Gauss's Law Exercise

Electricity and Magnetism Level pending

Define the following volume:

$x = r \cos \theta \\ y = r \sin \theta \\ 0 \leq z \leq 1 \\ 1 \leq r \leq 2 \\ 0 \leq \theta \leq 2 \pi$

Now consider the closed surface which is the boundary of the volume. It consists of four parts: a top washer, a bottom washer, an outer cylindrical portion, and an inner cylindrical portion. Suppose there is a charge $q$ at position $(x,y,z) = \Big( \frac{3}{2}, 0, \frac{2}{3} \Big)$ . Let the fluxes of the electric field through the four sub-surfaces be $(\phi_T,\phi_B,\phi_O,\phi_I)$ . Determine the following ratio:

$\frac{\phi_T \, \phi_B \, \phi_O \, \phi_I }{\phi_T + \phi_B + \phi_O + \phi_I }$

Details and Assumptions:
1) Electric permittivity $\epsilon_0 = 1$
2) $q = +10$
3) Use outward facing normal vectors for all sub-surfaces
4) All four fluxes are positive numbers

The answer is 2.29.

2 solutions

Karan Chatrath
Feb 17, 2020

As always, nice problem.

The position vector of the point charge is:

$\vec{r}_p = \frac{3}{2} \hat{i} + \frac{2}{3} \hat{k}$

Consider the outer cylindrical surface:

Position of a point on the outer surface is:

$\vec{r}_c = 2\cos{\theta} \hat{i} + 2\sin{\theta} \hat{j} + z\hat{k}$

$\vec{R} = \vec{r}_c - \vec{r}_p$

The electric field at this point on the surface due to the charge is:

$\vec{E} = \frac{kq\vec{R}}{\lvert \vec{R} \rvert^3}$

The outward normal surface area vector of this surface is:

$d\vec{S} = 2 \ d\theta \ dz\left(\cos{\theta} \hat{i} + \sin{\theta} \hat{j} \right)$

The elementary electric flux is:

$d\phi_O = \vec{E} \cdot d\vec{S} = -\frac{2160\,k\,\left(3\,\cos\left(\mathrm{\theta}\right)-4\right)}{{\left(36\,z^2-48\,z-216\,\cos\left(\mathrm{\theta}\right)+241\right)}^{3/2}}d\theta \ dz$

Repeat the exact process for the other three surfaces.

For the inner cylindrical surface:

$\vec{r}_c = \cos{\theta} \hat{i} + \sin{\theta} \hat{j} + z\hat{k}$ $d\vec{S} = - \ d\theta \ dz\left(\cos{\theta} \hat{i} + \sin{\theta} \hat{j} \right)$ $d\phi_I = \vec{E} \cdot d\vec{S} = \frac{1080\,k\,\left(3\,\cos\left(\mathrm{\theta}\right)-2\right)}{{\left(36\,z^2-48\,z-108\,\cos\left(\mathrm{\theta}\right)+133\right)}^{3/2}} \ d\theta \ dz$

For the top surface: $\vec{r}_c = r\cos{\theta} \hat{i} + r\sin{\theta} \hat{j} + \hat{k}$ $d\vec{S} = r \ dr \ d\theta \hat{k}$ $d\phi_T = \vec{E} \cdot d\vec{S} = \frac{720\,k\,r}{{\left(36\,r^2-108\,\cos\left(\mathrm{\theta}\right)\,r+85\right)}^{3/2}}\ dr \ d\theta$

For the bottom surface: $\vec{r}_c = r\cos{\theta} \hat{i} + r\sin{\theta} \hat{j} + \hat{k}$ $d\vec{S} = -r \ dr \ d\theta \hat{k}$ $d\phi_B = \vec{E} \cdot d\vec{S} = \frac{1440\,k\,r}{{\left(36\,r^2-108\,\cos\left(\mathrm{\theta}\right)\,r+97\right)}^{3/2}}\ dr \ d\theta$

Finally:

$\phi_O = \int_{0}^{1} \int_{0}^{2\pi} -\frac{2160\,k\,\left(3\,\cos\left(\mathrm{\theta}\right)-4\right)}{{\left(36\,z^2-48\,z-216\,\cos\left(\mathrm{\theta}\right)+241\right)}^{3/2}}d\theta \ dz$

$\phi_I = \int_{0}^{1} \int_{0}^{2\pi} \frac{1080\,k\,\left(3\,\cos\left(\mathrm{\theta}\right)-2\right)}{{\left(36\,z^2-48\,z-108\,\cos\left(\mathrm{\theta}\right)+133\right)}^{3/2}} \ d\theta \ dz$

$\phi_T = \int_{1}^{2} \int_{0}^{2\pi} \frac{720\,k\,r}{{\left(36\,r^2-108\,\cos\left(\mathrm{\theta}\right)\,r+85\right)}^{3/2}}\ d\theta \ dr$

$\phi_B = \int_{1}^{2} \int_{0}^{2\pi} \frac{1440\,k\,r}{{\left(36\,r^2-108\,\cos\left(\mathrm{\theta}\right)\,r+97\right)}^{3/2}}\ d\theta \ dr$

$k = \frac{1}{4\pi}$

Simplifications are straightforward but tedious and time-consuming. Wolfram Alpha tackles the above four monsters. The sum of the fluxes adds up to 10 which verifies Gauss' Law. The answer comes out to be $\boxed{2.292}$ .

@Karan Chatrath Sir sorry for that.I think sir you are right in one step math is there excluding that step all physics is there. Please forgive me. How you write that expression outward normal surface area vector , I can't able to understand?? Please

A Former Brilliant Member - 1 year, 3 months ago

Oh, you have no reason to be sorry. I was just sharing my thoughts on the point you raised. In fact, you raised a valid point. There are many who get intimidated by mathematics while learning physics. My suggestions are against getting bogged down by mathematical details or notations.

To answer your question, I have attached a couple of links below. Hope they help. Pay attention to the point in the problem statement that says 'outward-facing normal vectors'. This will directly give you the directions of the area vectors. For the cylindrical surfaces, the directions are radial-outward (away from the central axis of the cylinder) for the outer surface and radial-inward (towards the central axis of the cylinder) for the inner surface. For the top surface, the vector is along the positive Z direction and for the bottom, it is along the negative Z direction.

http://citadel.sjfc.edu/faculty/kgreen/vector/block3/jacob/node4.html

https://www.youtube.com/watch?v=OTrh0VmbDO0

Karan Chatrath - 1 year, 3 months ago

Steven Chase
Feb 17, 2020

@Karan Chatrath has provided a nice detailed solution. I will post my code as well, since our styles are a bit different. One thing I like about the "step by step" approach in code is that I don't have to keep track of complicated expressions. I have printed results for one million partitions per sub-surface, and for one hundred million partitions per sub-surface.

import math

N = 10000     

q = 10.0
e0 = 1.0

k = 1.0/(4.0*math.pi*e0)

xT = 3.0/2.0
yT = 0.0
zT = 2.0/3.0

dtheta = 2.0*math.pi/N
dr = (2.0-1.0)/N
dz = (1.0 - 0.0)/N

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

# Top

nx = 0.0
ny = 0.0
nz = 1.0

phiT = 0.0

r = 1.0

while r <= 2.0:

    theta = 0.0

    while theta <= 2.0*math.pi:

        x = r*math.cos(theta)
        y = r*math.sin(theta)
        z = 1.0

        dS = r*dr*dtheta

        Dx = x - xT
        Dy = y - yT
        Dz = z - zT

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

        ux = Dx/D
        uy = Dy/D
        uz = Dz/D

        E = k*q/(D**2.0)

        Ex = E * ux
        Ey = E * uy
        Ez = E * uz

        dot = Ex*nx + Ey*ny + Ez*nz

        phiT = phiT + dot*dS

        theta = theta + dtheta

    r = r + dr

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

# Bottom

nx = 0.0
ny = 0.0
nz = -1.0

phiB = 0.0

r = 1.0

while r <= 2.0:

    theta = 0.0

    while theta <= 2.0*math.pi:

        x = r*math.cos(theta)
        y = r*math.sin(theta)
        z = 0.0

        dS = r*dr*dtheta

        Dx = x - xT
        Dy = y - yT
        Dz = z - zT

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

        ux = Dx/D
        uy = Dy/D
        uz = Dz/D

        E = k*q/(D**2.0)

        Ex = E * ux
        Ey = E * uy
        Ez = E * uz

        dot = Ex*nx + Ey*ny + Ez*nz

        phiB = phiB + dot*dS

        theta = theta + dtheta

    r = r + dr

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

# Outer

phiO = 0.0

z = 0.0
r = 2.0

while z <= 1.0:

    theta = 0.0

    while theta <= 2.0*math.pi:

        x = r*math.cos(theta)
        y = r*math.sin(theta)

        dS = r*dtheta*dz

        nx = x/r
        ny = y/r
        nz = 0.0

        Dx = x - xT
        Dy = y - yT
        Dz = z - zT

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

        ux = Dx/D
        uy = Dy/D
        uz = Dz/D

        E = k*q/(D**2.0)

        Ex = E * ux
        Ey = E * uy
        Ez = E * uz

        dot = Ex*nx + Ey*ny + Ez*nz

        phiO = phiO + dot*dS

        theta = theta + dtheta

    z = z + dz

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

# Inner

phiI = 0.0

z = 0.0
r = 1.0

while z <= 1.0:

    theta = 0.0

    while theta <= 2.0*math.pi:

        x = r*math.cos(theta)
        y = r*math.sin(theta)

        dS = r*dtheta*dz

        nx = -x/r
        ny = -y/r
        nz = 0.0

        Dx = x - xT
        Dy = y - yT
        Dz = z - zT

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

        ux = Dx/D
        uy = Dy/D
        uz = Dz/D

        E = k*q/(D**2.0)

        Ex = E * ux
        Ey = E * uy
        Ez = E * uz

        dot = Ex*nx + Ey*ny + Ez*nz

        phiI = phiI + dot*dS

        theta = theta + dtheta

    z = z + dz

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

print N
print ""
print phiT
print phiB
print phiO
print phiI
print ""
num = phiT * phiB * phiO * phiI
denom = phiT + phiB + phiO + phiI

print denom
print (num/denom)

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

#1000

#3.20988469921
#2.12667606634
#3.84825504409
#0.908807889705

#10.0936236993
#2.36526389424

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

#10000

#3.17109071653
#2.11175697673
#3.82215361712
#0.895633013466

#10.0006343239
#2.2922568955

I personally prefer a code based approach. However, going the symbolic route was a nice change.

Out of curiosity, how long does it take for your code to run? Whenever I solve multiple integrals such as these on my machine, a numerical resoloution of $N=10000$ is a long wait.

Karan Chatrath - 1 year, 3 months ago

Yeah, it took probably 20 minutes to run with $N = 10000$ . So I just do something else for a while as it runs. I think Python is a particularly slow program too, but I don't care. It would be interesting to run in C and compare the speed

Steven Chase - 1 year, 3 months ago

I think the long run time has more to do with processing capabilities of a machine than the choice of language. I will try this approach with MATLAB and see how long it takes, later, to know better.

Karan Chatrath - 1 year, 3 months ago

Gauss's Law Exercise

The answer is 2.29.

2 solutions

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