Exam Prep: Electric Field - Is It Zero?

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The discussion centers on understanding the electric field between two charged sheets. It questions whether the electric field is zero when using a Gaussian surface that encloses both sheets. Participants clarify that the electric field between two oppositely charged sheets is not zero; rather, it is uniform and directed from the positive to the negative sheet. The conversation emphasizes the importance of recognizing the configuration of charged plates and their effect on the electric field. Ultimately, the electric field between the sheets is significant and should not be assumed to be zero.
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Homework Statement
Two infinite plane sheets with uniform surface charge densities
+ sigma and - sigma are placed parallel to each other with separation d.
In the region between the sheets, where does the total electric
field have the greatest magnitude?
Relevant Equations
concept Q.
Screen Shot 2022-03-01 at 9.39.04 PM.png


I'm having an exam soon so i want to make sure. Is the electric field here zero?? cause if i draw gauss surface covering both of them they should cancel out or am i wrong.
 
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You are asked about the field between the sheets. How is drawing a surface enclosing both going to help with that?
 
What does this structure remind you? I mean two opposite charge sheets (or plates if that rings a bell) and the E-field in between them...
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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