I Gauss's law of sphere using integral

AI Thread Summary
The discussion focuses on applying Gauss's law to a sphere with radius R and total charge q. The user correctly identifies the integral form of Gauss's law but struggles with setting up the surface integral, particularly the limits for the angles θ and φ in spherical coordinates. Clarification is provided that θ is the polar angle (0 to π) and φ is the azimuthal angle (0 to 2π), emphasizing their roles in defining the sphere's geometry. The user realizes their mistake in mixing up the bounds for these angles, which is crucial for obtaining the correct result. Understanding the proper setup and symmetry in the integral is essential for solving related problems effectively.
pinkfishegg
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Hey I was just practicing Gauss's law outside a sphere of radius R with total charge q enclosed. So I know they easiest way to do this is:

∫E⋅da=Q/ε
E*4π*r^2=q/ε
E=q/(4*πε) in the r-hat direction

But I am confusing about setting up the integral to get the same result

I tried
∫ 0 to pi ∫0 to 2pi E*r^2sin(Φ) dΦdΘ for the LHS

i got this from the front of Griffiths where it says:
dl=dr(r-hat)+sdφ d(θ-hat)+rsin(θ)dφ (φ-hat)

But i keep getting 0 when i take this integral sin i get -cos(2π)-cos(0)= -1-(-1)=0

So there's obviously something wrong with my setup. I need to integrate over θ and φ because they are on the surface but I'm not sure about the terms in front. Aren't they supposed to be they ones in front from the dl term?..

Trying to practice this to be able to answer harder questions :P
 
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The surface integral is not a ## d \vec{l} ## type integration. The surface differential is ## dA=r^2 sin(\theta) \, \, d \theta \, d \phi ## in polar coordinates. ## \theta ## gets integrated from ## 0 ## to ## \pi ##, and ## \phi ## from ## 0 ## to ## 2 \pi ##. The integrals are separable, and there is complete symmetry as ## E ## is assumed to be a constant, (depending only on ## r ##), independent of ## \theta ## and ## \phi ##. The result of these two integrals is ## 4 \pi ##.
 
ohhh so i got the bounds for φ and θ mixed up..
 
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pinkfishegg said:
ohhh so i got the bounds for φ and θ mixed up..

So Φ is from top to bottom and θ from left to right correct? why are those the bounds. It seems like it could go either way and you'd still get the entire sphere.
 
pinkfishegg said:
So Φ is from top to bottom and θ from left to right correct? why are those the bounds. It seems like it could go either way and you'd still get the entire sphere.
## \phi ## is the azimuthal angle, basically counterclockwise, and goes all the way around (360 degrees). ## \theta ## is the polar angle, and when it points all the way downward the angle is 180 degrees. When the point lies in the x-y plane, the polar angle is 90 degrees. ## \\ ## Additional item, is that spherical coordinates is not a system that you can draw about some origin, (at radius ## r ## away from some other origin), where for small angles ## \phi ## is left and right (x-direction), and ## \theta ## is the elevation angle (y-direction) . This type of coordinate system is also used at times for sources that have a narrow beam spread, but it is not spherical coordinates.
 
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Charles Link said:
## \phi ## is the azimuthal angle, basically counterclockwise, and goes all the way around (360 degrees). ## \theta ## is the polar angle, and when it points all the way downward the angle is 180 degrees. When the point lies in the x-y plane, the polar angle is 90 degrees.
ok i just looked at this: https://en.wikipedia.org/wiki/Spherical_coordinate_system. sometimes its opposite in math so that's why i mixed it up.
 
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Please see also my edited comment (Additional item) in my previous post.
 
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