2061x3_solutions_sp03

2061x3_solutions_sp03 - PHY2061 R. D. Field PHY 2061 Final...

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PHY2061 R. D. Field Final Exam Solutions Page 1 of 10 April 22, 2003 PHY 2061 Final Exam Solutions Problem 1 (10 points): Circle true or false for following (1 point each). (a) (True or False) Gravity is much stronger than the electrostatic force. (b) (True or False) An electric dipole produces an electric field that is not zero anywhere except at “infinity”. (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) In relativity the electric charge of a particle depends on its velocity. (f) (True or False) In relativity the electric field produced by a charged particle depends on its velocity. (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) It is possible for a proton to travel through water faster than light travels through water. (i) (True or False) The virtual 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 10 April 22, 2003 Problem 2 (15 points): Three identical point charges +Q are placed at three corners of a square with sides of length L as shown in the Figure . One of the corners of the square is empty. (Note: K = 1/(4 πε 0 )) Part A (4 points): What is the magnitude of the electric field, E , and the value of electric potential, V , at the center of the square ( at point 1 )? ( Take V = 0 at infinity and express your answers in terms of K, Q, and L .) Answer: 2 1 / 2 L KQ E = L KQ V / 2 3 1 = Solution: The electric field at Point 1 is the vector superposition of the fields from the three charges as follows = + + = = = + = y x L KQ E E E E y x L KQ E y x L KQ E y x L KQ E Q Q Q Q Q Q ˆ 2 1 ˆ 2 1 2 ˆ 2 1 ˆ 2 1 2 ˆ 2 1 ˆ 2 1 2 ˆ 2 1 ˆ 2 1 2 2 3 2 1 1 2 3 2 2 2 1 r r r r r r r with magnitude given by 2 1 1 / 2 L KQ E E = = r . The electric potential is the scalar sum of the electric potential from each of the three charged as follows: L KQ L KQ L KQ L KQ V 2 3 2 2 2 1 = + + = . Part B (4 points): What is the magnitude of the electric field, E , and the value of electric potential, V , at the empty corner of the square ( at point 2 )? ( Take V = 0 at infinity and express your answers in terms of K, Q, and L .) Answer: 2 2 1 2 / ) 2 ( L KQ E + = L KQ V / ) 2 ( 2 1 2 + = Solution: The electric field at Point 2 is the vector superposition of the fields from the three charges as follows + + = + + = = = = y x L KQ E E E E y L KQ E y x L KQ E x L KQ E Q Q Q Q Q Q ˆ ) 2 2 1 1 ( ˆ ) 2 2 1 1 ( ˆ ˆ 2 1 ˆ 2 1 2 ˆ 2 3 2 1 2 2 3 2 2 2 1 r r r r r r r 2 +Q +Q +Q L 1 L 1 2 3
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PHY2061 R. D. Field Final Exam Solutions Page 3 of 10 April 22, 2003 with magnitude given by 2 2 1 2 2 / ) 2 ( L KQ E E + = = r .
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2061x3_solutions_sp03 - PHY2061 R. D. Field PHY 2061 Final...

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