# Using Gauss's Law to Calculate Charge Density Function

• james weaver

#### james weaver

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.

#### Attachments

• hw3_1.pdf
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... 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.

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: ?

• kuruman
Well it wouldn't make any sense if a were in units of c/m^4, would it?

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.

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

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.

• TSny, hutchphd, james weaver and 1 other person
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.

• Steve4Physics