mat-sln-asn-hwk8-spr02

# mat-sln-asn-hwk8-spr02 - Solution for Homework 8 Computing...

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Unformatted text preview: Solution for Homework 8 Computing the Fields of Conductors and Dielectrics Solution to Homework Problem 8.1(Charge Density on Hubcap) Problem: A metal hubcap (a conducting metal disk) lays flat on the ground. The earth produces an electric field of E = 150 N C downward. What is the surface charge density on the upper sur- face of the hubcap? Select One of the Following: (a) (b-Answer)- ε E (c)- E/ε (d)- E/ 2 ε (e) Zeppo was the fourth Marx Brother Earth Hubcap Solution Using a Gaussian surface which is a cylinder with one end in the field and one end in the hubcap, we can write φ e = σA ε since the field lines enter the Gaussian surface, the flux is negative φ e =-| E | A . Putting it all together and solving for the charge density gives, σ =- ε | E | =- (8 . 85 × 10- 12 C 2 Nm 2 )(150 N C ) =- 1 . 3 × 10- 9 C / m 2 Earth Hubcap----- + + + + + Total Points for Problem: 3 Points Solution to Homework Problem 8.2(Non-Uniform Spherical System) Problem: A spherical system of charge has NON-UNIFORM volume charge density ρ = γr 3 and occupies the region r < a . Compute the electric field at all points in the region r < a . Select One of the Following: (a) vector E = 4 3 πγa 6 4 πε r 2 ˆ r . 1 (b-Answer) vector E = γr 4 6 ε ˆ r (c) vector E = ρ ε ˆ r . (d) vector E = γr 4 3 ε ˆ r Solution A Gaussian surface of radius r encloses a total charge Q enc = integraldisplay r 4 πr 2 ρ ( r ) dr = integraldisplay r 4 πr 2 γr 3 dr = 4 πγ integraldisplay r r 5 dr Q enc = 4 πγr 6 6 = 2 πγr 6 3 For spherical symmetry, Gauss’ Law becomes vector E = Q enc 4 πε r 2 ˆ r = 2 πγr 6 3 4 πε r 2 ˆ r vector E = γr 4 6 ε ˆ r Total Points for Problem: 3 Points Solution to Homework Problem 8.3(Difference Between Dielectric and Conductors) Problem: Explain how the induced charge on a conductor and the bound charge on a dielectric differ atomically. Select One of the Following: (a) The charge can move freely in a dielectric, but has limited movement in a conductor. (b-Answer) A charge can move freely in a conductor but has movement is limited in a dielectric. (c) An induced charge on a conductor is larger than the same induced charge in a dielectric. (d) The induced charge on a dielectric is stronger than the same induced charge on a conductor. (e) Both can have an induced charge move over them but in a conductor it is limited to the surface of the conductor. Solution Charge can move the entire length of a conductor, while movement in a dielectric is limited to atomic or molecular distances. Total Points for Problem: 3 Points Solution to Homework Problem 8.4(Hollow Metal Conducting Sphere with a Charge at the Center) Problem: Consider a neutral hollow metal sphere with a point charge of +2 Q placed at the center of the hollow....
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## This note was uploaded on 03/29/2010 for the course PHYS 2469 taught by Professor Stewat during the Spring '10 term at University of Arkansas Community College at Batesville.

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mat-sln-asn-hwk8-spr02 - Solution for Homework 8 Computing...

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