hw8 - Solution for Homework 8 Computing the Fields of...

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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)- E (c)- E/ (d)- E/ 2 (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 e = A since the field lines enter the Gaussian surface, the flux is negative e =-| E | A . Putting it all together and solving for the charge density gives, =- | 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 = r 3 and occupies the region r &lt; a . Compute the electric field at all points in the region r &lt; a . Select One of the Following: (a) vector E = 4 3 a 6 4 r 2 r . 1 (b-Answer) vector E = r 4 6 r (c) vector E = r . (d) vector E = r 4 3 r Solution A Gaussian surface of radius r encloses a total charge Q enc = integraldisplay r 4 r 2 ( r ) dr = integraldisplay r 4 r 2 r 3 dr = 4 integraldisplay r r 5 dr Q enc = 4 r 6 6 = 2 r 6 3 For spherical symmetry, Gauss Law becomes vector E = Q enc 4 r 2 r = 2 r 6 3 4 r 2 r vector E = r 4 6 r Total Points for Problem: 5 Points Solution to Homework Problem 8.3(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. What charge is present on the inside and outside surfaces of the hollow conducting sphere? Select One of the Following: (a) There is zero net charge on the inside and outside surfaces of the conducting sphere. (b-Answer) The net charge on the inner surface is- 2 Q , and the charge on the outer surface is +2 Q . (c) The net charge on the inner surface of the conductor is +2 Q , and the charge on the outer surface is- 2 Q . (d) There is +2 Q net charge on both the inside and outside surfaces. Solution (a) Charge Inside of a Conductor: The electric field in a conductor is zero, therefore a Gaussian surface enclosing the inner sphere wall and the hollow must enclose zero total charge. (b) Net Charge on Inside Surface of the Conductor: As the hollow contains a charge of +2 Q , the inner sphere wall must contain- 2 Q of charge to make the total charge within the Gaussian surface described equal to zero....
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hw8 - Solution for Homework 8 Computing the Fields of...

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