# Homework Help: Gauss Law and Electrostatic Field

1. Jul 27, 2009

### Sami Lakka

1. The problem statement, all variables and given/known data

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 $$\rho=\alpha*e^{-abs(x/b)}$$

2. Relevant equations

Q=triple integral of density

3. 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?

Last edited: Jul 27, 2009
2. Jul 27, 2009

### alphysicist

Hi Sami Lakka,

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.

3. Jul 27, 2009

### tms

4. Jul 27, 2009

### djeitnstine

5. Jul 27, 2009

### 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.

6. Jul 28, 2009

### Sami Lakka

7. Jul 28, 2009

### turin

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

8. Jul 28, 2009

### 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.

9. Jul 28, 2009

### 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.

10. Jul 28, 2009

### kuruman

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?

11. Jul 28, 2009

### Redbelly98

Staff Emeritus
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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.

Last edited by a moderator: May 4, 2017
12. Jul 29, 2009

### Sami Lakka

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
$$\rho(x)=\alpha$$ when -b < x < b
$$\rho(x)=0$$ when abs(x) >= b

where $$\alpha$$ and b are constants.

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

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!

13. Jul 29, 2009

### Redbelly98

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

14. Jul 29, 2009

### kuruman

There are two regions where the electric field is different and you need to give expressions for both (call them EI(x) and EII(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?

15. Jul 29, 2009

### turin

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.

16. Jul 29, 2009