from scipy import *
import matplotlib.pyplot as plt
def plot_elines(x_grid, y_grid, potential, field):
fig, ax = plt.subplots(figsize=(13, 13))
im_cs = ax.contour(x_grid, y_grid, potential, 18, cmap='inferno')
plt.clabel(im_cs, inline=1, fontsize=7)
ax.quiver(x_grid[::3, ::3], y_grid[::3, ::3],
field[0, ::3, ::3], field[1, ::3, ::3],)
ax.set_xlabel("$x$")
ax.set_ylabel("$y$")
plt.show()
# define q configuration (x,y positions)
charges = [
[1, 1],
]
xx, yy = meshgrid(linspace(-4, 4), linspace(-4, 4))
# potential function
e_pot = 0.
for idx, q in enumerate(charges):
dx, dy = xx-q[0], yy-q[1]
rr = hypot(dx, dy)
e_pot += 1/(4*pi) * 1./rr
e_field = gradient(-e_pot)
e_field /= hypot(e_field[0], e_field[1]) * 5
# why is this needed?
e_field[0] = e_field[0].T
e_field[1] = e_field[1].T
plot_elines(xx, yy, e_pot, e_field)
I have a query concerning the usage of the gradient functionality from numpy/scipy.
In my current visualization, I am displaying the electric field equipotential lines and the field vectors for a single positive charge. According to the formula:
E = -grad(V)
The direction of the field vectors (quiver) and equipotential lines (contour) should be orthogonal at all points in space, directed away from the positive charge. However, when calculating E using scipy's gradient function, I noticed that transposing the x-y grid output is necessary for correct results.
When comparing the two outputs with and without .T:
vs
Why is the transpose operation required? Is there a mistake in how I am plotting the data?
Thank you.