PHY132 Sp04 Midterm1 Solutions

University Physics with Modern Physics with Mastering Physics (11th Edition)

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PHY 132 Midterm 1 Feb 26, 2004 Version A Put FULL NAME, ID#, and EXAM VERSION on the front cover of the BLUE BOOKLET! To avoid problems in grading: do all problems in order , write legibly , and show all work for partial credit ! Note: make sketches where appropriate and show your axis conventions. Do not forget units, number of signifi- cant digits, and check your results for consistency. This exam will last 1.5 hr. Success! 1. Two point charges are positioned as follows: Q 1 = +3.6 μC at x = 8.0 cm, y =0.0 cm; and Q 2 = 6.4 μC at y = 6.0 cm and x =0. Show all work! (40 points) a. Calculate the components of the electric field E at point P ( x =8.0 cm, y =6.0 cm). Solution: (9 points) In P, the E x is given by Q 2 , and the E y by Q 1 : E x = 1/(4 πε 0 ) Q 2 /(8.0 cm) 2 = 9×10 9 × –6.4×10 –6 / 64×10 –4 = –9×10 6 N/C ( B : 9×10 6 N/C) E y = 1/(4 πε 0 ) Q 1 /(6.0 cm) 2 = 9×10 9 × 3.6×10 –6 / 36×10 –4 = 9×10 6 N/C ( B : –9×10 6 N/C) b. Calculate the electric potential energy of the system consisting of Q 1 and Q 2 . Solution: (11 points) The potential energy of the two charges Q 1 and Q 2 , with separation distance 10.0 cm, is simply U = Q 1 V Q 2 = 1/(4 πε 0 ) Q 1 Q 2 /(10.0 cm) = 9×10 9 ×3.6×10 –6 ×–6.4×10 –6 /10×10 –2 = –2.07 J c. Calculate the electric potential at point P Solution: (9 points) The potential in P is the scalar sum of the potentials in P of Q 1 and Q 2 : V = V Q 1 + V Q 2 = 1/(4 πε 0 ) [ Q 1 /(6.0 cm) + Q 2 /(8.0 cm)] = 9×10 9 ×[6.0×10 –5 – 8.0×10 –5 ] = –1.8×10 5 V ( B : +1.8×10 5 V) d. Calculate the electric force F (magnitude and direction) on a point charge q = 2.0 μC placed in point P. Solution: (11 points) In P, the F x is given by qE x and F y by qE y : F x = –2.0×10 –6 /(4 πε 0 ) Q 2 /(8.0 cm) 2 = +18 N and F y = –2.0×10 –6 /(4 πε 0 ) Q 1 /(6.0 cm) 2 = –18 N. ( B : F = (+36 N, –36 N) Thus: F = 18 2 = 25.5 N ( B : 50.9 N), direction: –45° down from the x -axis. 2. Consider two long conducting coaxial cylinders . The inner cylinder has a total linear charge density + λ , an inner radius a , and an outer radius b . The outer conducting cylinder has a total linear charge density λ , an inner radius c , and an outer radius d . Thus: a<b<c<d . Initially the space between the cylinders is filled with air ( K = 1.00). (60 points) a. Calculate the electric field at i ) r<a , ii ) a<r<b , iii ) b<r<c , iv ) c<r<d , and v ) r>d . Note that for proper credit for your answers you must give valid arguments, derivations, or proofs!
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