CODEWITHSUNDEEP
pythonmatplotlibmathphysicsnumerical-methods
Taha Elbad
Taha Elbad

Reputation: 11

How to find the electric field from the potential?

That's my script to find a potentiel in a ionization room:

# Programme de résolution de l'équation de Laplace
# par la méthode de Gauss-seidel

# importation des librairies
import numpy as np
import time
import matplotlib.pyplot as plt

# définition des paramétres physiques de l'expérience
Vi = 300.0           # le couvercle est à  300 V
V0 = 0.0             # les cotés sont au potentiel nul
x = 0
# définition de la grille de calcul
M = 50
N = 300                     # nombre de pas sur la grile (identique en Ox et Oy)
V = np.zeros([M,N])     # grille de calcul courante
Vnew = np.zeros([M,N])  # grille de stockage des calculs nouveaux

# critère de précision du calcul
EPS = 1e-3
# initialisation des compteurs        
ecart = 1.0
iteration = 0

# définition des conditions aux limites
V[0:8, 0:20] = x
V[8:15, 0:10] = x
V[35:50, 0:10] = x
V[42:50, 0:20] = x
V[0,20:300]  = Vi  # bord supérieur
V[0:8, 20]  = Vi  # vue que 1.6/0,2 = 8
V[41:50, 20] = Vi # bas de la bord gauche
V[8, 10:21] = Vi
V[15, 0:11] = Vi
V[34, 0:11] = Vi
V[41, 10:21] = Vi
V[8:15, 10] = Vi
V[34:41, 10] = Vi
V[15:35, 0] = Vi
V[:,-1] = Vi  # bord droit
V[-1,20:300] = Vi  # bord inférieur
V[23:27,15:299] = V0

# début du calcul - enregistrement de la durée
tdebut = time.time()

# boucle de calcul - méthode de Gauss-Seidel
while ecart > EPS:
  iteration += 1
  # sauvegarde de la grille courante pour calculer l'écart
  Vavant = V.copy()
  # calcul de la nouvelle valeur du potentiel sur la grille
  V[1:-1,1:-1]= 0.25*(Vavant[0:-2,1:-1] +V[2:,1:-1] + Vavant[1:-1,0:-2] + V[1:-1,2:])
  # on repose les même conditions
  V[0:8, 0:20] = x
  V[8:15, 0:10] = x
  V[35:50, 0:10] = x
  V[42:50, 0:20] = x
  V[0,20:300]  = Vi 
  V[0:8, 20]  = Vi  
  V[41:50, 20] = Vi 
  V[8, 10:21] = Vi
  V[15, 0:11] = Vi
  V[34, 0:11] = Vi
  V[41, 10:21] = Vi
  V[8:15, 10] = Vi
  V[34:41, 10] = Vi
  V[15:35, 0] = Vi
  V[:,-1] = Vi  
  V[-1,20:300] = Vi  
  V[23:27,15:299] = V0
  # critère de convergence
  ecart = np.max(np.abs(V-Vavant))

# fin du calcul - affichage de la durée
print ('Nombre iterations : ',iteration  )  
print ('Temps de calcul (s) : ',time.time() - tdebut)

v, u = np.gradient(V)

x = np.linspace(0, 1, N) # x-axis
y = np.linspace(0, 1, M) # y-axis

X, Y= np.meshgrid(x,y)

# Create figure
fig, axarr = plt.subplots(1, 1, figsize=(14, 8), dpi=300)

# Axes 1: Electric potential and field
im1 = axarr[0].contourf(x, y, V, 20)
axarr[0].streamplot(x, y, -u, -v, color="k")
axarr[0].set_title("Electric potential and field")
fig.colorbar(im1, orientation='horizontal', ax=axarr[0],
                  label=r"Electric potential, $V/V_0$")
axarr[0].set_xlabel("$x/L$")
axarr[0].set_ylabel("$y/L$")

but I have a problem in the affichage method the error is in line 90:

line 90, in <module>
    im1 = axarr[0].contour(x, y, V, 20)
TypeError: 'AxesSubplot' object is not subscriptable

I almost tried everything but I couldn't find a solution. How can I fix this error?

Upvotes: 1

Views: 795

Answers (1)

William Miller
William Miller

Reputation: 10328

matplotlib.pyplot.subplots returns either an array of matplotlib.axes.Axes instances, or a single matplotlib.axes.Axes instance. The latter only when the number of rows (the first positional argument) and the number of columns (second positional argument) are both 1. So the syntax

fig, ax = plt.subplots(1, 1)

creates a single Axes instance (also referred to as an AxesSubplot), therefore ax[0] is illegal - it would be valid only if multiple subplots were created by plt.subplots. The correct syntax in your code is 

fig, axarr = plt.subplots(1, 1, figsize=(14, 8), dpi=300)

im1 = axarr.contourf(x, y, V, 20)
axarr.streamplot(x, y, -u, -v, color="k")
axarr.set_title("Electric potential and field")
fig.colorbar(im1, orientation='horizontal', ax=axarr,
                  label=r"Electric potential, $V/V_0$")
axarr.set_xlabel("$x/L$")
axarr.set_ylabel("$y/L$")

Upvotes: 0

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