- #1

iceman_ch

- 33

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## Homework Statement

Plane X=0 has a uniform charge density, [tex]\rho[/tex]

_{s}while plane X=a has -[tex]\rho[/tex]

_{s}. Determine the electric field intensity in region X<0.

## Homework Equations

E=[tex]\frac{Q}{4\Pi\epsilon

_{0}R

^{2}}[/tex]a

_{R}

## The Attempt at a Solution

I already know the answer to this question. What I'm trying to do is figure out a small part of the math behind it. Let's start by looking at one plate instead of two. How can I prove mathematically that all of the horizontal components cancel out leaving behind the component pointing away from the plate. I understand why but I need to understand the math behind it. This makes working problems so much easy when I can understand the math behind it easily. So here goes what I tried to do.

First I need to define a point to examine my e-field at. r=(x,y,z) next i need to define the different parts of the equation above. Best way to do this is to look at very very small peices and then adding them up.

dE = [tex]\frac{dQ}{4\Pi\epsilon

_{0}R

^{2}}[/tex]

dQ = [tex]\rho[/tex]

_{s}dS

dS = dY dZ

R = ?

I want to put that R=r-r' and that r' would the location of the very very small plate which would be (0,y',z') which would make R=(x,(y-y'),(z-z')) The y and z would stay constant because they are the point that we are evaluating E at. This is were I get lost. I know technically that y and z can be ignored because they will add to zero but, how can I prove this. If I keep going with my attempt the next step would be to find the magnitude of R and the unit vectors of R. I'm going to leave these out for now since they are long to type. So now I would put everything together and integrate. This is where everything gets messy. I know that if I put in a value for y say 4 and then integrate (4-y') from 0 to 8 the answer is 0. How can I write this out when even though y is constant because I don't know what it is I can't really prove that it ends up being 0.