Problem.Set.3.soln.2008

Problem.Set.3.soln.2008 - Problem Set 3 (Jackson 6.20). 1....

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Problem Set 3 (Jackson 6.20). 1 . An example of the preservation of causality and finite speed of propagation in spite of the use of the Coulomg gauge is afforded by a unit strength dipole source that is flashed on and off at t = 0. The charge and current densities are ( r , t ) = ( x ) ( y ) ( z ) ( t ) J ( r , t )=− e 3 ( x ) ( y ) ( z ) ( t ) where a prime means differentiation with repsect to the argument. This dipole is of unit strength and it points in the negative z direction. ( a ) Show that the instantaneous Coulomb potential (6.23 in text) is ( r , t 1 4  o z r 3 ( t ) . ( b ) Show that the transverse current, J t ,i s J t ( r , t e 3 2 3 ( 3 ) ( r ) + 1 4 3 n ( n e 3 ) − e 3 r 3 ( t ) where n = r ̂ , a unit vector along the r direction and the . 2 3 factor multiplying the delta function comes from treating the gradient of z r 3 according to (4.20 in text.) ( c ) Show that the electric and magnetic fields are causal and that the electric field components are given by: E ( r , t ) = e 3 c r ′′ ( ct r ) + 1 r 2 ( r ct ) − 1 r ( r ct ) r cz r 3 ′′ ( r ct ) − 3 r ( r ct ) + 3 r 2 ( r ct ) Hint: While the answer in part b displays the transverse current explicitly, the less explicit form, J t ( r , t e 3 ( r ) ( t ) − ( t ) 1 4 z 1 r can be used with (6.47 in text) to calculate the vector potential and the fields for part c. An alternative method is to use the Fourier transforms in time of J t and A , the Green function (6.40) and its sperical wave expansion from Chapter 9. Solution: Discussion: The scalar potential will equal the that of a static point dipole multiplied by ( t ) . Its contibution to the electric field will be a dipole field multiplied by ( t ) . This contibution must be cancelled by a contribution from the vector potential. This requires that a term multiplied by ( t ) must be obtained from the vector potential. There is not a unique procedure by which this can be achieved. In the approach used here the time dervative of the vector potential will be carried out before the space integration is completed. Solution ( a ): In the coulomb gauge the scalar potential satisfies Poisson’s equation with the charge density as the source. The scalar potential is obtained using the Green’s function for Poisson’s equation
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( r , t ) = ∫∫∫ ( x ) ( y ) ( z ) ( t ) | r r | d 3 r = ( t ) z 1 r =− ( t ) z r 3 Gaussian units = 1 4  o ( t ) z 1 r SI units
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Problem.Set.3.soln.2008 - Problem Set 3 (Jackson 6.20). 1....

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