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Gauss's Law vs Coulomb's law problem

  • Thread starter yeshuamo
  • Start date
  • #1
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Homework Statement


A thin sheet in the shape of an annular semicircle has a positive surface charge density +σ as shown. What is the electric field at point P?

Here is an illustration of the problem:
http://postimg.org/image/630bpqwan/


Homework Equations


Gauss's Law:
φ=Qenclosed/ε0
φ=∫E⋅dA
Qenclosed=ε0*∫E⋅dA

Coulomb's Law:
E=(1/(4πε0))(Σq/(r2))

The Attempt at a Solution


If I draw a gaussian surface somewhere between the point P and the radius a, the charge enclosed would be zero. Can I assume that this would mean that the electric flux through that surface is zero, and, by extension, the electric field experienced at point P is zero, too?

If I'm wrong about the Gauss's law approach, should I instead use a Coulomb's law approach and integrate the electric field generated by the semicircular sheet, minus the electric field generated by the semicircular cavity?

Thank you.
 

Answers and Replies

  • #2
SammyS
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Homework Statement


A thin sheet in the shape of an annular semicircle has a positive surface charge density +σ as shown. What is the electric field at point P?

Here is an illustration of the problem:
http://postimg.org/image/630bpqwan/


Homework Equations


Gauss's Law:
φ=Qenclosed/ε0
φ=∫E⋅dA
Qenclosed=ε0*∫E⋅dA

Coulomb's Law:
E=(1/(4πε0))(Σq/(r2))

The Attempt at a Solution


If I draw a gaussian surface somewhere between the point P and the radius a, the charge enclosed would be zero. Can I assume that this would mean that the electric flux through that surface is zero, and, by extension, the electric field experienced at point P is zero, too?

If I'm wrong about the Gauss's law approach, should I instead use a Coulomb's law approach and integrate the electric field generated by the semicircular sheet, minus the electric field generated by the semicircular cavity?

Thank you.
The short answer is to use Coulomb's Law.

The flux can be zero through some surface without the electric field being zero anywhere on that surface.
 
  • #3
174
24
there is no symmetry that permits efficient use of gauss's law in this problem. the symmetry is much better suited for coulomb's law. you will have some components that cancel out due to the symmetry.
 
  • #4
39
6
The short answer is to use Coulomb's Law.

The flux can be zero through some surface without the electric field being zero anywhere on that surface.
there is no symmetry that permits efficient use of gauss's law in this problem. the symmetry is much better suited for coulomb's law. you will have some components that cancel out due to the symmetry.
The symmetry factor makes this sensible. Thank you! I found the total charge of the thin sheet from the product of σ by its area, and used that total charge in the Coulomb's equation for electric field. And in polar coordinates, you were right, things canceled.

Thank you.
 

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