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Solving problem of elctrodynamic - Assignment Example

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Solving problem of elctrodynamic

As before, the procedure for determining the Greens function is to split. The region of interest in to two parts (one on each side of the observation point), and separate solutions of the Laplace equation that satisfy the boundary conditions of every region, and then join the two solutions at the source point such that their values match up but the first derivative (in whichever dimension we chose sides) has a finite discontinuity.
The solution in the first region must be admissible down to which excludes the in term and the negative powers of p. However, these terms may be included in the solutions for region in second place. In individually regions, the solution must vanish at which excludes the cos terms
Henceforth when the electromagnetic disturbance has reached the origin, the particle has traveled as far as the electromagnetic disturbance did, but in the opposite direction, so it is now twice as far from the origin as it was when the disturbance we are just now feeling was generated.
The electric fields do not depend on the unprimed variables and come out of the integrals, which was the point of the Taylor sequences expansion. After a little manipulation, we recognize the integrals that are left as the dipole moment and quadrupole moments
d) If, instead of the semi-classical charge density used above, the electron in the 2p state was described by a circular Bohr orbit of radius , rotating with the transition frequency w0, what would the predicted power be? Express your answer in the same units as in part b and evaluate the ratio of the powers numerically.
b) We now want to calculate the fields from the potentials. This is mostly straight forward, albeit tedious, algebra. We need to remember that the dipole moment is evaluated at the retarded time, so that there is an implicit time dependence in p (t0)
a) Starting with the Proca Lagrangian density (12.91) and following the same procedure as for the electro ...Show more

Summary

We must place an image charge outside the sphere on the axis defined by the real charge q and the sphere’s center. Use a Cartesian coordinate system and set the x-axis to be the axis defined by the charge, its image,…
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