Magnitude of the electric field?

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The discussion emphasizes the importance of demonstrating effort when seeking help on physics problems, particularly regarding electric fields. It clarifies that the term "electric field" refers to a vector quantity rather than an equation. Participants are reminded to adhere to forum rules by attempting to solve problems independently before asking for assistance. Understanding the nature of electric fields is crucial for proper analysis in physics. Overall, the conversation highlights the need for a foundational grasp of concepts before seeking guidance.
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Homework Statement
Charge of uniform density 4.0nC / m is distributed along the x axis from x = - 2.0m to x = + 3.0 * m . What is the magnitude of the electric field at the point x = +5.0 m on the X axis?
Relevant Equations
Electric field
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According to PF rules, you need to show some effort and attempt at a solution before receiving help. Also, "Electric field" is not an equation, it is a vector.
 
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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