Flux of Electric Field through a cone

  • Thread starter Moara
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  • #1
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Homework Statement:
A conic surface is placed Inside a region filled with uniform Electric Field E. The field lines are perpendicular to the geratrix AB. The cone has height h and radius R of it's base. Find the electric flux through the lateral surface of the cone.
Relevant Equations:
Ø=E.S.cosx Øtot=qint/€ (gaus law)
Since there is no charge inside the cone, the total flux through its surface is zero, hence Ø(lateral surface)+∅(base surface)=0. But ∅(base surface)=E.πR².cosΩ, because electric Field is homogenous. But by the figure, Ω is just arctg(h/R).
So Ø(lateral surface)=-E.π.R².R/√(R²+h²).
This is not the answer according to the solution I saw. My point is, does anyone could show me where is my mistake?
15720384362294540504960180785928.jpg
 

Answers and Replies

  • #2
TSny
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Your work looks correct to me. Do you remember the form of the answer from the solution?
 
  • #3
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Here is the solution I saw. The sentence : " temos que" is equal to " we have that"
Screenshot_2019-10-25-21-14-52-850_com.android.browser.png
 
  • #4
TSny
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Looks like this is a different problem where they want the flux through the two shaded regions (the triangle and the semicircle). But I think there is a mistake in the solution where they inadvertently switched the two angles in the third line:
1572050553808.png


Anyway, I think your work is correct for the problem you stated.
 

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