Using Gauss's Law to Calculate Charge Density Function

AI Thread Summary
The discussion centers on the application of Gauss's Law to determine the charge density function, specifically r(z) = az, where z represents the perpendicular distance within a plane. Participants identify issues with unit consistency, particularly questioning the units of the constant a and its implications for the volume charge density (ρ). It is clarified that if a is in units of C/m^4, it aligns with the expected volumetric charge density when multiplied by z, which is in meters. The conversation emphasizes the importance of understanding how units relate to physical quantities, ultimately concluding that the units can indeed make sense in the context of the problem. The discussion highlights the necessity of careful unit analysis in physics calculations.
james weaver
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Homework Statement
find E field of infinite plane with thickness T and charge density function az
Relevant Equations
.
I've attached what I have so far. Used Gauss's law, everything seemed to make sense except the units don't work out in the end. The charge density function if given by: r(z)=az, where z is the perpendicular distance inside the plane.
 

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james weaver said:
... everything seemed to make sense except the units don't work out in the end.
Why do you think there is a problem with units?
What are the units of ##a##?
What are the units of ##aT^2##?

Also, you seem to have lost a '2' in an intermediate step in your working - but then re-incorporated it in the final answer.
 
Sorry, forgot to mention. I believe a is units of c/m^3. In which case my final answer is in coloumb-meters and not coloumbs. Still waiting on confirmation for the units.
 
james weaver said:
I believe a is units of c/m^3
Volume charge density (##\rho##) is measured in units of ##\text {Cm}^{-3}##.
##z## is measured in units of ##\text m##.
##\rho = az##
Therefore the units of ##a## are: ?
 
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Well it wouldn't make any sense if a were in units of c/m^4, would it?
 
james weaver said:
Well it wouldn't make any sense if a were in units of c/m^4, would it?
What do you get when you multiply C⋅m-4 (proposed units of a) with m (units of z)?
 
I would get c/m^3, which would indeed be volumetric charge density. But I don't understand how c/m^4 makes any physical sense.
 
james weaver said:
I would get c/m^3, which would indeed be volumetric charge density. But I don't understand how c/m^4 makes any physical sense.
Would units other than c/m^4 for the constant ##a## make more physical sense to you if they gave units other than c/m^3 for the volume charge density ##\rho##?
 
No I suppose not. I guess a would have to be in units of c/m^4? In which case my final answer for E would be correct. I just assumed c/m^4 made no sense since we live in 3d space
 
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james weaver said:
No I suppose not. I guess a would have to be in units of c/m^4? In which case my final answer for E would be correct. I just assumed c/m^4 made no sense since we live in 3d space
It makes perfect sense! Try this...

Velocity is measured in units of m/s. If velocity changes from 0m/s to 10m/s over the interval t=0s to t=5s then the (average) acceleration is:
##a_{average} = \frac {\Delta v}{\Delta t} = \frac {10m/s - 0m/s}{5s - 0s} = \frac {10m/s}{5s} = 2 (m/s)/s##

Note the meaning of the unit (m/s)/s - it is how much the velocity (in m/s) changes per second. Of course (m/s)/s is usually written as m/s².

Now suppose you have an infinite ‘slab’ of material, with a = 5C/m⁴.

5C/m⁴ means the same as (5C/m³)/m. The charge density changes by 5C/m³ per metre as you move through the slab in the z-direction.
 
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Steve4Physics said:
It makes perfect sense! Try this...

Velocity is measured in units of m/s. If velocity changes from 0m/s to 10m/s over the interval t=0s to t=5s then the (average) acceleration is:
##a_{average} = \frac {\Delta v}{\Delta t} = \frac {10m/s - 0m/s}{5s - 0s} = \frac {10m/s}{5s} = 2 (m/s)/s##

Note the meaning of the unit (m/s)/s - it is how much the velocity (in m/s) changes per second. Of course (m/s)/s is usually written as m/s².

Now suppose you have an infinite ‘slab’ of material, with a = 5C/m⁴.

5C/m⁴ means the same as (5C/m³)/m. The charge density changes by 5C/m³ per metre as you move through the slab in the z-direction.
Ahhh, that's very helpful. Thank you.
 
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