Physics Solution Manual

# Physics Solution Manual - 5. Determine the value of Q enc...

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1. Flat surface (a) Area perpendicular to E Electric flux = EA . (b) Area vector makes an angle φ with the vector E , Electric flux = EA cos( φ ) = E . A 2. Non-Flat surface Divide area into infinitesimal areas and integrate Electric flux = E . d A Electric flux

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Gauss’s Law problems: 1. Carefully draw a figure - location of all charges, direction of electric field vectors E 2. Draw an imaginary closed Gaussian surface so that the value of the magnitude of the electric field is constant on the surface and the surface contains the point at which you want to calculate the field. 3. Write Gauss Law and perform dot product E . dA 4. Since you drew the surface in such a way that the magnitude of the E is constant on the surface, you can factor the | E | out of the integral.

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Unformatted text preview: 5. Determine the value of Q enc from your figure and insert it into Gauss's equation. 6. Solve the equation for the magnitude of E . Under electrostatic conditions, any excess charge resides entirely on the surface of a solid conductor. Gaussian surface E = 0 Application of Gauss’s Law to conductors: The solution of this problem lies in the fact that the electric field inside a conductor is zero and if we place our Gaussian surface inside the conductor (where the field is zero), the charge enclosed must be zero (+ q – q) = 0. Symmetries Planar Spherical Cylinderical Electric field between two oppositely charged parallel plates. Some Useful results: The electric field of a uniformly charged INSULATING sphere....
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## This note was uploaded on 05/30/2011 for the course PHYS 150 taught by Professor Anz-meador during the Spring '08 term at Embry-Riddle FL/AZ.

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Physics Solution Manual - 5. Determine the value of Q enc...

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