PHYS
S13Phys2BaLec23B

# Since this problem has spherical symmetry it doesnt

• Notes
• 19

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Since this problem has spherical symmetry (it doesn’t matter which direction you travel outward), use a spherical Gaussian surface. Next, apply Gauss’s Law. How much charge is located inside the Gaussian sphere? +q.

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Gauss’ Law Answer Next turning to the definition of electric flux. Since the electric field is constant in magnitude at all points on the Gaussian surface and dA is always parallel to E, we can pull it out of the integral. Taking the surface integral:

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Gauss’ Law Answer Equate the two terms: This corresponds to our original Coulomb’s Law and validates the results.
Clicker Question 23B-1 For a gaussian surface through which the net flux is zero, the following four statements could be true. Which one of the statements must be true? A) No protons are inside the surface. B) The net charge inside the surface is zero. C) The electric field is zero everywhere on the surface. D) No electrons are inside the surface. E) No neutrons are inside the surface.

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Clicker Question 23B-2 Consider a conducting sphere (which may or may not be neutral). The electric field inside this conductor (in electrostatic equilibrium) is: A) zero. B) radial (in direction). C) symmetrically outward from a central axis. D) uniform (with constant nonzero magnitude and direction).
For Next Time (FNT) Finish the homework for Chapter 22. Keep reading Chapter 23.
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• Spring '08
• schuller
• Physics, Electric charge, Surface, Gauss’ Law

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