2061x3_solutions_fa03

2061x3_solutions_fa03 - PHY2061 R. D. Field PHY 2061 Final...

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PHY2061 R. D. Field Final Exam Solutions Page 1 of 9 December 9, 2003 PHY 2061 Final Exam Solutions Problem 1 (10 points): Circle true or false for following (2 points each). (a) (True or False) The vector function z x y y x z z y x E ˆ ˆ ˆ ) , , ( + = r represents a possible electrostatic field with magnitude given by 2 2 2 ) , , ( z y x z y x E + + = . Note that 0 ˆ 2 = × y E r r and hence this vector function cannot represent an electrostatic field since all electrostatic fields have 0 = × E r r . (b) (True or False) The vector function z x y y x z z y x B ˆ ˆ ˆ ) , , ( + = r represents a possible magnetic field with magnitude given by 2 2 2 ) , , ( z y x z y x B + + = . Note that 0 1 = B r r and hence this vector function cannot represent a magnetic field since all magnetic fields have 0 = B r r . (c) (True or False) The electric field is zero on the surface of all conductors in static equilibrium. (d) (True or False) It is possible to construct both electric and magnetic dipole field configurations, but it is not possible to construct a magnetic monopole field. (e) (True or False) Lines of constant electric potential are always perpendicular to electric field lines. (f) (True or False) In relativity, energy and momentum are conserved, but mass is not conserved. (g) (True or False) When light moves from one medium to another both its speed and wavelength change, but its frequency remains the same. (h) (True or False) Very little diffraction occurs when light of wavelength λ passes through a single slit of width W provided λ << W. (i) (True or False) The real image created by a concave mirror is always inverted. (j) (True or False) Convex mirrors can produce both real and virtual images.
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PHY2061 R. D. Field Final Exam Solutions Page 2 of 9 December 9, 2003 Problem 2 (18 points): Suppose that a non-uniform electric field is present and is given by r ar r E ˆ ) ( = r , where a = 50 V/m 2 and r is the radial distance from the origin, 2 2 2 z y x r + + = . The electric field points radially away from the origin and has a magnitude given by E(r) = ar . (Note that K = 8.99x10 9 Nm 2 /C 2 , ε 0 = 8.85x10 -12 C 2 /Nm 2 ) Part A (6 points): How much electric charge ( in nanoC ) is contained with a spherical shell centered at the origin with inner radius R 1 = 1 m and outer radius R 2 = 2 m ? Answer: 38.9 Solution: There are two ways to do this problem. One way is to calculate the net flux emanating from the spherical shell as follows: ) ( 4 4 ) ( 4 ) ( 3 1 3 2 2 1 1 2 2 2 1 2 R R a R R E R R E A d E A d E A d E R R shell = = + = = Φ π r v r v r v and we know from Gauss’s Law that nC C Nm m m m V K R R a R R a Q enclosed 9 . 38 / 10 99 . 8 ) ) 1 ( ) 2 )(( / 50 ( ) ( ) ( 4 2 2 9 3 3 2 3 1 3 2 3 1 3 2 0 0 = × = = = Φ = πε ε . A second method is to calculate the charge density according to a z E y E x E E z y x 0 0 0 3 ρ = + + = = v r where I used E x = x, E y = y, E z = z. Now we calculate the charge in the shell using ) ( 4 3 3 1 3 2 0 0 2 1 R R a dV a dV Q R R = = = , which of course gives the same answer (note that the charge density is constant).
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2061x3_solutions_fa03 - PHY2061 R. D. Field PHY 2061 Final...

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