Maxwell's Capacitor argument for a time independent Current

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The discussion revolves around the concept of "time independent current," which is clarified to mean a constant current that does not vary with time, contrasting with time-dependent currents like AC. Participants seek to understand this terminology and its implications in the context of capacitors. The conversation highlights confusion over the term "time independent," suggesting that "constant current" may be a more accurate description. The distinction between constant and time-dependent currents is emphasized, particularly in relation to how they interact with capacitors. Overall, the thread seeks clarity on the definitions and examples of these types of currents.
Shreya
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Homework Statement
Why can't we apply the. Maxwells argument to a time independent Current applied to a capacitor, to conclude the existence of a displacement Current inside a capacitor?
Maxwells argument considered a pot like surface between the plates of a capacitor with its open end outside. And argued that since the current enclosed is 0, B must be 0 just inside the capacitor, which is not true.
Relevant Equations
Ampere Maxwell Law
Is it because the current applied to a capacitor will never be time independent? Please help me out🙏
 
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Sorry, what is a "time independent current"? Can you define that, or give some examples?
 
berkeman said:
Sorry, what is a "time independent current"? Can you define that, or give some examples?
What I meant was a constant current - one which is not a function of t. A time dependent current would be AC Current. I am not sure if the name time independent is correct.
 
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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