THE UNIVERSITY OF BRITISH COLUMBIA

*Physics 401 *
Assignment #
**10: **

** RETARDED **

POTENTIALS

*SOLUTIONS:*

Wed. 15 Mar. 2006 - finish by Wed. 22 Mar.

POTENTIALS

Wed. 15 Mar. 2006 - finish by Wed. 22 Mar.

- (p. 426, Problem
**10.8**) -**Retarded Gauge:**Confirm that the RETARDED POTENTIALS satisfy the LORENTZ GAUGE condition,

**ANSWER**:*Following the**hint*, we first show

*where , denotes derivatives with respect to , and denotes derivatives with respect to : The identity*

*and the (hopefully by now familiar) results*

*Adding together Eqs. (6) and (7) gives Eq. (3).*

Next, noting that depends on both explicitly and through , whereas it depends on only through , we confirm that

*Derivatives of with respect to (on which it does not depend explicitly) mix in the**time*derivative through the*implicit*dependence of*t*_{r}on . That is,

*because, for a given , , and .*

However, depends*explicitly and implicitly*upon , and must locally satisfy the EQUATION OF CONTINUITY (*i.e.*charge conservation) at any instant of time in terms of the source coordinates , so we have

*because .*

Finally we use this to calculate the divergence of in Eq. (10.19):

*The DIVERGENCE THEOREM tells us that*

*Now, if the closed surface encloses**all*the charges and currents in the source volume, over the whole surface and the surface integral is zero, leaving

*or .* - (p. 427, Problem
**10.10**) -**Weird Loop:**

- Calculate the retarded vector potential at the center.
**ANSWER**:*Choose the origin at the same place as the field point: the centre. Thus and . The source region is uncharged, so**V*= 0.

*where . Now, by symmetry there is as much current going "up" as "down" at the same and**t*_{r}, so the components cancel. This leaves

*where*

*or .* - Find the electric field at the center.
**ANSWER**:*Since**V*=0 we have just

. - Why does this (neutral) wire produce an
*electric*field?**ANSWER**:*Because the vector potential is changing with time, "Doh!" I think this is meant as a retroactive hint in case you got hung up on the preceding question.* - Why can't you determine the
*magnetic*field from this expression for ?**ANSWER**:*Finding requires knowledge of the dependence of on ; but we have calculated only at one point in space! If you want a differentiable you will have a far more difficult calculation to perform.*

- Calculate the retarded vector potential at the center.
- (p. 434, Problem
**10.13**) -**Circulating Charge:**A particle of charge*q*moves in a circle of radius*a*at constant angular velocity . [Assume that the circle lies in the plane, centered at the origin, and that at time*t*=0 the charge is at (*a*,0), on the positive*x*axis.] Find the LIÉNARD-WIECHERT POTENTIALS for points on the*z*axis.**ANSWER**:*In general,*

*where means that the quantities in the square brackets are to be evaluated at the retarded time . Relative to the origin, .*

For a point on the*z*axis, and so , independent of time. We also have and . Thus and where . Then , leaving

and

. - (p. 441, Problem
**10.19**) -**Sliding String of Charges:**An infinite, straight, uniformly charged string, with charge per unit length, slides along parallel to its length at a constant speed*v*.- Calculate the electric field a distance
*d*from the string, using Eq. (10.68):

where .**ANSWER**:*Suppose the field point is a perpendicular distance**s*from the string; measure*z*from the nearest point on the string, as shown in the diagram. Equation (10.68), in which we do*not*need to evaluate anything at a retarded time, gives the contribution to from a single charge*q*. We need to superimpose such contributions from all charge elements at positions down the string: for each of these we use :

*For each element**dz*at*z*there is an equal element*dz*at -*z*; thus the "horizontal" components cancel, leaving only the*x*component of , namely . Meanwhile, since , ; and since , . So and

*Let so that and :*

just as for a line charge at rest! - Find the
*magnetic*field of this string, using Eq. (10.69):

where .**ANSWER**:*Well, and , so this is trivial:*^{1}

where . (Again, the same result as for a steady current in magnetostatics.)

- Calculate the electric field a distance

2006-03-21