- #1

jisbon

- 476

- 30

- Homework Statement
- There is an infinite charged plate in yz plane with surface charge density ##\sigma = 8*10^8 C/m^2## and negatively charged particle at coordinate (4,0,0) Find magnitude of efield at coordinate (4,4,0)

- Relevant Equations
- E= E1+E2

So I figured to get e-field at point (4,4,0), I need to find the resultant e-field from the negatively charged particle and the plate

##E_{resultant}=E_{particle}+E_{plate}##

##E_{particle}=\frac{kq}{d^2}=\frac{(9*10^9)(-2*10^-6)}{4^2}=-1125N/C##

Now for the plate is where I'm confused.

If this was a wire, it would have been okay for me since I only need to deal with one dimension.

Since what they requested was a plate in yz plane, does this means that my:

##\sigma=dy*dy*x?## where ##dy## is the 'slice' I take and x is the width of the plate? Is that accurate?

If it is true, then to find the e-field created by that slice at the point,

##dE=\frac{kdq}{R^2}##

##dE=\frac{k\sigma *x*dy}{a^2+y^2}##

I know that the vertical components of the resultant e-field will cancel out because there are same amount of segments on top and below the point.

So need to find ##dE_{x}##, which = ##dEcos\theta##, where ##\theta## is shown:

So ##dE_{x} = dEcos\theta = (\frac{k\sigma *x*dy}{a^2+y^2}) (\frac{a}{\sqrt{y^2+a^2}})##,

Now the problem is I can't integrate this to find my resultant e-field because I do not know what the value of x is. If this was a wire in a plane it will have been solvable for me, but now I'm kind of stuck.

Any clues/help? Thanks :)