Phy102&amp;07-II-ms1

# Phy102&amp;amp;07-II-ms1 - Kuwait University Physics 102...

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Kuwait University Physics Department Physics 102 First Mid Term Examination March 22, 2007 11.00 am – 12.30 pm Name………………………………… Student No.………………. Instructors: Drs. Abdel Karim, K. Bhatia, E. Davis, A. Farhan, Y. Kokay, P. Lajko, H. Manaa, M. Marafi, S. Razee, M. Sharma. Note : The points for each problem are given on right margin. The conceptual questions carry 1 point each. (Fundamental constants ) k = 2 2 9 C / m . N 10 0 . 9 4 1 × = πε ο (Coulomb constant) ε o = 8.85 × 10 -12 C 2 / (N · m 2 ) (Permittivity of free space) μ 0 = 4 π × 10 -7 T .m/A (Premeability of free space) e = 1.60 × 10 -19 C (Elementary unit of charge) N A = 6.02 × 10 23 (Avogadro’s number) g = 9.8 m/s 2 (Acceleration due to gravity) m e = 9.11 × 10 -31 kg (Electron mass) m p = 1.67 × 10 -27 kg (Proton mass) Prefixes of units m = 10 -3 μ = 10 -6 n = 10 -9 p = 10 -12 k = 10 3 M = 10 6 G = 10 9 T = 10 12 For use by Instructors only Prob. 1 2 3 4 5 6 7 8 Total Marks Ques. 1 2 3 4 5 6 7 8 Total Marks

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1. A point charge q = 20 μ C is attached to a horizontal spring of 30cm length and spring constant k=10N/m. If a charge Q is placed at 50cm from the origin O the spring is stretched by Δ x = 10cm. Calculate the charge Q. a. 0.55 nC b. 5.5 × 10 3 C c. 0.55 μ C d. 180 μ C e. 55 nC Solution : Kx = () 2 10 . 0 m kQq () ( )( ) () Q m m 2 6 9 2 10 . 0 10 20 10 9 10 N/m. 10 × × = Q = 5.5 × 10 -8 C = 55 nC 2. Two point charges –2q and q are placed at x = 2cm and y = 3cm axis as shown in fig. Calculate the x and y coordinates of a 3 rd charge –2q so that the net electric field at origin is zero. a. x = 13.1cm, y = 0.81cm b. x = 1.9cm, y = 0.4cm c. x = 1.8cm, y = 1.9cm d. x = 0.1cm, y = 12.5cm e. x = 2cm, y = -3cm Solution : E Y = () 2 02 . 0 2
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## Phy102&amp;amp;07-II-ms1 - Kuwait University Physics 102...

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