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Charge enclosed Guassian surface

by craka
Tags: charge, enclosed, guassian, surface
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craka
#1
Nov9-08, 08:52 PM
P: 20
1. The problem statement, all variables and given/known data
Question states "Total electric flux from a cubical box 0.28m on a side is 1.84x10^3 Nm^2/C. What is the charge enclosed by the box


2. Relevant equations
Gauss's Law

[tex]
\begin{array}{l}
\phi _E = \vec E \bullet \vec A \\
\oint {\vec E \bullet d\vec A = \frac{{Q_{enclosed} }}{{\varepsilon _0 }}} \\
[/tex]


3. The attempt at a solution


[tex]
\begin{array}{l}
Q_{enclosed} = (4\pi r^2 )\varepsilon _0 \phi _E \\
\varepsilon _0 = 8.85 \times 10^{ - 12} \\
\phi _E = 1.84 \times 10^3 \\
r = 1.2 \times 0.28 = 0.14 \\
Q_{enclosed} = 4.01 \times 10^{ - 9} Nm^2 /C \\
\end{array}
[/tex]

But the answer is given as 1.63*10^-8 Nm^2/C

What I have I done wrong here?
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G01
#2
Nov9-08, 09:22 PM
HW Helper
G01's Avatar
P: 2,688
I noticed the following mistakes at first glance:

1) You have a cubical box, yet you are using the surface area of a sphere: [itex]4\pi r^2[/itex]

You need to use the surface area of this cube, which is the sum of the area of each side.

2)The correct answer you give has the wrong units. The units should be C, not Nm^2/C. Check that answer to make sure it corresponds to this problem.

Are either of these the problem?
craka
#3
Nov9-08, 09:41 PM
P: 20
Sorry I do realise the units should be in Coulombs I just in autopilot when I was typing units in sorry.

As far as the Areas goes, that what I thought initially aswell.

When I calculated the surface area initially I did so for a cube.
Thus 6x^2 where x was side dimension of cube. ie the 0.28m

When I calculate that my final answer for charge is 7.66*10^-9 Coulombs

This is still not the answer given, which is 1.63*10^-8 Coulombs.
Am I wrong or the answer in the text?

borgwal
#4
Nov9-08, 10:43 PM
P: 369
Charge enclosed Guassian surface

You are given the *flux*, the area is totally irrelevant for this question.
craka
#5
Nov10-08, 12:52 AM
P: 20
Thanks borgwal, that made it pretty damn easy.

Qencl= permitivity of free space * electric flux
borgwal
#6
Nov10-08, 01:08 AM
P: 369
Yep, sometimes the easiest questions are the hardest :-)
G01
#7
Nov10-08, 12:08 PM
HW Helper
G01's Avatar
P: 2,688
Heh! It's always the little mistakes that get you. I don't know how I missed that. Sometimes I look at these forums right before bed, and I miss things I would otherwise see. Sorry about that!
craka
#8
Nov11-08, 12:02 AM
P: 20
No worries. It's just great that you people put your time in a forum such as this.
Thanks


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