How can I find "dx" in a straight line of electric charge?(Picture)

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The discussion revolves around understanding the relationship between dx and dteta in the context of a straight line of electric charge. The user initially struggles with the equation dx=R/cos^2(teta) dteta and seeks clarification on its derivation. After engaging with others, the user realizes their previous misconceptions about the topic. Ultimately, they express gratitude for the assistance received, indicating a resolution to their confusion. The conversation highlights the importance of grasping the underlying concepts in physics equations.
MatinSAR
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Homework Statement
How can I find "dx" in a straight line of electric charge?(Details in Picture)
Relevant Equations
r=R/cos(teta)
x=rtan(teta)
1646478251956.png

This is the question ... I have it's solution ...
1646478322100.png

1646478364142.png
My problem : I can't understand why dx=R/cos^2(teta) dteta
I have thought many hours but I couldn't find it's reason ... Can anyone please help with this ?!
 
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If ##x = R\tan{\theta}##, then ##dx/d\theta= \dots##
 
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ergospherical said:
If ##x = R\tan{\theta}##, then ##dx/d\theta= \dots##
I got it ... I have learned this in a wrong way but now I know what was wrong ... Thanks a lot 🌷
 
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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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