Gauss' Law and electric flux

Joules23

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:
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Kurdt

Staff Emeritus
Gold Member
What are you supposed to do? Find Q? ε is a constant usually denoted $$\epsilon_0$$.

Joules23

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:

antonantal

yes im supposed to find Q .. and $$\epsilon_0$$, accordning to my book is the permittivity of free space, and really doesnt say much else about it
$$\epsilon$$ 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 $$\epsilon_0$$. 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:

Joules23

is that right? i thought i was supposed to get zero?

antonantal

is that right? i thought i was supposed to get zero?
Why would you get zero?

Joules23

I was reading this, and it seems similar to my problem Last edited:

antonantal

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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