Electric potential of a dipole moment

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SUMMARY

The electric potential of a dipole moment, specifically for a water molecule with a dipole moment of 6.29E-30 Cm, can be calculated at a distance of 1.43 nm along the dipole axis. The relevant formula for the potential at a point P is given by V(r) = qk(1/(r-d/2) - 1/(r+d/2)), where q is the charge, k is Coulomb's constant, and d is the distance between the charges. The dipole moment P is defined as P = qd, and the potential is proportional to this dipole moment. Understanding the derivation of the potential formula is crucial for solving related problems.

PREREQUISITES
  • Understanding of electric potential and dipole moments
  • Familiarity with Coulomb's law and its applications
  • Knowledge of basic calculus for manipulating equations
  • Ability to interpret physical concepts in electrostatics
NEXT STEPS
  • Study the derivation of the electric potential formula for point charges
  • Learn about the behavior of electric fields generated by dipoles
  • Explore the implications of dipole moments in molecular chemistry
  • Investigate the applications of dipole potential in real-world scenarios
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Students studying electrostatics, physicists working with molecular interactions, and educators teaching concepts related to electric potential and dipole moments.

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Homework Statement


The dipole moment of a water molecule is 6.29E-30Cm. What is the electric potential's magnitude 1.43nm from a water molecule along the axis of the dipole?


Homework Equations


p = qr


The Attempt at a Solution


I don't have values for q or r. I just know the magnitude of the dipole moment and the distance away from the molecule in which I have to figure out the electric potential. I don't know where to go from here.

And even if given q, how would I figure out V.
 
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Find the electric potential of two equal and opposite charges d distance apart symbolically, at distance x from the centre of the dipole, and then take into account that x>>d.

ehild
 
That doesn't really help me.
 
You know the formula for the potential of a point charge? A point charge q is at x1=d/2, and an other one is at x2=-d/2. What is the potential at the point with coordinate x?

ehild
 
Has anyone figured out how to do this question? I have it too but I can't find the procedure anywhere online for it.
 
TheExibo said:
Has anyone figured out how to do this question? I have it too but I can't find the procedure anywhere online for it.
Do you know what is the electric potential of a point charge at distance D from it?
You have two charges, arranged along the x axis, both at distance d/2 from the origin. What are their potential at the point P, distance x from the origin?
upload_2017-10-31_5-39-15.png
 

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ehild said:
Do you know what is the electric potential of a point charge at distance D from it?
You have two charges, arranged along the x axis, both at distance d/2 from the origin. What are their potential at the point P, distance x from the origin? View attachment 214070

How do I determine what d/2 is?
 
TheExibo said:
How do I determine what d/2 is?
You do not need to determine it. The dipole momentum is given, and P=qd. At the end, you will find that the potential is proportional to qd, that is, the dipole moment.
 
ehild said:
You do not need to determine it. The dipole momentum is given, and P=qd. At the end, you will find that the potential is proportional to qd, that is, the dipole moment.

So I have P/(2q)=d/2 so far. I have also made the equations V=qk/(r-d/2) and V=qk/(r+d/2) and r is the distance from the origin to the point at which potential is measured.

What is next? I don't understand how to substitute them since d/2, V, and q are unknown.
 
  • #10
Nevermind. I have found that the equation V=kp/r^2 works, although I don't know how this equation was derived.
 
  • #11
TheExibo said:
So I have P/(2q)=d/2 so far. I have also made the equations V=qk/(r-d/2) and V =qk/(r+d/2) and r is the distance from the origin to the point at which potential is measured.

What is next? I don't understand how to substitute them since d/2, V, and q are unknown.
The two charges have opposite signs. The positive charge causes V+=qk/(r-d/2) potential at distance r from the origin, and the negative charge contributes to the potential by V-=qk/(r+d/2). The potential at P is the sum of V+ and V-: ##V(r)=qk\left(\frac{1}{r-d/2}-\frac{1}{r+d/2}\right)##. Bring the fractions to common denominator and use that d/r << 1.
 

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