E Field at Point P w/ Line Charge & Closed Surface

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The discussion focuses on calculating the electric field at point P due to a line charge and determining the electric flux through various surfaces surrounding the charge. It emphasizes that electric flux is independent of distance and only depends on the amount of charge within a closed surface, as explained by Gauss's Law. The flux through surface S3 is given as (D*L)/(8*Eo), leading to the conclusion that surfaces S1 and S3 have equal flux due to symmetry. Consequently, surface S2 shares (3/4) of the total flux, calculated based on the total charge inside the closed surface. The conversation also touches on the distinction between electric field lines and electric flux.
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1.In the diagram below, find the E field at point P.What is E when L is infinity
2.A closed surface is made up of three surfaces,S1 being circular , S2 being cylindrical, and S3 being conical.A line charge of length L and density (D) is placed symmetrically with respect to S2 as shown.If the flux through S3 is (D*L)/(8*Eo),find the flux through S2
 
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I never get bored of saying this.

It is homework. Fill in the form. Do some math. If you are wrong someone will help.
 
Hi, do you still remember what Dr. Cheng taught about the nature of electric flux and Gauss's Law? It is INDEPENDENT of the distance from the charges but only dependent on the AMOUNT of charges inside the closed surface. So when you look at S3, it has the same amount of flux as the base surface as S1. Because they are symmetrical, they are the same. Remember the total charge inside the whole surface is equal to lambda times l, so the TOTAL flux is equal to (lambda)(l)/epsilon zero). You see S3 shares (1/8) of the total flux, and so as S1, so S2 shares (1-(1/8)-(1/8))( (lambda)(l)/epsilon zero) ) flux, which is equal to (3/4) of the total flux. Always remember flux is like air blown out from the charge. Good luck on your assignment!

Sorry, I confused the concept of electric field and flux, Always remember "electric field lines" is like air blown out from positive charges, and electric flux then is what the amount of air is in terms of cubic meters per second.

P.S. We are perhaps in the same class.
 
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Narold said:
Hi, do you still remember what Dr. Cheng taught about the nature of electric flux and Gauss's Law? It is INDEPENDENT of the distance from the charges but only dependent on the AMOUNT of charges inside the closed surface. So when you look at S3, it has the same amount of flux as the base surface as S1. Because they are symmetrical, they are the same. Remember the total charge inside the whole surface is equal to lambda times l, so the TOTAL flux is equal to (lambda)(l)/epsilon zero). You see S3 shares (1/8) of the total flux, and so as S1, so S2 shares (1-(1/8)-(1/8))( (lambda)(l)/epsilon zero) ) flux, which is equal to (3/4) of the total flux. Always remember flux is like air blown out from the charge. Good luck on your assignment!
are u student or the helper?Nice to meet u
 
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