September 10th, 2017, 01:55 PM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 120 Thanks: 2  Proofing Gauss Law problem..
Gauss's Law says that the net charge, Q, enclosed by a closed surface, S, is Q = ϵ0 ∫∫ E * N dS where E is an electric field and ϵ0 (the permittivity of free space) is a known constant; N is oriented outward. Use Gauss's Law to find the charge contained in the cube with vertices (+/1, +/1, +/1) if the electric field is E = < x, y, z >. Official Answer: 24ϵ0 My solution $\displaystyle < x, y, z > * < 4, 0, 0 > + < x, y, z > * < 4, 0, 0 > + < x, y, z > * < 0, 4, 0 > + < x, y, z > * < 0, 4, 0 > + < x, y, z > * < 0, 0, 4 > + < x, y, z > * < 0, 0, 4 > $ I think I am on the right direction, but not sure how to continue... 
September 10th, 2017, 05:48 PM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,410 Thanks: 715 
Gauss's law says $\displaystyle \int_S E\cdot n ~dS = \int_V \nabla \cdot E ~dV$ $\nabla \cdot E = 1+1+1 = 3$ Integrating a constant over the box is just the constant times the volume of the box. the box measures 2 units to a side so has volume $2^3 = 8$ so $Q = \epsilon_0 ( 8 )(3) = 24\epsilon_0$ Last edited by romsek; September 10th, 2017 at 05:50 PM. 
September 11th, 2017, 02:56 PM  #3 
Senior Member Joined: Jan 2017 From: Toronto Posts: 120 Thanks: 2 
I got it! $\displaystyle ϵ0 \int_{1}^{1} \int_{1}^{1} < 1, u, v > * < 1, 0, 0 > ~dvdu + ϵ0 \int_{1}^{1} \int_{1}^{1} < 1, u, v > * < 1, 0, 0 > ~dvdu + ϵ0 \int_{1}^{1} \int_{1}^{1} < u, v, 1 > * < 0, 0, 1 > ~dvdu + ϵ0 \int_{1}^{1} \int_{1}^{1} < u, v, 1 > * < 0, 0, 1 > ~dvdu + ϵ0 \int_{1}^{1} \int_{1}^{1} < u, 1, v > * < 0, 1, 0 > ~dvdu + ϵ0 \int_{1}^{1} \int_{1}^{1} < u, 1, v> * < 0, 1, 0 > ~dvdu $ = (4 + 4 + 4 + 4 + 4 + 4) ϵ0 
September 11th, 2017, 03:07 PM  #4 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,410 Thanks: 715 
This looks like a direct evaluation of the surface integral. The problem asks you to apply Gauss's law. 
September 11th, 2017, 04:12 PM  #5 
Senior Member Joined: Jan 2017 From: Toronto Posts: 120 Thanks: 2 
What is the $\displaystyle \nabla $ stands for? $\displaystyle \int_S E\cdot n ~dS = \int_V \nabla \cdot E ~dV $ $\displaystyle \nabla \cdot E = 1+1+1 = 3 $ Last edited by zollen; September 11th, 2017 at 04:14 PM. 
September 11th, 2017, 06:06 PM  #6  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,410 Thanks: 715  Quote:
$\nabla \cdot E = \dfrac{\partial E_x}{\partial x}+\dfrac{\partial E_y}{\partial y}+\dfrac{\partial E_z}{\partial z}$ You really should have seen this before getting into Gauss's law.  
September 11th, 2017, 06:11 PM  #7 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,649 Thanks: 681 
$\displaystyle \nabla$ is the "vector differential operator", $\displaystyle \frac{\partial}{\partial x}\vec{i}+ \frac{\partial}{\partial y}\vec{j}+ \frac{\partial}{\partial z}\vec{k}$. Since there are three different kinds of vector multiplication there are three different ways $\displaystyle \nabla$ can be applied. The scalar product: $\displaystyle \nabla f$ (also called "grad f" or "gradient f") is $\displaystyle \frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}+ \frac{\partial f}{\partial z}\vec{k}$. You can think of this as the scalar f multiplying the vector $\displaystyle \nabla$ 2) the dot product: $\displaystyle \nabla\cdot \vec{f}$ (also called "div f" or "divergence f") is $\displaystyle \frac{\partial f_x}{\partial x}+ \frac{\partial f_y}{\partial y}+ \frac{\partial f_z}{\partial z}$. Notice that $\displaystyle \vec{f}= f_x\vec{i}+ f_y\vec{j}+ f_z\vec{k}$ is a vector valued function and $\displaystyle \nabla f$ is a scalar valued function. 3) The cross product: $\displaystyle \nabla\times \vec{f}$ (also called "curl f") is $\displaystyle \left(\frac{\partial f_y}{\partial z} \frac{\partial f_z}{\partial y}\right)\vec{i}$$\displaystyle + \left(\frac{\partial f_x}{\partial z} \frac{\partial f_z}{\partial x}\right)\vec{j}$$\displaystyle + \left(\frac{\partial f_y}{\partial x} \frac{\partial f_x}{\partial z}\right)\vec{k}$. Here $\displaystyle \vec{f}$ is a vector valued function and $\displaystyle \nabla\times f$ is a vector valued function. 

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gauss, law, problem, proofing 
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