# hw4 - Physics 21 Fall 2009 Solution to HW-4 MC22-2 A...

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Unformatted text preview: Physics 21 Fall, 2009 Solution to HW-4 MC22-2 A nonuniform electric field is directed along the x axis at all points in space. This magnitude of the field varies with x , but not with respect to y or z . The axis of a cylindrical surface, 0.80 m long and 0.20 m in diameter, is aligned parallel to the x axis. The electric fields E 1 and E 2 , at the ends of the cylindrical surface, have magnitudes of 2000 N/C and 1000 N/C respectively, and are directed as shown. Find the charge enclosed by the cylindrical surface shown in the figure. From Gauss’ Law, the charge enclosed will be related to the surface integral of the electric ﬂux. Since E is always in the x direction, there is no contribution to the surface integral from the curved sides of the surface. Letting r be the radius of the cylinder, we have Q = Z E · d A = πr 2 ( E 2 − E 1 ) . Hence Q = [(1000 − 2000)N / C] π (0 . 10 m) 2 = − . 278 nC . 22-3 You measure an electric field of magnitude E = 1 . 36 × 10 6 N/C at a distance of r = 0 . 163 m from a point charge. (a) What is the electric ﬂux through a sphere at that distance from the charge? (b) What is the magnitude of the charge? Since E is perpendicular to d A at every point of the surface of the sphere, Z E · d A = 4 πr 2 E = 4 . 54 × 10 5 Nm 2 / C . The magnitude the field is E = 1 4 π | q | r 2 , so | q | = 4 π r 2 E = 4 . 01 × 10- 6 C MC22-4,5 Two hollow conducting spheres have a common center O . The dimensions of the spheres are as shown. A charge of − 200 nC is placed on the inner conductor and a charge of +80 nC is placed on the outer conductor. The innercharge of +80 nC is placed on the outer conductor....
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hw4 - Physics 21 Fall 2009 Solution to HW-4 MC22-2 A...

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