# Using Gauss's Law to Calculate Charge Density Function

• james weaver
In summary, there is an issue with the units in the given problem, as the units for the charge density function are not consistent with the units for the constant a. However, it is shown that c/m^4 does make physical sense and the meaning of the unit (m/s)/s is explained to clarify this. The charge density changes by 5C/m³ per metre as you move through the slab in the z-direction.
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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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: ?

kuruman
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

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.

TSny, hutchphd, james weaver and 1 other person
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.

Steve4Physics

## What is Gauss's Law?

Gauss's Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface. It is one of Maxwell's equations and is used to calculate the electric field due to a distribution of charge.

## How is Gauss's Law used to calculate charge density function?

Gauss's Law can be used to calculate the charge density function by integrating the electric field over a closed surface surrounding the charge distribution. The charge density function is equal to the enclosed charge divided by the volume enclosed by the surface.

## What is the relationship between charge density function and electric field?

The charge density function and electric field are directly related through Gauss's Law. The electric field at a point is proportional to the charge density at that point, with the proportionality constant being the permittivity of free space.

## Can Gauss's Law be used for any charge distribution?

Yes, Gauss's Law can be used for any charge distribution as long as the charge is enclosed by a closed surface. This includes point charges, line charges, and surface charges.

## What are some practical applications of using Gauss's Law to calculate charge density function?

Gauss's Law is used in many practical applications, including designing electrical circuits, calculating the electric field inside a capacitor, and analyzing the electric potential of a conducting sphere. It is also used in engineering and physics research to study the behavior of electric fields in different scenarios.

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