ps01sol

# ps01sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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PS01 Solution - 1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Fall 2008 Problem Set 1 Solutions Problem 1: Two Vectors Given two vectors, ± A = (2 ˆ i - 2 ˆ j + ˆ k ) and ± B = (4 ˆ i - 7 ˆ j +4 ˆ k ) , evaluate the following: (a) ± A +3 ± B ; ± A ± B = (2 ˆ i - 2 ˆ j + ˆ k ) + 3 (4 ˆ i - 7 ˆ j ˆ k ) = 14 ˆ i - 23 ˆ j + 13 ˆ k (b) 4 ± A - ± B ; 4 ± A - ± B = 4(2 ˆ i - 2 ˆ j + ˆ k ) - (4 ˆ i - 7 ˆ j ˆ k )=4 ˆ i - ˆ j (c) 4 ± A · ± B ; Since ˆ i · ˆ i = ˆ j · ˆ j = ˆ k · ˆ k =1 and ˆ i · ˆ j = ˆ j · ˆ k = ˆ k · ˆ i =0 , the dot product is ± A · ± B = (2)(4) + ( - 2)( - 7) + (1)(4) = 26 (d) ± A × ± B ; With ˆ i × ˆ j = ˆ k, ˆ j × ˆ k = ˆ i, ˆ k × ˆ i = ˆ j the cross product ± A × ± B is given by ± A × ± B = ± ± ± ± ± ˆ i ˆ j ˆ k 2 - 21 4 - 74 ± ± ± ± ± = - ˆ i - 4 ˆ j - 6 ˆ k (e) What is the angle between and ? The dot product of ± A and ± B is ± A · ± B = | ± A || ± B | cos θ where is the angle between the two vectors. With: A = | ± A | = (2) 2 +( - 2) 2 + (1) 2 = 9=3 B = | ± B | = (4) 2 - 7) 2 + (4) 2 = 81 = 9 , and using the result from part (c), we obtain cos θ = ± A · ± B | ± A || ± B | = 26 3 × 9 . 963 θ = 15 . 6

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PS01 Solution - 2 Problem 1: Two Vectors continued (f) Find two unit vectors that are perpendicular to and . The cross product (or ) is perpendicular to both and . Therefore, from the result of part (d), the unit vectors may be obtained as ˆ n = ± ± A × ± B | ± A × ± B | = ± ( - ˆ i - 4 ˆ j - 6 ˆ k ) ± ( - 1) 2 +( - 4) 2 - 6) 2 = ± 1 53 ( - ˆ i - 4 ˆ j - 6 ˆ k ) Problem 2: Electrostatic Force Consider three point charges (A, B, C) located as shown in figure at right (where d is 9.0 cm). A has positive charge 1.0 μ C while B & C both have negative charge 3.0 μ C. Calculate the resultant electric force on A. Be sure to specify both the magnitude and direction. Let d ca = 1.5 d be the distance between A & C. The displacement from C to A is . The displacement from B to A is . Using Coulomb’s law, the force exerted on A is where are the charges on A, B and C respectively. Since q B = q C q BC = - 3 . 0 μC this becomes: Note that this is easier than the more typical method of finding the magnitudes first and then multiplying by the sine and cosine of angles to get the components. Now we can plug in numerical values: It is sufficient to leave the answer as this (both the magnitude and direction are indicated) but if you wanted to be explicit, you can calculate. The magnitude of is and the angle with respect to the + x axis is
PS01 Solution - 3 Problem 3: Charges Three charges equal to –Q , –Q and +Q are located a distance a apart along the x axis (see sketch).

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ps01sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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