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

AI Thread Summary
The discussion revolves around calculating the electric field and charge between two parallel metal plates separated by 40mm with a potential difference of 10V. The challenge arises from the absence of a specified surface area for the plates. It is established that the electric field can be calculated using E = V/d, yielding a value of 0.25 kV/m. Additionally, using Gauss' Law, the charge per unit area can be determined, with the correct formula for the electric field being E = σ/ε₀, accounting for both plates. The clarification about the factor of 2 in the electric field equation is noted, emphasizing the importance of superposition when considering both plates.
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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


?
 
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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.
 
no surface area, bra`

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

so how do I use that?
 
Last edited:
SuckIt said:

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?
 
The electric field inside a parallel plate capacitor:

buffordboy23 said:
E = \frac{\sigma}{2\epsilon_{0}}

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

<br /> E = \frac{\sigma}{\epsilon_{0}} <br />

for points internal to the plates.
 
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