Parallel plates, calculate charges and E. fields without surface area?

[SuckIt]

Homework Statement

two metal plates 10mm thin are held a distance 40mm apart with a P.D. 10V across them. +ve plate on left. In between the plates is a vacuum (free space basically). Calculate the charges and electric fields within this system.

Homework Equations

Q=VC
C = εo.A/d
No surface area given, so how to do it?

E = V/d can be done but is that what is asked for?

The Attempt at a Solution

?

 

[Delphi51]

E = V/d gives you the electric field between the plates and you can use Gauss' Law to get the charge per unit area on each plate.

 

[SuckIt]

no surface area, bra`

ok so gauss' law is...
$$\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0}$$

so how do I use that?

 

[buffordboy23]

Homework Statement

two metal plates 10mm thin are held a distance 40mm apart with a P.D. 10V across them. +ve plate on left. In between the plates is a vacuum (free space basically). Calculate the charges and electric fields within this system.

Homework Equations

Q=VC
C = εo.A/d
No surface area given, so how to do it?

E = V/d can be done but is that what is asked for?

So you can determine the electric field. Recall for parallel plates that

$$E = \frac{\sigma}{2\epsilon_{0}}$$

Sigma is the surface charge density. In other words the amount of charge per area. How can you use these ideas for your problem?

 

[mikelepore]

The electric field inside a parallel plate capacitor:

$$E = \frac{\sigma}{2\epsilon_{0}}$$

My book doesn't have that 2 in the denominator.

 

[buffordboy23]

The electric field inside a parallel plate capacitor:
My book doesn't have that 2 in the denominator.

Thank you. You are correct. The equation I gave was for the field due to a single plate. Since there are two plates superposition gives

$$E = \frac{\sigma}{\epsilon_{0}}$$

for points internal to the plates.