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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

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.

Page1 / 10

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online