# Homework Help: Electric Field Intensity and Infinite Sheet Charges

1. Feb 27, 2015

### BruteCoder

Hey,
Just wanted to thank everyone for their time to help solve people problems, I can generally find all my answers here but I made an account now because i cant seem to find a related problem to this.

In general the textbook for these types of problems make the plane z=something. In this case i am able to use cylindrical coordinate system and solve the problem with ease. However i don't think I can apply that same principle that easily here. The answer for a) is 0.

1. The problem statement, all variables and given/known data
Plane x = 0 has a uniform density ρs, while plane x = a has -ρs. Determine the electric field intensity in regions a) x<0
b) 0<x<a
c) x>a

2. Relevant equations
dQ = ρsdS
E = Q(r-r')/4*pi*ε0*|r-r'|3

3. The attempt at a solution
E
= E1 + E2, where E1 is the electric field due to plane x = 0 and E2 is the electric field due to plane x = a:
a)
Because both are planes:
dS = dydz
For E1:
Q = ∫ ∫ ρs dydz, where both limits are from -infinity to infinity
r = -xâx + yây + zâz ,because x<0
r' = 0âx + y'ây + z'âz
r - r' = -xâx + (y-y')ây + (z-z')âz
|r-r'| = (-x2 + (y-y')2 + (z-z')2)1/2
E1 = 1/(4*π*ε0) ∫ ∫ ρs(-xâx + (y-y')ây + (z-z')âz)/ (x2 + (y-y')2 + (z-z')2)3/2(dydz), where both limits are from -infinity to infinity

At this point I am not sure how to solve these limits, I do know that for infinite sheet charges the only direction that will remain is normal to plane so in this case âx. Am i able to say that (y-y') and (z-z') are equal to zero because regardless where the observation point may be be placed there is a corresponding point on the plane?
If so the equation simplifies too:
E1 = 1/(4*π*ε0) ∫ ∫ ρs(-âx / (x)2(dydz), where both limits are from -infinity to infinity

But the answer to this would be infinity? and I don't understand how we could prove that adding this to E2 would result in 0.

Any help would be appreciated.

Last edited: Feb 27, 2015
2. Feb 27, 2015

### BvU

Ave, Brutus, welcome to PF

I take it $\rho_s$ is a surface charge density ? So you have something like a parallel plate configuration ? Does it say in the exercise text that the plates are conducting ? The the field must be $\perp$ the plates, right ?

Your second equation isn't really very handy in such a case. A plane like x = 0 extends all the way to infinity and y or z components of the field tend to cancel.

3. Feb 27, 2015

### BruteCoder

Yes ρs is surface charge density they didn't give a value for it however it was meant to stay as a variable. No it does not say the plates are conducting I actually copied the question word by word to try to avoid confusion. I thought their may be an easier equation to use, but all textbook examples use the coulomb law equation alone. Like I mentioned though their examples always have the plane z =, in this case cylindrical coordinates make it very easy to solve the problem as dS = [PLAIN]http://mathworld.wolfram.com/images/equations/CylindricalCoordinates/Inline3.gifd[PLAIN]http://mathworld.wolfram.com/images/equations/CylindricalCoordinates/Inline4.gifd[PLAIN]http://mathworld.wolfram.com/images/equations/CylindricalCoordinates/Inline3.gif [Broken] and the limits would be 0-2π for d http://mathworld.wolfram.com/images/equations/CylindricalCoordinates/Inline4.gif and 0-infinty d [PLAIN]http://mathworld.wolfram.com/images/equations/CylindricalCoordinates/Inline3.gif. [Broken] And its also apparent that due to symmetry the http://mathworld.wolfram.com/images/equations/CylindricalCoordinates/Inline3.gif component cancels out, the integral can then be solved with simple substitution. Then they arrive at the equation E = ρs/2ε0âz. They also say that this can be applied to any infinite sheet charge by E = ρs/2ε0ân. Using this it easy to see that the answer is 0 as:
E1 = ρs/2ε0-âx, it's negative because the original equation assumes x>than sheet charge
E2 = -ρs/2ε0-âx
However deriving this from the values in the question alone seem very difficult.

Last edited by a moderator: May 7, 2017
4. Feb 28, 2015

### BvU

This isn't going right at all. Check the link. And check here, where Gauss law is used efficiently.

I can't really decipher E1 = ρs/2ε0-âx . You mean $\vec E_1 = -{\rho_s\over 2\epsilon_0}\; \hat x$ ? And this is for part a) where x < 0 ? (In that case: that is correct).

And then "it's negative because the original equation assumes x>than sheet charge" is gobbledygook: x has a dimension of length, how can it possibly be > than some charge ? What original equation ? E = Q(r-r')/4*pi*ε0*|r-r'|3 ? That doesn't make any assumption at all on $\vec r - \vec r'$ except that it's $\ne$ 0 !

Now the other regions ?

And perhaps afterwards, as an understanding-tester: does it matter whether the plates are conducting or not ?

5. Feb 28, 2015

### BruteCoder

Yes i mean $\vec E_1 = -{\rho_s\over 2\epsilon_0}\; \hat x$, sorry i'm not to good at knowing how to make the symbols correctly. When I meant x>than sheet charge, I should have specified that I meant the sheet charge x position. By the equation i meant the one derived $\vec E_1 = {\rho_s\over 2\epsilon_0}\; \hat z$, it had z>0 and the infinite sheet charge was at z = 0. So it was positive when they took the Z component of $\vec r - \vec r'$. The other regions:

b)
$\vec E_1 = {\rho_s\over 2\epsilon_0}\; \hat x$,
$\vec E_2 = -{\rho_s\over 2\epsilon_0}\;- \hat x$
$\vec E_F = {\rho_s\over\epsilon_0}\;\hat x$
c)
$\vec E_1 = {\rho_s\over 2\epsilon_0}\; \hat x$
$\vec E_1 = -{\rho_s\over 2\epsilon_0}\; \hat x$,
$\vec E_F = 0 \hat x$

Wow after seeing Gauss law being applied I now understand how it can be very useful. It seems it does not matter if the plates are conducting but I am not sure now that its even relevant. Does not two separation of charges mean that they conduct inside of the gap regardless if its explicitly mentioned or not? Lastly the original way i went about solving this problem, is it 'possible' to arrive at the same equation or is it near impossible?

Thanks again.

6. Feb 28, 2015

### BvU

Correct on two counts: it doesn't matter and it's not relevant. The sheets are supposed to be thin and the only thing that counts is the presence of the (surface) charge.

Re original approach: I think it could work. The y and z components go away if you see that $\int_{-\infty}^\infty = \int_{-\infty}^0 + \int_0^\infty$ and the two terms are equal and opposite. x component might well come out to be $2\pi$ if you change from $dydz$ to $\rho d\rho d\phi$.