vec{F}[/itex] = k[itex]\stackrel{q1q2}{r^{2}}[/itex]

The Attempt at a Solution



6.022e23*(0.505) = 3.041e23C <-- electrons
6.022e23*(0.495) = 2.981e23C <-- protons
3.041e23 - 2.981e23 = 6e21C <-- the difference between the two charges
[itex]\vec{F}[/itex][itex]_{e}[/itex] = (9e9Nm[itex]^{2}[/itex]/C[itex]^{2}[/itex])(6e21C)[itex]^{2}[/itex] / 2.25m[itex]^{2}[/itex]
[itex]\vec{F}[/itex][itex]_{e}[/itex] = 1.44e53 N
 

Answers and Replies

  • #2
CAF123
Gold Member
2,889
88
  • #3
33
0


Given that the force of repulsion is similar to the gravitational force exerted on an object with a mass the size of earth, you should expect F ~ 1025N.

There is an identical thread here:
https://www.physicsforums.com/showthread.php?t=431698
The answer gives me this error: "Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error."
 
  • #4
3,732
414


What is the answer that you put in the box?
 
  • #5
33
0


What is the answer that you put in the box?
I put in F~10 25N.
 
  • #6
6,054
390


How do you get the number of protons?
 
  • #7
3,732
414


I put in F~10 25N.
I would consider trying to enter 26 in the box. Depending on how you estimate, you may get 10^26. And the "weight" of the Earth is also of the order 10^26 (6x10^25).
 

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