Question

In Griffiths' section 4.2.1, we saw that the potential of a polarized object with dipole moment p = Pdt' in each element of volume dr' is: 1 I ñ :P(r) da! V(r = - 47 60 - — 22 Polarized Object p = Pdt (0,0,0) We performed 3 transformations to show that this potential was equivalent to the potential of a surface charge distribution plus the potential of a volume charge distribution and we called those bound charges: Ob=P:ñ Pb= -V.P a) First Transformation V(x) = ste / P. 045)dr' Show that indeed = v'(1), remembering that the separation vector is: n=r-r' = (x – X')+ (y - y')ý + (z – z'). (4 points)

Answer 1

ANSWER :

Given that :

Dipole moment :

the electric dipole moment is a measure of the separation of positive and negative electrical charges within a system, that is, a measure of the system's overall polarity. The electric field strength of the dipole is proportional to the magnitude of dipole moment.

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