Coulomb's law for a one-dimensional source:¶
Using the notation of Griffiths, the electric field at observation point ${\bf r}$ due to a one-dimensional charge distribution is
$$ {\bf E}({\bf r}) = \frac{1}{4\pi\epsilon_0}\, \int\frac{{\bf s}}{s^3}\, \lambda({\bf r}^\prime)\, d\ell^\prime, \tag{1} $$where $\lambda$ Is the linear charge density of the 1-d source, $d\ell^\prime$ is an infinitesimal length along the 1-d source, and ${\bf s}$ is a stand-in for Griffiths' script-r vector from source point $dq=\lambda\, d\ell^\prime$ at position ${\bf r}^\prime$ to the observation point at ${\bf r}$:
\begin{equation} {\bf s} \equiv {\bf r} - {\bf r}^\prime. \tag{2} \end{equation}Equation (1) is just Griffiths Eq. (2.6), with the substition
$$ \frac{\hat{\bf s}}{s^2} \longrightarrow \frac{\bf s}{s^3}. \tag{3} $$For 1-d sources it is appropriate to parametrize the path in terms of a single scalar variable which I call $t$:
$$ {\boldsymbol \ell}^\prime(t) = x^\prime(t)\, \hat{{\bf x}} + y^\prime(t)\, \hat{{\bf y}} + z^\prime(t)\, \hat{{\bf z}}, \tag{4} $$so $$ {\bf s} \longrightarrow {\bf r} - {\boldsymbol \ell}^\prime(t), \tag{5} $$
and the linear charge density becomes
$$ \lambda \longrightarrow \lambda(t). \tag{6} $$We also have
$$ d{\boldsymbol \ell}^\prime = \left( \frac{dx^\prime}{dt}\, \hat{\bf x} + \frac{dy^\prime}{dt}\, \hat{\bf y} + \frac{dz^\prime}{dt}\, \hat{\bf z}\right)\, dt \tag{7} $$which gives
$$ d\ell^\prime = \sqrt{\left(\frac{dx^\prime}{dt}^2 + \frac{dy^\prime}{dt}^2 + \frac{dz^\prime}{dt}^2\right)}\, dt \tag{8} $$It is convenient to work with dimensionless parameters defined in terms of some length $R$ and charge $Q$ characterizing the source:
$$ {\bf r}^\ast \equiv \frac{\bf r}{R} \quad\quad {\boldsymbol \ell}^\ast \equiv \frac{\boldsymbol \ell}{R}\quad\quad {\bf s}^\ast \equiv \frac{\bf s}{R}\quad\quad \lambda^\ast = \frac{\lambda}{Q/R}\quad\quad \mbox{and} \quad\quad{\bf E}^\ast \equiv \frac{{\bf E}}{Q/(4\pi \epsilon_0 R^2)}. \tag{9} $$Coulomb's law for 1-d sources written in terms of these parameters is
\begin{eqnarray*} {\bf E}^\ast &=& \int \frac{{\bf s}^\ast}{{s^\ast}^3}\lambda^\ast\, d{l^\prime}^\ast \\ &=& \int_{t_1}^{t_2} \frac{{\bf s}^\ast}{{s^\ast}^3} \lambda^\ast(t)\, \sqrt{\left(\frac{d{x^\prime}^\ast}{dt}^2 + \frac{d{y^\prime}^\ast}{dt}^2 + \frac{d{z^\prime}^\ast}{dt}^2\right)}\, dt \tag{10} \end{eqnarray*}This is really three separate integrals, one for each component.
All calculations in what follow will be in terms of dimensionless variables, asterisks will
be implicit. The dimensional electric
field can be recovered by multiplying the numerical results below by the factor $Q/(4\pi\epsilon_0 R^2)$.
This notebook uses the quad_vec method from the integrate submodule of scipy
for numerical integration. There are other choices, but this works well (and I like to use scipy
tools when they're available).
Marty Ligare, November 2022, March 2024
import numpy as np
from scipy import integrate
import numdifftools as nd
import matplotlib as mpl
import matplotlib.pyplot as plt
# %matplotlib notebook # Use this for notebooks before Notebook 7
# If used in Notebook 7, graphics calls return "Javascript Error: IPython is not defined"
#
# Use following for Notebook 7. See https://github.com/jupyter/notebook/issues/7115
%matplotlib widget
mpl.style.use('classic')
# M.L. modifications of matplotlib defaults using syntax of v.2.0
# More info at http://matplotlib.org/2.0.0/users/deflt_style_changes.html
# Changes can also be put in matplotlibrc file, or effected using mpl.rcParams[]
plt.rc('figure', figsize=(6, 4.5)) # Reduces overall size of figures
plt.rc('axes', labelsize=16, titlesize=14)
# Adjusts supblot parameters for new size
plt.rc('figure', autolayout=True)
def integrand(t, x, y, z):
'''
Integrand of Eq. (10) above.
Inputs are 1) the parameter "t" used in Eqs. (4-8) above, and the cartesian coordinates x, y, and z of the obesrvation
point. Individual cartesion coordinats are used (rather than an array representing a vector for ease of passing of 'args'
for the numerical integration done by scipy's integrate.quad_vec().
Return is an infinitesimal vector field element in represented in a numpy array (with cartesian coordinates as its elements).
'''
r = np.array([x,y,z]) # numpy array representing observation point vector
s = r - l(t) # numpy array representing vector from source point to observation point.
dE = s/np.dot(s,s)**(3/2)*lam(t)*np.sqrt(np.dot(dl(t),dl(t)))
return dE # infinitesimal field vector returned as numpy array
def e(r):
'''
Numerical integration of Eq. (10) above.
Input is a numpy array representing the observation point vector.
Return is the vector field represented in a numpy array (with cartesian coordinates as its elements).
'''
#x = r[0]
#y = r[1]
#z = r[2]
a = integrate.quad_vec(integrand, t1, t2, args=tuple(r))
return(a[0])
def ex(x, y, z):
'''
Isolation of the x-component of the vector field returned by function e(r).
This is useful when creating meshgrids necessary for use in matplotlib plotting of field vectors.
'''
r = np.array([x,y,z])
return e(r)[0]
def ey(x, y, z):
'''
Isolation of the y-component of the vector field returned by function e(r).
This is useful when creating meshgrids necessary for use in matplotlib plotting of field vectors.
'''
r = np.array([x,y,z])
return e(r)[1]
def ez(x, y, z):
'''
Isolation of the z-component of the vector field returned by function e(r).
This is useful when creating meshgrids necessary for use in matplotlib plotting of field vectors.
'''
r = np.array([x,y,z])
return e(r)[2]
Circle of charge; extending Griffiths Problem 2.5¶
This is the same source as considered in the notebook coulomb_1D_circle_1. The difference is the way that the differential line element is determined. In coulomb_1D_circle_1 the derivatives are taken 'by hand' and then rendered into code. In this notebook, coulomb_1D_circle_2, the derivatives are determined numerically using the nd.Dervative() function from the numdifftools package.¶
Characterizing the source¶
Charge $Q$ uniformly distributed on a ring of radius $R$ in the $x$-$y$ plane. An obvious choices for the
characteristic length and charge is $R$ and $Q$.
One way to parametrize of charge (in dimensionless variables) is
giving
$$ d{\boldsymbol \ell} = \left[-\pi \sin(\pi t) dt\, \hat{\bf x} + \pi \cos(\pi t)\, \hat{{\bf y}} \right]\, dt. \tag{12} $$The dimensionless charge density is
$$ \lambda^\ast = \frac{\lambda}{Q/R} = \frac{Q/2\pi R}{Q/R} = \frac{1}{2\pi}. \tag{13} $$NOTE: In this example it is easy to determine $d{\boldsymbol \ell}$. In a later example I will show how
to find the derviative numerically with the numdifftools package. (For an an intro
to numdifftools see https://www.eg.bucknell.edu/~phys310/jupyter/numdiff.html, or, in Jupyter notebook form, https://www.eg.bucknell.edu/~phys310/jupyter/numdiff.ipynb.)
# limits for the paramter t; see Eqs. (11&12) above
t1 = 0
t2 = 2
def l(t):
'''
Parameterization of linear charge distribution position vector
See Eq. (11) above.
'''
lx = np.cos(np.pi*t)
ly = np.sin(np.pi*t)
lz = 0
return np.array([lx, ly, lz])
def dl(t):
'''Differential vector line element.
See Eq. (12) above.
Use numdifftools to calculate components of dfferential line element vector.
'''
dl = nd.Derivative(l)
return np.array([dl(t)[0], dl(t)[1], dl(t)[2]])
def lam(t):
'''
Linear charge density.
See Eq. (13) above.
'''
return 1/2/np.pi
r = np.array([0,0,1])
e(r)
array([7.96454916e-16, 1.36002321e-15, 3.53553391e-01])
Computational non-dimensional ${\bf E}$:
$$ {\bf E}^\ast \simeq 0.353555\, \hat{\bf z} $$Computational dimensional field: $$ {\bf E} \simeq \frac{Q}{4\pi\epsilon_0 R^2} \times {\bf E}^\ast = \frac{Q}{4\pi\epsilon_0 R^2}\times 0.35355\, \hat{\bf z} $$
Analytical result (solution of Griffiths problem 2.5): $$ {\bf E}(0,0,z) = \frac{1}{4\pi\epsilon_0}\, \frac{\lambda (2\pi R) z}{(R^2 + z^2)^{3/2}} \, \hat{\bf z} = \frac{1}{4\pi\epsilon_0 R^3}\, \frac{Q z}{(1 + (z/R)^2)^{3/2}} \, \hat{\bf z} $$
Non-dimensional $z^\ast = 1$ $\longrightarrow$ $z = R$, so the analytical result for this point is
$$ {\bf E}(0,0,R) = \frac{Q}{4\pi\epsilon_0 R^2}\, \frac{1}{{2}^{3/2}} \, \hat{\bf z} \simeq \frac{Q}{4\pi\epsilon_0 R^2}\, 0.353555\, \hat{\bf z} $$Visualizing the vector field.¶
z = np.linspace(1,3,3)
x = np.linspace(-2,2,5)
X, Z = np.meshgrid(x,z)
vec_ex = np.vectorize(ex)
vec_ez = np.vectorize(ez)
U, W = vec_ex(X, 0, Z), vec_ez(X, 0, Z)
plt.figure()
plt.quiver(X, Z, U, W)
plt.axis('equal')
plt.axhline(0)
plt.axvline(0)
plt.xlabel('$x$')
plt.ylabel('$z$')
plt.title('Electric field in $y=0$ plane from circle of charge')
xs = np.linspace(-1, 1, 2)
ys = 0*xs
plt.plot(xs, ys, color='r', lw = 3)
plt.ylim(-.5,4)
plt.grid()
Version Information¶
version_information is from J.R. Johansson (jrjohansson at gmail.com)
See Introduction to scientific computing with Python:
http://nbviewer.jupyter.org/github/jrjohansson/scientific-python-lectures/blob/master/Lecture-0-Scientific-Computing-with-Python.ipynb
for more information and instructions for package installation.
If version_information has been installed system wide (as it has been on Bucknell linux computers with shared file systems), continue with next cell as written. If not, comment out top line in next cell and uncomment the second line.
%load_ext version_information
#%install_ext http://raw.github.com/jrjohansson/version_information/master/version_information.py
version_information numpy, scipy, numdifftools, matplotlib
| Software | Version |
|---|---|
| Python | 3.11.7 64bit [GCC 11.2.0] |
| IPython | 8.20.0 |
| OS | Linux 4.9.0 9 amd64 x86\_64 with glibc2.28 |
| numpy | 1.26.4 |
| scipy | 1.11.4 |
| numdifftools | 0.9.41 |
| matplotlib | 3.8.0 |
| Sun Apr 07 19:50:35 2024 EDT | |