Doubt regarding derivation of bound charges in dielectric

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SUMMARY

The discussion centers on the derivation of bound charges in dielectrics as presented in Griffiths' textbook. The formula for the potential due to a dipole is given as V(r) = ∫(x·P(r'))/X² dτ', where X = r - r' and x is the unit vector in the direction of X. A point of contention arises regarding the expression ∇'(1/X) and whether it should be represented as (1/X²)−hat{r'}. The suggestion is made to clarify this derivation using Cartesian coordinates for better understanding.

PREREQUISITES
  • Understanding of Griffiths' "Introduction to Electrodynamics" concepts
  • Familiarity with vector calculus, particularly gradient operations
  • Knowledge of dipole moments and their mathematical representation
  • Basic proficiency in Cartesian coordinate systems
NEXT STEPS
  • Review the derivation of dipole potential in Griffiths' "Introduction to Electrodynamics"
  • Study vector calculus operations, focusing on gradients and their applications
  • Explore the concept of bound charges in dielectrics and their physical implications
  • Practice converting expressions between Cartesian and spherical coordinates
USEFUL FOR

This discussion is beneficial for physics students, particularly those studying electromagnetism, as well as educators and researchers focusing on dielectric materials and their properties.

nuclear_dog
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In Griffiths, for deriving the bound charges for a given polarization P , the formula used is the general formula for dipoles .i.e ( equation 4.9)
{Here the potential at r is calculated due to the dipole at r' )

V(r) = ∫\frac{x.P(r')}{X^2}d\tau'

Here X = r - r' , and x = unit vector in the direction of X

Then it is written that \frac{x}{X^2} = \nabla'(1/X).

since X = (r-r') , and ∇' = (∂/∂r')\widehat{r'} ...

Shouldn't ∇'(1/X) be (1/X^2)\widehat{r'} ?
 
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It might be best to work it out in Cartesian coordinates where all coordinates are written explicitly. See if you can fill in the details of the derivation outlined in the figure.
 

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Thanks , I can see that in Cartesian coordinates .
 

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