Does the concept of dipole moment of charged molecule exist or not?

  • Thread starter linwenzi
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  • #1
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Main Question or Discussion Point

Generally the concept of dipole moment is used in neutral system, ie the total positive charge equals to the total negtive charge.
Could the concept and also the calculation be used in charged system?

For example, NaOH, (Na+)(O-2)(H+), we can calculate its dipole moment.
Could I calculate the dipole moment of (OH)- alone?

If yes, is there a authoritative definition in some book or papers?

Many thanks!
 

Answers and Replies

  • #2
DrDu
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Of course the dipole moment can be calculated also for charged systems. However, its value depends on the choice of origin. Usually one uses the center of charge as origin to fix the value.
 
  • #3
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It's just a series expansion of the electric potential in 1/distance; the presence of a monopole term doesn't mean there might not also be dipole or quadrupole terms, just that the monopole term is likely to dominate in the far field. Griffiths has a good section on this and should be in any physics library.
 
  • #4
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Of course the dipole moment can be calculated also for charged systems. However, its value depends on the choice of origin. Usually one uses the center of charge as origin to fix the value.
Many thanks for your reply.

I searched some papers that claim the dipole moment of charged molecule depends on the origin and molecular orientation as you said.
Unfortunately I cannot find any research papers or books telling the detials of calculation of charged molecules as you said to use the center of charge as origin .

If possible, could you give some references?

Thanks again!
 
  • #5
DrDu
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The dipole moment is
## \mathbf{d}=\sum_i (\mathbf{r}_i-\mathbf{R})q_i ## where ##q_i## is the charge of the i-th particle in the molecule and ##\mathbf{r}_i## its position. ##\mathbf{R}## is the center around which a multipole moment expansion is done. Changing the center ##\mathbf{R}\to \mathbf{R}+\Delta \mathbf{R} ## changes the dipole moment by ## \Delta \mathbf{d}=\sum_i (-\Delta \mathbf{R})q_i =-\Delta \mathbf{R} Q##, where Q is the total charge of the molecule.
The book by Griffiths "Introduction to Electrodynamics" or "Classical Electrodynamics" by Jackson should contain ample information of multipole expansions.
 
  • #6
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It's just a series expansion of the electric potential in 1/distance; the presence of a monopole term doesn't mean there might not also be dipole or quadrupole terms, just that the monopole term is likely to dominate in the far field. Griffiths has a good section on this and should be in any physics library.
Thank you very much for your reply!
 
  • #7
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The dipole moment is
## \mathbf{d}=\sum_i (\mathbf{r}_i-\mathbf{R})q_i ## where ##q_i## is the charge of the i-th particle in the molecule and ##\mathbf{r}_i## its position. ##\mathbf{R}## is the center around which a multipole moment expansion is done. Changing the center ##\mathbf{R}\to \mathbf{R}+\Delta \mathbf{R} ## changes the dipole moment by ## \Delta \mathbf{d}=\sum_i (-\Delta \mathbf{R})q_i =-\Delta \mathbf{R} Q##, where Q is the total charge of the molecule.
The book by Griffiths "Introduction to Electrodynamics" or "Classical Electrodynamics" by Jackson should contain ample information of multipole expansions.
It's very helpful.
Thank you very much for your help!
 

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