Electric Field and Potential of Thin Charged Metal Plates: Analysis and Sketch

[stunner5000pt]

Two thin long charged metal plates are placed at x = 0, and x = d (that means they are plaed vertically, parallel to the Y Axis. Thus the distance between them is d. Assume that V(x=0) = 0, and that the thickness of the plates is lesser than d.

a) Find E(x) and V(x) when 0 < x < d.

First of all let the surface charge density $$\sigma = \frac{Q}{A}$$

If i use Gauss Law then the Electric field $$E = \frac{\sigma}{\epsilon_{0}}$$ since the plates are conducting surfaces.
To find $$V(x) = \int E \cdot dx = \frac{\sigma}{\epsilon_{0}} \cdot dx = \frac{\sigma}{\epsilon_{0}} x$$

i'm not really sure if i derived that expression correctly for V(x) which i why i need your help, please.

b) Find E(x) and V(x) if x > d
E(x>d) = zero or kq / (x+d)^2
V(x>d) = zero as well? or simply integrate E over x??

c) Find E(x) and V(x) if x<d
same as the previous one?? i.e. both E and V are zero?? Not really sure here either?

d) Sketch E(x) and V(X)
I m not really sure about my expressions in a so if i found those out properly, with your help i can easily do this one!

Thank you in advance for your help!

 

[stunner5000pt]

so is ther anyone out there who can help me!

i am cofnused about hte first part whether E = kq/r^2 or E = constnat

but then the graphing part would be useless no?

And wha about the outsides of the plates?

 

[Tide]

Your E field and potential for the middle region (0 < x < d) are correct!

For x > d the electric field will be zero (it's discontinous due to the charge) and likewise for x < 0. However, the potential will be continuous across those surface.