Electric Field and Plate story

Electricity and Magnetism Level 3

Find the electric field in the z z direction due to a plate of area charge density λ \lambda at a distance a a above the plate.. ∠ B = 40 ° \angle B=40° If your answer comes in the form of E ( k ⃗ ) = α λ ϵ 0 E(\vec{k})=\alpha \frac{\lambda}{\epsilon_{0}}
Find 100 α = ? 100 \alpha=?
Details and Assumptions
1) The triangle is isosceles .
2) A B = B C = 100 AB=BC=100
3) a = 0.5 a=0.5
4) ϵ 0 \epsilon_{0} is electric permittivity .
5) The triangle is in x y xy plane
6) Distance a a is measured from B B and is perpendicular to x y xy plane.
7) Cosider B as origin. Right side + x +x axis and in forward + y +y axis.


The answer is 5.555.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Steven Chase
May 16, 2020

I double-integrated numerically over the left side of the triangle, and then doubled the result. It took some extra computation to get good convergence, since the distance above the plane is so small relative to the dimensions of the plane. 

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import math

Num = 20000    # spatial resolution parameter

lam = 1.0      # area charge density
e0 = 1.0       # electric permittivity
k = 1.0/(4.0*math.pi*e0)  # coulomb constant

# Consier Point A to be the origin

theta = (70.0/180.0)*math.pi  # Angle A

AB = 100.0 # length of AB

x0 = AB*math.cos(theta)  # position of test point relative to Point A
y0 = AB*math.sin(theta)
z0 = 0.5

m = y0/x0     # slope of line AB in xy plane

dx = x0/Num   # x coordinate step

#########################################

# Initialize field and position

Ex = 0.0
Ey = 0.0
Ez = 0.0

x = dx
z = 0.0

# Integrate over left side of triangle to get Ez
# Then double the value

while x <= x0:   # sweep x from left side to middle of triangle

    ymax = m*x   # calculate ymax based on x position and slope

    dy = ymax/Num  # y coordinate step

    y = 0.0

    dA = dx*dy     # infinitesimal area
    dq = lam*dA    # infinitesimal charge

    while y <= ymax:

        Dx = x0 - x  # displacement vector from charge patch to test point
        Dy = y0 - y
        Dz = z0 - z

        D = math.sqrt(Dx**2.0 + Dy**2.0 + Dz**2.0)

        ux = Dx/D  # unit vector
        uy = Dy/D
        uz = Dz/D

        dE = k*dq/(D**2.0)  

        dEx = dE * ux # field contribution
        dEy = dE * uy
        dEz = dE * uz

        Ex = Ex + dEx
        Ey = Ey + dEy
        Ez = Ez + dEz

        y = y + dy

    x = x + dx


#########################################

Ez = 2.0*Ez

print Num
print Ez
print (100.0*Ez)

#########################################

>>> 
1000
0.048210166259
4.8210166259
>>> ================================ RESTART ================================
>>> 
2000
0.0588974248212
5.88974248212
>>> ================================ RESTART ================================
>>> 
4000
0.0566300243295
5.66300243295
>>> ================================ RESTART ================================
>>> 
5000
0.0563482288595
5.63482288595
>>> ================================ RESTART ================================
>>> 
10000
0.0547553255221
5.47553255221
>>> ================================ RESTART ================================
>>> 
20000
0.0549975993195
5.49975993195
>>>

1 Helpful 1 Interesting 1 Brilliant 0 Confused

@Steven Chase are you posting a question today? Please

A Former Brilliant Member - 1 year ago

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Perhaps, if I can think of something

Steven Chase - 1 year ago

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@Steven Chase in the meantime enjoy and post the solution of my new problem.

A Former Brilliant Member - 1 year ago

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@A Former Brilliant Member – Indeed, I'm working on it now

Steven Chase - 1 year ago

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@Steven Chase – @Steven Chase I have uploaded new one now. BTW do you copy your code from python or write it here in solutions whole code.??

A Former Brilliant Member - 1 year ago

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@A Former Brilliant Member – I just copy the code from Python

Steven Chase - 1 year ago

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