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Unformatted text preview: PHY2049 Spring 2010 Profs. P. Avery, A. Rinzler, S. Hershfield Exam 1 Solution 1. Four parallel infinite sheets of charge spaced by 4 cm between adjacent sheets are shown edgewise in the figure. Their surface charge densities are as indicated with = 2 . C/m 2 . What is the magnitude of the electric field (in N/C) at point P located midway between two sheets as shown? (Hint: Electric fields are vectors that add as such.) A B C D P + g g 2 G + 3 g g 4 G Answer: 4 . 5 10 5 Solution: Sheet A produces an electric field of magnitude / (2 o ) going away from it, which means the electric field is pointing towards the bottom of the page at point P. Sheet B produces an electric field of magnitude 2 / (2 o ) going towards it, which implies that the electric field is pointing towards the bottom of the page at point P. Similarly, at point P the electric fields due to sheets C and D are 3 / (2 o ) toward the top of the page and 4 / (2 o ) toward the bottom of the page. Adding these four fields together produces a net electric field of 4 / (2 o ) toward the bottom of the page. 2. 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 16 from the electric field strength at the surface? Answer: R/ 16 and 4 R Solution: Applying Gausss law, the flux through a sphere of radius r < R is = ( vector E n )4 r 2 = Q enc o = 1 o 4 3 r 3 vector E n = 1 3 o r. The flux through a sphere of radius R > r is = ( vector E n )4 r 2 = Q enc o = 1 o 4 3 R 3 vector E n = 1 3 o R 3 r 2 . The electric field at the surface, r = R , is vector E n = R/ (3 o ). To determine where the electric field is 1 / 16 of the value at the surface, set vector E n for the r < R and r > R cases equal to (1 / 16) R/ (3 o ) and solve for r . 3. A non-uniform electric field given by vector E = (5 . 5 i- 2 . 1 j + (4 . 6 z 2- 3) k ) N/C pierces a cube with sides 3 m, as shown in the figure. The cube has its rear corner at the origin. What is the total charge inside the cube? y z x Answer: +3 . 3 nC Solution: By Gausss law the net flux is equal to Q enc / o . The flux through each face of the cube is ( vector E n )(3 m ) 2 , where the normal vector points outwards. For the front face the normal component of the electric field is ( vector E n ) = 5 . 5 N/C, while for the back face it is ( vector E n ) =- 5 . 5 N/C. These two fluxes will cancel in computing the net flux. Similarly, the flux for the right and left faces will cancel. Only the fluxes through the top and bottom faces do not cancel. The flux through the top face ( z = 3) is (4 . 6 3 2- 3)3 2 , and the flux through the bottom face is- (4 . 6 2- 3)3 2 . Combine these to find the charge enclosed is o 4 . 6 3 4 . 4. Members of a space mission discover a mysterious alien artifact consisting of a solid conducting sphere of radius 4 cm4....
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