Help understanding the units in Gauss's law

In summary, the conversation revolved around a homework question involving finding the charge Q located inside a rectangular box with given electric flux values for each surface. The person asking the question was unsure of how to use Gauss's law to find the charge and was confused about the units. Through guidance, they were able to correctly calculate the charge by using the correct constant, εo, and understanding the units involved.
  • #1
goldenwest
8
0

Homework Statement



Well, here's the homework question:

A charge Q is located inside a rectangular box. The electric flux through each of the six surfaces of the box is: phi1=+1500Nm^2/C, phi2=+2200Nm^2/C, phi3=+4600Nm^2/C, phi4=-1800Nm^2/C, phi5=-3500Nm^2/C, and phi6=-3400Nm^2/C. What is Q?

My guess is that I will have to use Gauss's law (electric flux = enclosed charge over electric constant) to find out the charge for each number and then find the mean of all the charges? But what's really confusing me is that Gauss's law's units don't seem to cancel out properly. Please show me why I'm an idiot.

Homework Equations


confusion1.gif

seems like it should be
confusion2.gif

...why isn't it?

The Attempt at a Solution


My attempt at a solution would be
+1500 * 8.99E9 = 1.35E13
+2200 * 8.99E9 = 1.98E13
+4600 * 8.99E9 = 4.14E13
-1800 * 8.99E9 =-1.62E13
-3500 * 8.99E9 =-3.14E13
-3400 * 8.99E9 =-3.06E13

The mean of those numbers = -5.83E11 = Q?
 
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  • #3
Okay so [tex](\sum\phi)\epsilon = Q[/tex]? Q = -2.16E13?
 
  • #4
goldenwest said:
Okay so [tex](\sum\phi)\epsilon = Q[/tex]? Q = -2.16E13?

Are you sure that's the sum of the Φ ?
 
  • #5
Oops, -3.60E12... but am I doing it correctly now?
 
  • #6
goldenwest said:
Oops, -3.60E12... but am I doing it correctly now?

So correctly that it looks like you got it right.
 
  • #7
Okay, thank you very very very very much. I'm still confused about why the units work out that way - could you explain that to me?
 
  • #8
ε is in units of C²/Nm²

Φ is in units of Nm²/C

Q is in units of C

Edit: Sorry. I posted the units of Coulombs Constant which is the inverse.
 
Last edited:
  • #9
Yes, I know. I know this is probably a stupid question... please bear with me.

[tex]\stackrel{Q}{\overline{\epsilon}} = \stackrel{C}{\overline{Nm^{2}C^{-2}}} = \stackrel{C^{3}}{\overline{Nm^{2}}}[/tex]

Why is that wrong?
 
  • #10
Sorry. I posted the units of Coulombs Constant which is the reciprocal of the units of ε.

I corrected my error.

Electrical flux has SI units of volt metres (V m), or, equivalently, Newton metres squared per coulomb (N m² C−1).
http://en.wikipedia.org/wiki/Electric_flux
 
  • #11
LowlyPion said:
Edit: Sorry. I posted the units of Coulombs Constant which is the inverse.

Oooooooooook, now I understand. Thank you very much for all your help!
 
  • #12
So was I using the wrong constant when solving for the problem... meaning I got the answer wrong? I was using 8.99E9. Should I be using 8.85E-12?
 
  • #13
goldenwest said:
So was I using the wrong constant when solving for the problem... meaning I got the answer wrong? I was using 8.99E9. Should I be using 8.85E-12?

Yes. I missed that too I guess. I should have checked it.

εo is what you want, and not k.
 
  • #14
Thanks again, you have been extremely helpful.
 

Related to Help understanding the units in Gauss's law

What is Gauss's law?

Gauss's law is a fundamental law in physics that describes the relationship between the electric field and electric charge. It states that the electric flux through a closed surface is proportional to the enclosed electric charge.

What are the units used in Gauss's law?

The units used in Gauss's law depend on the system of units being used. In SI units, the unit of electric charge is coulombs (C) and the unit of electric field is newtons per coulomb (N/C). In Gaussian units, the unit of electric charge is statcoulombs (statC) and the unit of electric field is statvolts per centimeter (statV/cm).

How do I understand the units in Gauss's law?

The units in Gauss's law can be understood by using the equation E∙A = q/ε0, where E is the electric field, A is the surface area, q is the enclosed electric charge, and ε0 is the permittivity of free space. By manipulating this equation, you can see that the units of electric field are N/C or statV/cm, the units of surface area are m2 or cm2, and the units of electric charge are C or statC.

Why is understanding the units in Gauss's law important?

Understanding the units in Gauss's law is important because it allows for the application of the law in various systems of units and helps in making accurate calculations. It also helps in understanding the relationship between electric charge and electric field and their corresponding units.

What are some examples of using Gauss's law in real life?

Gauss's law has many practical applications in our daily lives, such as calculating the electric field around a charged particle, designing electrical circuits, and determining the electric field inside a capacitor. It is also used in industries like electronics, telecommunications, and power generation.

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