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Unformatted text preview: PHY2049 Spring 2009 Profs. D. Acosta, A. Rinzler, S. Hershfield Exam 1 Solutions 1. What is the flux through the right side face of the shown cube if the electric field is given by vector E =- 2 x i + 3 y j and the cube has a side length of 2? Answer:- 16 Solution: The flux is contintegraltext n vector EdA , where n is directed out of the surface. For the right face this means that the normal is n = i and x = 2 so i vector EA =- 2 2 A =- 16. 2. Two large nonconducting plates each with surface area 5 m 2 are parallel to each other and separated by 1 cm. If the total charge distributed uniformly on one plate is +5 C, and the total charge distributed uniformly on the other plate is- 7 . 5 C, what is the magnitude of the electric field midway between the two plates? Answer: 1 . 4 10 5 N/C Solution: The electric field for a single charged plate has magnitude | | / (2 o ). The field goes away from the positively charged plate and towards the negatively charged plate. Thus, for this problem the electric fields of the two plates add constructively between the plates: | vector E | = (5 C + 7 . 5 C ) / (2 o 5 m 2 ). 3. A charged particle is held at the center of two concentric conducting spherical shells. A cross section is shown in the figure. If the charged particle at the center has charge +2 C, and the two conducting shells A and B have charges- 3 C and +4 C deposited on them, respectively, what is the charge on the inner surface of shell B? Answer: +1 C Solution: The electric field inside the conducting shells is zero. Thus, the charge enclosed in a Gaussian surface inside shell B is zero by Gausss law. The charge enclosed is 2 C +- 3 C + q in = 0 so that q in = +1 C . 4. A long charged cylinder with a radius of R 1 = 3 cm has a charge density per unit length of 4 C/m. It is surrounded by a concentric charged cylindrical shell of radius R 2 = 5 cm with a charge density per unit length of- 3 C/m. What is the magnitude of the electric field at a radial distance of 6 cm from the centerline? Answer: 3 . 10 5 N/C Solution: A cylindrical Gaussian surface with radius 6 cm encloses both the inner and outer cylinder. The flux through the surface is 2 rL | E | . The charge enclosed is (4 C- 3 C ) L . Using Gausss law, the magnitude of the electric field is | E | = (1 C ) / (2 r o ). 5. In the figure shown, what is the net electric potential at the origin of the coordinate system due to the four other charged particles if...
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This note was uploaded on 02/26/2012 for the course PHYSICS 302K taught by Professor Irenepolycarpou during the Spring '09 term at University of Texas at Austin.
- Spring '09