Exercise 6.   Find the electrostatic potential  [Graphics:Images/ElectrostaticsModHome_gr_246.gif]  in the infinite strip   [Graphics:Images/ElectrostaticsModHome_gr_247.gif],  

that satisfies the following boundary values,  (shown in Figure 11.44)  

                    [Graphics:Images/ElectrostaticsModHome_gr_248.gif]  

Hint.  Use  [Graphics:Images/ElectrostaticsModHome_gr_249.gif].

Solution 6.

See text and/or instructor's solution manual.

Answer.   Map the infinite strip onto the right half plane slit along the ray  [Graphics:../Images/ElectrostaticsModHome_gr_250.gif]  with the mapping  [Graphics:../Images/ElectrostaticsModHome_gr_251.gif],  

then construct   [Graphics:../Images/ElectrostaticsModHome_gr_252.gif].   

Solution.   The transformation  [Graphics:../Images/ElectrostaticsModHome_gr_253.gif]  maps the infinite strip  [Graphics:../Images/ElectrostaticsModHome_gr_254.gif]  

onto the right half plane slit along the ray  [Graphics:../Images/ElectrostaticsModHome_gr_255.gif].     

                     [Graphics:../Images/ElectrostaticsModHome_gr_256.gif]          [Graphics:../Images/ElectrostaticsModHome_gr_257.gif]

                      The mapping   [Graphics:../Images/ElectrostaticsModHome_gr_258.gif].  

 

Now construct the intermediate solution  [Graphics:../Images/ElectrostaticsModHome_gr_259.gif]  in the w-plane that has the boundary values

                    [Graphics:../Images/ElectrostaticsModHome_gr_260.gif]  
        
Notice that the intermediate solution   [Graphics:../Images/ElectrostaticsModHome_gr_261.gif]   will agree with these boundary values (and also [Graphics:../Images/ElectrostaticsModHome_gr_262.gif] for [Graphics:../Images/ElectrostaticsModHome_gr_263.gif]).  

Therefore the solution in the infinite strip   [Graphics:../Images/ElectrostaticsModHome_gr_264.gif]   is   

                    [Graphics:../Images/ElectrostaticsModHome_gr_265.gif],  

                    [Graphics:../Images/ElectrostaticsModHome_gr_266.gif].  

Observation.  There is an equipotential  [Graphics:../Images/ElectrostaticsModHome_gr_267.gif] for [Graphics:../Images/ElectrostaticsModHome_gr_268.gif]  that is the preimage of the segment  [Graphics:../Images/ElectrostaticsModHome_gr_269.gif]  where we have  [Graphics:../Images/ElectrostaticsModHome_gr_270.gif].

 

We are done.   

 

Aside.  For illustration purposes we can graph the function   [Graphics:../Images/ElectrostaticsModHome_gr_271.gif].   

                     [Graphics:../Images/ElectrostaticsModHome_gr_272.gif]

                     A contour graph of the function   [Graphics:../Images/ElectrostaticsModHome_gr_273.gif]  

                     where   [Graphics:../Images/ElectrostaticsModHome_gr_274.gif]   for   [Graphics:../Images/ElectrostaticsModHome_gr_275.gif].  

 

We are really really done.   

 

                     [Graphics:../Images/ElectrostaticsModHome_gr_276.gif]

                     A graph of the function   [Graphics:../Images/ElectrostaticsModHome_gr_277.gif],   

                    [Graphics:../Images/ElectrostaticsModHome_gr_278.gif]  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) 2008 John H. Mathews, Russell W. Howell