Two-Dimensional Electrostatics


11.6 Two-Dimensional Electrostatics

    A two-dimensional electrostatic field is produced by a system of charged wires, plates, and cylindrical conductors that are perpendicular to the z plane. The wires, plates, and cylinders are assumed to be so long that the effects at the ends can be neglected, as mentioned in Section 11.4.  This assumption results in an electric field  [Graphics:Images/ElectrostaticsMod_gr_1.gif]  that can be interpreted as the force acting on a unit positive charge placed at the point [Graphics:Images/ElectrostaticsMod_gr_2.gif].  In the study of electrostatics the vector field  [Graphics:Images/ElectrostaticsMod_gr_3.gif]  is shown to be conservative and is derivable from a function  [Graphics:Images/ElectrostaticsMod_gr_4.gif],  called the electrostatic potential, expressed as  


    If we make the additional assumption that there are no charges within the domain D,  then Gauss' Law for electrostatic fields implies that the line integral of the outward normal component of   [Graphics:Images/ElectrostaticsMod_gr_6.gif]  taken around any small rectangle lying inside D is identically zero.  A heuristic argument similar to the one for steady state temperatures with  [Graphics:Images/ElectrostaticsMod_gr_7.gif]  replaced by   [Graphics:Images/ElectrostaticsMod_gr_8.gif]  will show that the value of the line integral is  

This quantity is zero, so we conclude that  [Graphics:Images/ElectrostaticsMod_gr_10.gif]  is a harmonic function.  If we let  [Graphics:Images/ElectrostaticsMod_gr_11.gif]  be the harmonic conjugate, then  


is the complex potential (not to be confused with the electrostatic potential).

The curves  [Graphics:Images/ElectrostaticsMod_gr_13.gif]  are called the equipotential curves, and the curves  [Graphics:Images/ElectrostaticsMod_gr_14.gif]  are called the lines of flux.  If a small test charge is allowed to move under the influence of the field   [Graphics:Images/ElectrostaticsMod_gr_15.gif],  then it will travel along a line of flux.  Boundary value problems for the potential function [Graphics:Images/ElectrostaticsMod_gr_16.gif] are mathematically the same as those for steady state heat flow, and they are realizations of the Dirichlet problem where the harmonic function is [Graphics:Images/ElectrostaticsMod_gr_17.gif].


Example 11.18.  Consider two parallel conducting planes that pass perpendicular to the z plane through the lines  [Graphics:Images/ElectrostaticsMod_gr_18.gif],  which are kept at the potentials  [Graphics:Images/ElectrostaticsMod_gr_19.gif] and [Graphics:Images/ElectrostaticsMod_gr_20.gif], respectively.  Then according to the result of Example 11.1, (see Section 11.1), the electrical potential is



Explore Solution 11.18.


Example 11.19.  Find the electrical potential  [Graphics:Images/ElectrostaticsMod_gr_34.gif]  in the region between two infinite coaxial cylinders  [Graphics:Images/ElectrostaticsMod_gr_35.gif],  which are kept at potentials  [Graphics:Images/ElectrostaticsMod_gr_36.gif],  respectively.  


Solution.  The transformation  


maps the annular region between the circles  [Graphics:Images/ElectrostaticsMod_gr_39.gif]  onto the infinite strip  [Graphics:Images/ElectrostaticsMod_gr_40.gif]  in the w plane, as shown in Figure 11.36.  The potential  [Graphics:Images/ElectrostaticsMod_gr_41.gif]  in the infinite strip has the boundary values  

            [Graphics:Images/ElectrostaticsMod_gr_42.gif]   and   [Graphics:Images/ElectrostaticsMod_gr_43.gif]   for all  v.

If we use the result of Example 11.18, the electrical potential  [Graphics:Images/ElectrostaticsMod_gr_44.gif]  is  


Because  [Graphics:Images/ElectrostaticsMod_gr_46.gif],  we can use this equation to conclude that the potential  [Graphics:Images/ElectrostaticsMod_gr_47.gif]  is  


The equipotentials  [Graphics:Images/ElectrostaticsMod_gr_49.gif]  are concentric circles centered on the origin, and the lines of flux are portions of rays emanating from the origin.  If  [Graphics:Images/ElectrostaticsMod_gr_50.gif],  then the situation is as illustrated in Figure 11.36.

Figure 11.36  The electrical field in a coaxial cylinder, where [Graphics:Images/ElectrostaticsMod_gr_51.gif].

Explore Solution 11.19.


Example 11.20.   Find the electrical potential  [Graphics:Images/ElectrostaticsMod_gr_64.gif]  produced by two charged half-planes that are perpendicular to the z plane and pass through the rays  [Graphics:Images/ElectrostaticsMod_gr_65.gif]  where the planes are kept at the fixed potentials  



Solution.  The result of Example 10.13, (see Section 10.4), shows that the function  


is a conformal mapping of the z plane slit along the two rays  [Graphics:Images/ElectrostaticsMod_gr_69.gif]  onto the vertical strip  [Graphics:Images/ElectrostaticsMod_gr_70.gif]. The new problem is to find the potential  [Graphics:Images/ElectrostaticsMod_gr_71.gif]  that satisfies the boundary values  


From Example 11.1, (see Section 11.1),


As in the discussion of Example 11.17, (see Section 11.5),the solution in the z plane is


Several equipotential curves are shown in Figure 11.37.

Figure 11.37  The electric field produced by two charged half-planes
                    that are perpendicular to the complex plane.

Explore Solution 11.20.


Example 11.21.  Find the electrical potential  [Graphics:Images/ElectrostaticsMod_gr_89.gif]  in the disk  [Graphics:Images/ElectrostaticsMod_gr_90.gif]  that satisfies the boundary values



Solution.  The mapping  [Graphics:Images/ElectrostaticsMod_gr_93.gif]  is a one-to-one conformal mapping of D onto the upper half-plane  [Graphics:Images/ElectrostaticsMod_gr_94.gif]  with the property that [Graphics:Images/ElectrostaticsMod_gr_95.gif] is mapped onto the negative u axis and [Graphics:Images/ElectrostaticsMod_gr_96.gif] is mapped onto the positive u axis.  The potential [Graphics:Images/ElectrostaticsMod_gr_97.gif] in the upper half-plane that satisfies the new boundary values


is given by  

(11-29)            [Graphics:Images/ElectrostaticsMod_gr_99.gif].  

A straightforward calculation shows that  


We substitute the real and imaginary parts,  [Graphics:Images/ElectrostaticsMod_gr_101.gif]  and  [Graphics:Images/ElectrostaticsMod_gr_102.gif]  from this equation, into Equation (11-29) to obtain the desired solution:   


The level curve  [Graphics:Images/ElectrostaticsMod_gr_104.gif]  in the upper half-plane is a ray emanating from the origin, and the preimage  [Graphics:Images/ElectrostaticsMod_gr_105.gif]  in the unit disk is an arc of a circle that passes through the points  [Graphics:Images/ElectrostaticsMod_gr_106.gif].  Several level curves are illustrated in Figure 11.38.

            Figure 11.38  The potentials  [Graphics:Images/ElectrostaticsMod_gr_107.gif]  and  [Graphics:Images/ElectrostaticsMod_gr_108.gif]

Explore Solution 11.21.


Extra Example 1.  Find the electrical potential  [Graphics:Images/ElectrostaticsMod_gr_120.gif]  in the crescent-shaped region that lies inside the disk [Graphics:Images/ElectrostaticsMod_gr_121.gif]  and outside the circle [Graphics:Images/ElectrostaticsMod_gr_122.gif]  that satisfies the following boundary values shown in Figure 11.41:



Solution.  The result of Example 10.7, (see Section 10.2), shows that the function  


is a conformal mapping of the crescent-shaped region that lies inside the disk [Graphics:Images/ElectrostaticsMod_gr_126.gif]  and outside the circle [Graphics:Images/ElectrostaticsMod_gr_127.gif]  onto the horizontal strip  [Graphics:Images/ElectrostaticsMod_gr_128.gif].  

Figure 11.41  The electrical potential [Graphics:Images/ElectrostaticsMod_gr_129.gif] inside [Graphics:Images/ElectrostaticsMod_gr_130.gif]  and outside [Graphics:Images/ElectrostaticsMod_gr_131.gif].  


Revisit and Explore Solution 10.7.


    The new problem in the w plane is to find the potential  [Graphics:Images/ElectrostaticsMod_gr_182.gif]  that satisfies the boundary values  

In the w plane the solution is


A straightforward calculation shows that  


We substitute the imaginary part,  [Graphics:Images/ElectrostaticsMod_gr_186.gif]  from this equation, into  [Graphics:Images/ElectrostaticsMod_gr_187.gif]  to obtain the desired solution:   


Explore Extra Solution 1.


Extra Example 2.  Find the electrical potential  [Graphics:Images/ElectrostaticsMod_gr_195.gif]  in the semi-infinite strip [Graphics:Images/ElectrostaticsMod_gr_196.gif] that has the boundary values shown in Figure 11.42:



Solution.  The result of Example 10.13, (see Section 10.4), shows that the function  


is a conformal mapping of the semi-infinite strip  [Graphics:Images/ElectrostaticsMod_gr_200.gif]  onto the upper half-plane  [Graphics:Images/ElectrostaticsMod_gr_201.gif].

    The new problem in the w plane is to find the electrical potential  [Graphics:Images/ElectrostaticsMod_gr_202.gif]  that satisfies the boundary values  


This is a three-value Dirichlet problem in the upper half-plane defined by  [Graphics:Images/ElectrostaticsMod_gr_204.gif].  For the w plane, the solution in Equation (11-5) becomes  


Here we have  [Graphics:Images/ElectrostaticsMod_gr_206.gif]  and  [Graphics:Images/ElectrostaticsMod_gr_207.gif],  which we substitute into the above equation for  [Graphics:Images/ElectrostaticsMod_gr_208.gif]  to obtain  


Now make the substitutions  [Graphics:Images/ElectrostaticsMod_gr_210.gif]  and  [Graphics:Images/ElectrostaticsMod_gr_211.gif]  to get the solultion  [Graphics:Images/ElectrostaticsMod_gr_212.gif]  in the z plane  



Figure 11.42  The electrical potential [Graphics:Images/ElectrostaticsMod_gr_215.gif] in the semi-infinite strip [Graphics:Images/ElectrostaticsMod_gr_216.gif].  

Explore Extra Solution 2.


Exercises for Section 11.6.  Two-Dimensional Electrostatics


Library Research Experience for Undergraduates

Dirichlet Problem

Poisson Integral


Steady State Temperature

Complex Potential




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(c) 2012 John H. Mathews, Russell W. Howell