Why is My Polarized Plate Field Calculation Incorrect?

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The discussion revolves around a misunderstanding in calculating the polarized plate field, specifically regarding the relationship between polarization (P), permittivity (ε), and the electric field (E). The user presents their calculation but expresses confusion over its correctness. A key point raised is the potential confusion between the applied electric field and the local field, which can lead to incorrect results. Clarifying these concepts is essential for accurate calculations in electrostatics. Understanding the distinction between applied and local fields is crucial for resolving the issue.
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Homework Statement
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Relevant Equations
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I am a little confused why my answer is wrong... Briefly, my answer is as follow: $$P = \epsilon_{0} X E = (\epsilon-\epsilon_{0}) E; \space \space \space \space \space \space \space E = P/(\epsilon-\epsilon_{0})$$
 
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Herculi said:
Homework Statement:: ...
Relevant Equations:: ...

View attachment 282969

I am a little confused why my answer is wrong... Briefly, my answer is as follow: $$P = \epsilon_{0} X E = (\epsilon-\epsilon_{0}) E; \space \space \space \space \space \space \space E = P/(\epsilon-\epsilon_{0})$$
Disclaimer: all I know about this topic is from two minutes' web search.
Are you perhaps confusing the applied field with the local field?
 
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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