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• ## Study Guides

 1 answerLast reply by: walid hamarnehSun Feb 16, 2014 1:24 PMPost by Werner Heisenberg on January 1, 2014In case 2 (r

### Application of Gauss's Law, Part 1

• For a sphere of charge, the electric field outside the sphere, with a total charge Q and uniform charge density rho, is the same as the field due to a point charge Q at the center of the sphere. In other words, to an observer outside the sphere of uniform charge density (or a charge density that depends only on r, the distance from the center) the sphere appears to be a point charge at the center.
• For a sphere of charge, with uniform charge density rho, the electric field inside the sphere, at a distance r from the center, is (rho) * r / (3*epsilon_0).
• Gauss’s law may be used to find the electric field inside a spherical cavity with a sphere of charge. It may also be used to find the E-field both inside and outside a sphere of charge with a charge density that varies with r, the distance from the center.

### Application of Gauss's Law, Part 1

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• When is Gauss Law Useful? 0:18
• Need a Surface S
• Gaussian Surface
• Sphere of Charge 10:11
• Charge Density is Uniform
• Case 1: R>A
• Any Direction On Cone Is Same
• Case 2: R<A
• Point R Within the Surface
• Concentric Cavity 31:11
• Inside Circle and Outside Circle
• R>A
• R<B
• Radius Dependent Charge Density 37:39
• Sphere
• Total Charge: Q
• Spherical Shell
• Finding Electric Field R>A
• R<A
• Example 1: Charged Sphere
• Example 2: Charged Spherical Cavity