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Gauss' Law and electric flux

  • Thread starter Joules23
  • Start date
33
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1. Homework Statement
A charge Q is located inside a rectangular box. The electric flux through each of the six surfaces of the box is: Φ1=+1500 Φ2=+2200 Φ3=+4600 Φ4=-1800 Φ5=-3500 Φ6=-5400.
(unit: N x m^2/C)
What is Q?


2. Homework Equations

ΦE =Q/ε

3. The Attempt at a Solution
Add up all the Φ's to get -2400, then use ε0Φ=Q

(-2400)(8.85x10^-12) = -2.124 x 10^-8 C
 
Last edited:

Kurdt

Staff Emeritus
Science Advisor
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What are you supposed to do? Find Q? ε is a constant usually denoted [tex] \epsilon_0[/tex].
 
33
0
yes im supposed to find Q

Hows this:
Add up all the Φ's to get -2400, then use ε0Φ=Q

(-2400)(8.85x10^-12) = -2.124 x 10^-8 C
 
Last edited:
218
0
yes im supposed to find Q .. and [tex] \epsilon_0[/tex], accordning to my book is the permittivity of free space, and really doesnt say much else about it
[tex]\epsilon[/tex] is the permittivity of the medium enclosed by the surface which, in this case is the rectangular box. If the medium is air than it is equal to the permittivity of free space denoted by [tex]\epsilon_0[/tex]. For more info about permittivity see http://en.wikipedia.org/wiki/Permittivity


You should start by thiking what does Gauss' law state? What does each quantity in the formula of the law represent?

EDIT: looks like you solved it
 
Last edited:
33
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is that right? i thought i was supposed to get zero?
 
33
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I was reading this, and it seems similar to my problem

Picture1.png
 
Last edited:
218
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In this problem the total flux through the surface of the box is 0, so the electric charge inside it will be 0.

In your problem the total flux is non-zero so the charge inside will be non-zero.

This is exactly what Gauss' law is all about.
 

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