My Math Forum Proofing Gauss Law problem..

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 September 10th, 2017, 01:55 PM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 169 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: USA Posts: 1,781 Thanks: 919 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 $Q = \epsilon_0 ( 8 )(3) = 24\epsilon_0$ Thanks from SenatorArmstrong and zollen 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: 169 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: USA Posts: 1,781 Thanks: 919 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
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What is the $\displaystyle \nabla$ stands for?

$\displaystyle \int_S E\cdot n ~dS = \int_V \nabla \cdot E ~dV$

$\displaystyle \nabla \cdot E = 1+1+1 = 3$

Quote:
 Originally Posted by romsek This looks like a direct evaluation of the surface integral. The problem asks you to apply Gauss's law.

Last edited by zollen; September 11th, 2017 at 04:14 PM.

September 11th, 2017, 06:06 PM   #6
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Quote:
 Originally Posted by zollen What is the $\displaystyle \nabla$ stands for? $\displaystyle \int_S E\cdot n ~dS = \int_V \nabla \cdot E ~dV$ $\displaystyle \nabla \cdot E = 1+1+1 = 3$
It's called the Divergence of E, or $\nabla \cdot E$

$\nabla \cdot E = \dfrac{\partial E_x}{\partial x}+\dfrac{\partial E_y}{\partial y}+\dfrac{\partial E_z}{\partial z}$

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: 2,966 Thanks: 807 $\displaystyle \nabla$ 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. Thanks from romsek and zollen

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