Electric field between parallel plates

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The discussion focuses on the electric field between parallel plates of a capacitor, emphasizing that each plate carries opposite charges, with one plate having a surface charge density of σ and the other -σ. It raises questions about the nature of charge distribution on the plates and the presence of negative charges in a static electric field scenario. The electric field outside a single conducting plate is described by the equation E = σ/ε, while the field between two plates is derived using the superposition principle. The conversation highlights a fundamental misunderstanding regarding the application of these principles to capacitors. Overall, the interaction of electric fields in capacitors is crucial for understanding their behavior in static situations.
lys04
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Homework Statement
The electric field outside a conducting plate of charge is given by sigma/epsilon right? Then why not for a capacitor, since that is 2 conducting plates, is the electric field 2sigma/epsilon using superposition principle?
Relevant Equations
E=sigma/epsilon
^^
 
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Here’s a cross -section through (due to space limitations, part of) an infinite conducting plate with a (say) positive charge:
___________________________
___________________________

Here are some questions to consider:
- are both surfaces charged?
- if only one (of the two) surfaces is charged, what determines which one?
- in a static situation, every ‘field line’ starts on a positive charge and ends on a negative charge; where are the negative charges here?
 
lys04 said:
Homework Statement: The electric field outside a conducting plate of charge is given by sigma/epsilon right? Then why not for a capacitor, since that is 2 conducting plates, is the electric field 2sigma/epsilon using superposition principle?
Relevant Equations: E=sigma/epsilon
Generally, the two plates of a capacitor are oppositely charged. Right?

So, if one plate has a surface charge density, ##\sigma##, then the other plate has surface charge density, ##-
\sigma##,

Now use superposition to determine the electric field outside the plates as well as between the plates.
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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