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Unformatted text preview: PHY2049 Spring 2011 Profs. P. Avery, S. Hershfield Exam 1 Solution 1. A solid conducting sphere of 15 cm radius is located at the center of a spherical conducting shell, whose inner radius is 21 cm and outer radius 42 cm. At a distance of 50 cm from the center of these concentric conductors, the electric field is 2 . 1 10 5 V/m and is pointing outward. The shell has a net charge of +3 . 1 C. What is the net charge on the sphere at the center in C? Answer: +2 . 7 Solution: A spherical Gaussian surface of radius 50 cm encloses a charge of 4 (0 . 5 m ) 2 E o = 5 . 8 C . Since the shell has a net charge of 3 . 1 C , the sphere has a net charge of 5 . 8- 3 . 1 = 2 . 7 C . 2. A long, straight wire is uniformly charged with a linear charge density of 5.0 C/m. The wire runs along the axis of a cylinder of radius 2.0 cm and length 30.0 cm. What is the total electric flux through the cylinder surfaces in N m 2 /C? Answer: 1 . 69 10 5 Solution: From Gausss Law, the electric flux is equal to the charge enclosed divided by o . The charge enclosed is (5 C/m )(0 . 3 m ) = 1 . 5 C so the total electric flux is (1 . 5 C ) / o . 3. The figure shows a uniformly charged, nonconducting spherical shell of inner radius a and outer radius 2 a . If the electric field at the outer radius is E , what is the electric field at point P with radius r = 1 . 5 a ? 2 a a P Answer: . 6 E Solution: Here we apply Gausss law twice: once for a spherical surface of radius 2 a and once for a spherical surface of radius 1 . 5 a . Let the magnitude of the electric field at 2 a be E , and the magnitude of the electric field at 1 . 5 a be E . The charge density is denoted by . contintegraldisplay vector E ndA = 4 (2 a ) 2 E = Q enc / o = (4 / 3) ( (2 a ) 3- a 3 ) / o contintegraldisplay vector E ndA = 4 (1 . 5 a ) 2 E = Q enc / o = (4 / 3) ( (1 . 5 a ) 3- a 3 ) / o Divide the second equation by the first: (1 . 5) 2 E 2 2 E = (1 . 5) 3- 1 2 3- 1 , which implies that E = 0 . 603 E . 4. A thick conducting spherical shell has inner radius r and outer radius R , as shown in the figure. A point charge of- Q is located at the center of the sphere and a charge of + q is placed on the conducting shell. The charge on the outer surface of the conducting shell is: r R- Q +q Answer: q- Q Solution: A Gaussian surface just larger than R will enclose a charge of q- Q , while a Gaussian surface just smaller than R will enclose no net charge because the electric field is zero inside a conductor. Consequently, the charge on the outer surface of the shell must be q- Q . 5. Suppose we have an insulating spherical ball of uniform charge density and radius R . At what radius or radii from the center of the sphere is the electric field strength reduced by a factor of 4 from the electric field strength at the surface?center of the sphere is the electric field strength reduced by a factor of 4 from the electric field strength at the surface?...
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- Summer '08