What is the electric force of the molecule on the proton?

Click For Summary
SUMMARY

The electric force exerted by an ammonia molecule (NH3) on a proton located 2.50 nm away is calculated using the formula F = (-kpe)/(d^3), where k is Coulomb's constant, p is the dipole moment (5.0 x 10^-30 C*m), and d is the distance (2.5 x 10^-9 m). The initial calculation yielded a force of 5.0 x 10^-13 N, which was deemed incorrect due to insufficient precision. It is essential to convert results into atomic units for better clarity in atomic scale systems.

PREREQUISITES
  • Understanding of electric dipole moments
  • Familiarity with Coulomb's law and electric force calculations
  • Knowledge of atomic units and their significance in molecular physics
  • Proficiency in basic algebra for manipulating equations
NEXT STEPS
  • Learn about atomic units and how to convert between SI units and atomic units
  • Study the implications of electric dipole moments in molecular interactions
  • Explore advanced calculations involving forces at the atomic scale
  • Investigate the role of precision in scientific calculations and its impact on results
USEFUL FOR

Students in physics, particularly those studying molecular interactions, as well as researchers and educators focused on atomic-scale phenomena and electric forces.

Bigworldjust
Messages
53
Reaction score
0

Homework Statement



An ammonia molecule (NH3) has a permanent electric dipole moment 5.0 x 10^-30cm. A proton is 2.50nm from the molecule in the plane that bisects the dipole.

What is the electric force of the molecule on the proton?


Homework Equations



F = QE = eE_yj = (-kpe)/(d^3)j

The Attempt at a Solution



((9*10^9 N*m^2/C^2)(5.0*10^-30 C*m)(1.6*10^-19 C))/(2.5*10^-9)^3

This gives me an answer of 5.0*10^-13 N, but that seems to be wrong. Anyone know where I went wrong here? Thank you!
 
Physics news on Phys.org
1. You need to use more digits. My answer is about 10% off yours.
2. Why does this seem wrong? Is it because it's very small? Try converting this to atomic units i.e. eV/nm (electron charge*1V/nm) or calculate the resulting acceleration assuming no other forces (in m/s^2 or nm/ns^2).

The thing about atomic scale systems is that using human scale units is not appropriate. N,m,s,kg just don't make sense as good units, and it makes large quantities seem small and small quantities large.
 
frogjg2003 said:
1. You need to use more digits. My answer is about 10% off yours.
2. Why does this seem wrong? Is it because it's very small? Try converting this to atomic units i.e. eV/nm (electron charge*1V/nm) or calculate the resulting acceleration assuming no other forces (in m/s^2 or nm/ns^2).

The thing about atomic scale systems is that using human scale units is not appropriate. N,m,s,kg just don't make sense as good units, and it makes large quantities seem small and small quantities large.
Ah, okay I see what you mean. I finally got it, thanks!
 

Similar threads

Replies
1
Views
3K
  • · Replies 9 ·
Replies
9
Views
8K
  • · Replies 6 ·
Replies
6
Views
10K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
9K
Replies
1
Views
9K
  • · Replies 26 ·
Replies
26
Views
3K
Replies
3
Views
14K