_008 - 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(Field in Dielectric Slab) Problem: A dielectric slab with a dielectric constant, = 2 , is placed in a uniform electric field parallel to the normal of the slab with magnitude E = 12 N C . What is the magnitude of the field inside the dielectric? Select One of the Following: (a) 0 (b) 24 N C (c-Answer) 6 N C (d) 12 N C Solution In a dielectric the field, vector E (the field without the dielectric), is reduced by a factor of, the dielectric constant, vector E = vector E substitute, vector E = 12 N C 2 evaluate, vector E = 6 N C Total Points for Problem: 1 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 < a . Compute the electric field at all points in the region r < a . Select One of the Following: (a) vector E = 4 3 a 6 4 r 2 r . (b-Answer) vector E = r 4 6 r (c) vector E = r . (d) vector E = r 4 3 r Solution 1 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(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----- + + + + + 2 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. 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....
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_008 - Solution for Homework 8 Computing the Fields of...

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