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For a point charge between two grounded conductor planes forming a 60° angle, the potential and electric field generated by point charge for Yukawa’s potential (e^{-μr}/*r*) and Coulomb’s potential (1/*r*) are modeled and simulated. The expression for the potential that generalizes the cases discussed in López-Mariño, M. and Trujillo Caballero, J. (2017) Point Charges and Conducting Planes for Yukawa’s Potential and Coulomb’s Potential. *Journal of Electromagnetic Analysis and Applications*, 9, 135-146.
https://doi.org/10.4236/jemaa.2017.910012 is presented. Graphs for the potential and electric field for both cases are showed using Maple, that of Coulomb and that of Yukawa for different values of *μ* . The purpose of this work is to offer students a practical guide for problem analysis of electrostatics using Maple’s capabilities as a computational tool.

In a recent work [

In [

Our goal in this work is to analyze the case of a point charge between two grounded and conducting planes forming a 60˚ angle for Yukawa’s potential and Coulomb’s potential and to discuss the relationship between the expression of the electric potential and the cases discussed in [

The motivation for this work coincides with the results of the study carried out by López et al. [

The parts of this work are the following: Section 2, where we discuss the case of a point charge located between two grounded conductor planes forming a 60˚ angle and the distribution of the real charge and image ones; Section 3, where we obtain the function of the electric potential and show both the potential and electric field for different values of μ; Section 4, where we offer our final comments.

Our goal for this section is to establish the distribution of the actual, Q, charge and the image charges in the region of the conducting planes, as shown in

To find the function of potential between the conducting planes, we must first make sure that each one of them has a potential V = 0 . To analyze this aspect, we refer to

We can observe that one of the conductors is placed on the horizontal while the second plane is located at an α from the aforementioned one. The charge is placed at a θ angle from the horizontal plane and at a distance r from C. Note that

θ < α , and, in our case, α = 2 θ . In addition, we have the restriction 360 ˚ α ∈ Z .

In order to make the potential of the plane located at an α angle equal to zero, it is necessary to place an image charge −Q at a 3θ angle from the horizontal plane. Since this does not guarantee a potential on the horizontal plane equal to zero, an image charge −Q is placed at a −θ angle. As a result, potential in the inclined plane is no longer zero, making it necessary to place a Q charge at 5θ angle. But now, the horizontal plane does not have a potential equal to zero, so a Q charge at a −3θ angle is placed. Finally, it is guaranteed that the potential of both planes becomes zero by placing a charge −Q at a θ+ π angle. All charges are at a distance r from the origin of the plane.

The configuration of the planes, the actual Q charge and their images are shown in

q 0 = Q , position θ (1)

q 1 = − Q , position 3 θ = 2 ( α − θ ) + θ (2)

q 2 = Q , position 5 θ = 2 ( α − θ ) + 3 θ (3)

q 3 = − Q , position 7 θ = 4 ( α − θ ) + 3 θ (4)

q 4 = Q , position 9 θ = 4 ( α − θ ) + 5 θ (5)

q 5 = − Q , position 11 θ = 6 ( α − θ ) + 5 θ (6)

In this section, we have determined the location of the Q charge and the image charges. In the next section, we will present the expression of the electrical potential and we will show the graphs of the potential and the electric field generated by the set of charges.

To calculate the total potential produced by the charges in the space between the planes, we find the actual Q charge in ( r cos θ , r sin θ ) and a point P ( x , y ) , as shown in

By the problem’s geometry, we have magnitude of r q p given by

| r q p | = ( x − h − r cos θ ) 2 + ( y − t − r sin θ ) 2 (7)

and if we consider a = x − h and b = y − t , and keep in mind that α = 2 θ , it is possible to rewrite (7) as

| r q p | = ( a − r cos θ ) 2 + ( b − r sin θ ) 2 . (8)

Taking in to account the superposition principle and the previous expressions, we can express the total electric potential produced by the charges (real and image ones) as

V = ∑ i = 0 5 ( − 1 ) i Q 4 π ε 0 ( a − r cos φ i ) 2 + ( b − r sin φ i ) 2 , (9)

where

φ i = [ i + ( i % 2 ) ] ( α − θ ) + [ i + [ ( i + 1 ) % 2 ] ] θ . (10)

For our discussed case, with the following values α = π 3 , θ = π 6 , we can find the positions of the real charge and image charges, Equations (1)-(6).

To graphically illustrate the total potential, the equipotential curves and the electric field in the region where the actual charge is located, the following values

are considered: r = 2 , q = 4 π × 8.85 × 10 − 12 Coul , ( h , t ) = ( 0 , 0 ) , α = π 3 , θ = π 6 .

First, considering the case of Coulomb,

Then, secondly, for the Yukawa’s case,

In this section, we have presented the expression of the electrical potential for the configuration of charges in

From the superposition of the potentials of each of the charges, we obtained the expression for the total potential, (9), where φ i , (10), provides the locations of the image charges from the location of the actual charge and the value for α . For the case we have discussed in this work, we can verify this if we consider α = 60 ˚ and the position of the actual charge as θ = 30 ˚ , since we find the locations of the image charges of the system: φ 1 = 90 ˚ , φ 2 = 150 ˚ , φ 3 = 210 ˚ , φ 4 = 270 ˚ , φ 5 = 330 ˚ . On the other hand, if we take as the location of the actual charge θ = 90 ˚ and α = 180 ˚ , we find that the only possible image charge is at φ 1 = 270 ˚ , which is the case of a charge in the region of a conducting plane. Also, if we choose the position of the actual charge as θ = 45 ˚ and α = 90 ˚ , we obtain the locations of the three necessary image charges: φ 1 = 135 ˚ , φ 2 = 225 ˚ , φ 3 = 315 ˚ and this result corresponds to the case of a charge between two perpendicular conducting planes. These last two cases were recently discussed [

As physics professors, we are interested in continuing to support student learning through modeling and simulation with the help of computational tools, such as Maple, among others. It is clear that a software capable of performing numerical and symbolic calculations in addition to graphing can be helpful to develop sensitivity analysis in prototypes. All of the above, as additional benefits to one discussed in this paper: thoroughly understanding electromagnetic theory.

We would like to express our gratitude to Professors N. Aquino from Universidad Autónoma Metropolitana, Iztapalapa and G. Maldonado from Tecnológico de Monterrey, Campus Central de Veracruz, for careful reading of the manuscript and for valuable comments on this work.

The authors declare no conflicts of interest regarding the publication of this paper.

López-Mariño, M.A., López, O.O. and Regueiro, J.M.P. (2021) Point Charge between Two Grounded and Conducting Planes Forming a 60˚ Angle and for Yukawa’s Potential and Coulomb’s Potential. Journal of Electromagnetic Analysis and Applications, 13, 57-66. https://doi.org/10.4236/jemaa.2021.134004

It can be used Yukawa potential ( e − μ r / r ) or Coulomb potential (1/r)

Procedure for the calculation of the potential and electric field of a charge Q between two planes forming an angle of 60˚

> restart;

> with(plots):

> with(plottools):

> with(linalg):

> with(StringTools):

> poteimgs := proc(cpoint, centerp, angplan, r, q, typec)

> local charget, i, n, chargex, phi, k, epsilon, angplanx, rxy;

> charget := 0;

> n := abs(((360)/convert(angplan[

> epsilon := 8.85*10^(-12);

> k := 1/(4*Pi*epsilon);

> angplanx := `if`(type(r, list) or type(r, vector) or type(r, matrix), invfunc[tan](`if`(traperror(r[

> rxy := `if`(type(r, list) or type(r, vector) or type(r, matrix), sqrt(r[

> chargex := (i) -> (((-1)^i)*k*q)/sqrt((cpoint[

> phi := proc(i) (i + modp(i, 2))*(angplan[

> if (abs(evalf(angplan[

> for i from 0 to n do charget := charget + chargex(i); end do:

> `if`(UpperCase(convert(typec, string)) = "CAMPO" ,-grad(charget, [x, y, z]), charget);

> else printf("El angulo al que se encuentra la carga no debe ser mayor al ángulo de abertura de los planos..."); end if;

> end proc:

The stntax with which the potential in calculated is

> v(x, y) := poteimgs([x, y], [coordcx, coordcy], [alphax, thetax], rx, qx, potencial);

and for the field

> campo(x, y) := poteimgs([x, y], [coordcx, coordcy], [alphax, thetax], rx, qx, campo);

To obtain the graphs of the potential and the field, it is necessary to execute the following instructions

> plotCP3d := proc(cpoint, centerp, angplan, r, q, typex)

> local plotsx, plotxyz, ik, elcamp, charc, charge;

> elcamp := seq(implicitplot3d(y - (ik*0.2) = x*tan(angplan[

> charc := sphere([`if`(type(rx, list) or type(rx, vector) or type(rx, matrix), rx[

> charge := plots[display](charc, scaling=constrained, style = patchnogrid, axes=boxed, color = COLOR(RGB, 0.75, 0.15, 0.25)), implicitplot3d(y = x*tan(angplan[

> plotsx[

> plotsx[

> plotxyz := `if`(UpperCase(convert(typex, string)) = "POTENCIAL", [plotsx[

> display(plotxyz, axes = boxed, view = [0..Pi, 0..Pi, 0..`if`(UpperCase(convert(typex, string)) <> "POTENCIAL", Pi, 4*Pi)]);

> end proc:

To show the potential graph, use

> plotCP3d([x, y, z], [coordcx, coordcy], [alphax, thetax], rx, qx, potencial);

and for the field, use

> plotCP3d([x, y, z], [coordcx, coordcy], [alphax, thetax], rx, qx, campo);

To graph the potenctial and the field together, run

> plotCP3d([x, y], [coordcx, coordcy], [alphax, thetax], rx, qx, "");