Gauss Law and Electrostatic Field

[Sami Lakka]

Homework Statement

Use Gauss' law and symmetry to find electrostatic as a function of position for an infinite plane of charge. Let the charge lie in the yz-plane and denote the charge per unit area by \rho=\alpha*e^{-abs(x/b)}

Homework Equations

Q=triple integral of density

The Attempt at a Solution

First compute the Q. After the integration I plan to apply Gauss law to get the electric field.

I first tried to convert the density to cylindrical coordinates and the tried to integrate the density using right circular cylinder. The triple integral was

\int\int\int\alpha*e^{-abs(r*cos(O)/b}*r*dr*dO*dz
The integration limits were r = [0,r], dO = [0,2*pi] and dz = [0,1]

However the integration becomes very difficult. Is there an easier way?

 

[alphysicist]

Hi Sami Lakka,

Homework Statement

Use Gauss' law and symmetry to find electrostatic as a function of position for an infinite plane of charge. Let the charge lie in the yz-plane and denote the charge per unit area by \rho=\alpha*e^{-abs(x/b)}

Homework Equations

Q=triple integral of density

The Attempt at a Solution

First compute the Q. After the integration I plan to apply Gauss law to get the electric field.

I first tried to convert the density to cylindrical coordinates and the tried to integrate the density using right circular cylinder. The triple integral was

\int\int\int\alpha*e^{-abs(r*cos(O)/b}*r*dr*dO*dz
The integration limits were r = [0,r], dO = [0,2*pi] and dz = [0,1]

However the integration becomes very difficult. Is there an easier way?

Here you are converting from Cartesian (x,y,z) coordinates to cylindrical (r,theta,z) coordinates. But there is no reason why the cylindrical z-axis has to be the same as the Cartesian z-axis.

In this case, the plane is in the y-z (cartesian) directions, and it changes in the x-direction, so try making the new z-axis the same direction as the original x-axis.

 

[tms]

Homework Statement

Use Gauss' law and symmetry to find electrostatic as a function of position for an infinite plane of charge. Let the charge lie in the yz-plane and denote the charge per unit area by \rho=\alpha*e^{-abs(x/b)}

How can the density of charge restricted to the y-z plane be a function of x? I think there must be some mistake in the statement of the problem.

 

[djeitnstine]

How can the density of charge restricted to the y-z plane be a function of x? I think there must be some mistake in the statement of the problem.

The charge lies on the yz plane but the density varies in x. Since the charge is planar where do the charge field lines extend? (Only in the x)

Sami Lakka If the charge is planar why use cylindrical coordinates? Use a cartesian system and the integration becomes absurdly simple.

 

[turin]

tms has a valid point. The expression for charge density (not field lines) is given as an x-dependent expression, which contradicts a previous statement in the problem that the charge lays in the yz-plane. Furthermore, the field lines will curve somewhat in the axial direction as well if the charge density is not uniform in the plane. Sami Lakka needs to rephrase the problem statement more precisely.

 

[Sami Lakka]

Homework Statement

Use Gauss' law and symmetry to find electrostatic as a function of position for an infinite plane of charge. Let the charge lie in the yz-plane and denote the charge per unit area by \rho=\alpha*e^{-abs(x/b)}

How can the density of charge restricted to the y-z plane be a function of x? I think there must be some mistake in the statement of the problem.

I made a mistake regarding the problem statement. It is an infinite slab of charge parallel to the yz plane.

 

[turin]

Sami Lakka, you still need to correct the expression for the charge density. Right now, it depends on x.

 

[kuruman]

If it is a slab, then it must have a thickness from some lower limit of x to some higher limit of x, say -d < x < +d. Is that the case? Otherwise the problem does not make sense as stated.

 

[turin]

A slab of charge with zero thickness makes no less sense than a point charge, which we use in physics all the time. A slab of charge can be modeled as a plane of charge with zero thickness, depending on the problem. This is not the issue that makes the problem nonsensical.

 

[kuruman]

A slab of charge with zero thickness makes no less sense than a point charge, which we use in physics all the time. A slab of charge can be modeled as a plane of charge with zero thickness, depending on the problem. This is not the issue that makes the problem nonsensical.

I agree, but the problem implies that the slab has finite thickness in the x-direction. If this were not the case and the slab is infinitely thin, then the correct volume charge density should be written in terms of a Dirac delta function, \rho(x)=a*e^-abs(x/b)*\delta(x). In either case, one can formally integrate over the volume to find the enclosed charge or whatever else is needed.

What does not make sense to me is what the problem wants. Find the electrostatic ... what?

 

[Redbelly98]

Let the charge lie in the yz-plane and denote the charge per unit area by https://www.physicsforums.com/latex_images/22/2286535-0.png

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The blue and red statements both imply an infinitesimally thin sheet of charge. But then the x-dependent equation contradicts that.

We need an exact, word-for-word statement of the problem before any useful help can be given.

 

[Sami Lakka]

The blue and red statements both imply an infinitesimally thin sheet of charge. But then the x-dependent equation contradicts that.

We need an exact, word-for-word statement of the problem before any useful help can be given.

Here is the problem statement from word to word. The problem is in Div, Grad, Curl and all that, p. 55, problem II-11

a) Use Gauss' law and symmetry to find the electrostatic field as a function of position for an infinite uniform plane of charge. Let the charge lie in the yz-plane and denote the charge per unit area by \sigma

b) Repeat part (a) for an infinite slab of charge parallel to the yz-plane whose density is given by
<br /> \rho(x)=\alpha when -b < x < b
\rho(x)=0 when abs(x) >= b


where \alpha and b are constants.

c) Repeat part (b) with <br /> \rho(x)=\alpha*e^{-abs(x/b)}<br />


I'm doing the c) part. The correct answer is E=\alpha*b/\epsilon_{0}*(1-e^{-abs(x)/b})\textbf{i} for x > 0 and -E for x < 0

Yesterday I actually managed to get the right result by changing to cylindrical coordinates where the cylinder runs through x-axis instead of the normal z-axis (thanks alphysicist!).
Then the triple integral is Q=\int\int\int\alpha*e^{-abs(x/b)}*r*dr*dO*dx
I used following integration limits x=[0,x], r=[0,r], O=[0,2*pi]. Then I multiplied the integral by two (because of the integration limit for x). The flux for that cylinder is 2*E(x)*pi*r² as only the cylinder's ends contribute to the flux. However I'm not sure about my reasoning.

Sorry for not posting the complete problem definition right from the start!

 

[Redbelly98]

Triple integrals aren't necessary here. Just integrate over x, using some fixed hypothetical area A that remains parallel to the yz-plane.

 

[kuruman]

Here is the problem statement from word to word. The problem is in Div, Grad, Curl and all that, p. 55, problem II-11

a) Use Gauss' law and symmetry to find the electrostatic field as a function of position for an infinite uniform plane of charge. Let the charge lie in the yz-plane and denote the charge per unit area by \sigma

b) Repeat part (a) for an infinite slab of charge parallel to the yz-plane whose density is given by
<br /> \rho(x)=\alpha when -b < x < b
\rho(x)=0 when abs(x) >= b


where \alpha and b are constants.

c) Repeat part (b) with <br /> \rho(x)=\alpha*e^{-abs(x/b)}<br />


I'm doing the c) part. The correct answer is E=\alpha*b/\epsilon_{0}*(1-e^{-abs(x)/b})\textbf{i} for x > 0 and -E for x < 0

Yesterday I actually managed to get the right result by changing to cylindrical coordinates where the cylinder runs through x-axis instead of the normal z-axis (thanks alphysicist!).
Then the triple integral is Q=\int\int\int\alpha*e^{-abs(x/b)}*r*dr*dO*dx
I used following integration limits x=[0,x], r=[0,r], O=[0,2*pi]. Then I multiplied the integral by two (because of the integration limit for x). The flux for that cylinder is 2*E(x)*pi*r² as only the cylinder's ends contribute to the flux. However I'm not sure about my reasoning.

Sorry for not posting the complete problem definition right from the start!

There are two regions where the electric field is different and you need to give expressions for both (call them E_I(x) and E_II(x): (I) outside the slab (x > b) and (II) inside the slab (x < b). If x is inside the slab, to find the enclosed charge you add all the contributions from zero to x as you have already done. If x is outside the slab, there is no more charge density beyond x = b, so what does this say about your limits of integration in x?

 

[turin]

If x is outside the slab, there is no more charge density beyond x = b, so what does this say about your limits of integration in x?

Sami Lakka is working part (c). There is no "outside the slab" for part (c), because the charge distribution fills all of space. For |x|>b, the charge distribution is ρ<αe, but not zero.

 

[turin]

The flux for that cylinder is 2*E(x)*pi*r² as only the cylinder's ends contribute to the flux. However I'm not sure about my reasoning.

Well, what is your reasoning? I don't disagree with your conclusion; we just can't help you with your reasoning unless you tell us what it is.

Do you understand what direction the electric field is pointing through each of the distict, well-defined bounding surfaces of the integration region? The reasoning is easier to understand in Cartesian coordinates (as djeitnstine somehow realized before we even knew what the problem statement was).

 

[kuruman]

Sami Lakka is working part (c). There is no "outside the slab" for part (c), because the charge distribution fills all of space. For |x|>b, the charge distribution is ρ<αe, but not zero.

Interesting ... The instructions in part (c) ask us to repeat part (b) which mentions a slab. Having fixed the idea of a slab in my head from part (b), I interpreted the instructions in (c) to mean "replace the non-zero constant charge density with the exponential form and leave the zero density alone". To clinch the "no slab" interpretation in part (c), the problem's author should have added "for all x" after the exponential form.