Common 2 Problem Solutions and Notes

# Common 2 Problem Solutions and Notes - Notes and Solved...

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Unformatted text preview: Notes and Solved Problems for Common Exam 2 3. GAUSS LAW Key concepts: Gaussian Surface, Flux, Enclosed Charge Gauss Law is equivalent to the Coulomb law but sometimes more useful. Gauss law considers a flux of an electric field thru some hypothetical surface (called a Gaussian surface) The flux of the field E thru the surface S is formally an integral EdS Φ = ∫ r r The Gauss law relates the flux of the electric field thru the Gaussian surface with the total charge enclosed by this surface total q EdS ε Φ = = ∫ r r The usefulness of the Gauss law is seen when calculating electric fields near symmetrical objects. For these situations, the electric field can for example be a constant on the surface of the integration and can be taken out of the integral defined above. The integral can then often be done easily (it is just the area of the Gaussian surface) and one can immediately find and expression for the electric field on the surface. The units of electric flux: 2 / Nm C Gauss’s Law helps us understand the behavior of electric fields inside the conductors. Since the conductors are the objects where the electrons move freely the electric field must be zero everywhere inside the conductor. The reason: if a conducting object is placed in a non-zero external electric field the charge inside the conductor will move towards its surface until the electric field created by this displaced charge completely compensates the external electric field. The total electric field inside the conductor is therefore zero. If that were not so, free charges would move due to the field and make it so. The Gauss law also helps us understand the distribution of electric charge placed into a conductor. Since the electric field inside every conductor is always zero, there cannot be extra electric charge located inside of one. Therefore, any extra charge placed into the conductor will move until it is distributed only on the surface of the conductor. This can be proven by considering any Gaussian surface lying completely inside the conductor, since electric field inside the conductor is zero, there is no charge which is enclosed inside this Gaussian surface. Typical problems related to Gauss Law: Problem 12 . Find the flux through a spherical Gaussian surface of radius a = 1 m surrounding a charge of 8.85 pC. A) 1 x 10-16 N m 2 /C B) 1 x 10-12 N m 2 /C C) 1 x 10-8 N m 2 /C D) 1 x 10-4 N m 2 /C E) 1 N m 2 /C Solution. The flux thru the Gaussian surface is the charge located inside the surface. Therefore: 12 12 2 / 8.85*10 /8.85*10 1 / q Nm C ε-- Φ = = = Answer E Problem 13. A positive charge Q= 8 mC is placed inside the cavity of a neutral spherical conducting shell with an inner radius a and an outer radius b. Find the charges induced at the inner and outer surfaces of the shell....
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## This note was uploaded on 09/16/2011 for the course PHYSICS 121 taught by Professor Fayngold during the Spring '10 term at NJIT.

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Common 2 Problem Solutions and Notes - Notes and Solved...

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