ps04sol

ps04sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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Problem Set 4 Solutions p. 1 of 7 Fall 2009 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Fall 2009 Problem Set 4 Solutions Problem 1: Two Vectors Given two vectors, ˆˆ ˆ (4 3 5 ) =− + Ai j k r and ˆ ( 744 =+ + Bi ) j k r , evaluate the following: (a) ; 2 + AB r r ˆˆˆ 22 ( 4 i 3 j 5 k ) ( 7 4 4 ) 1 5 2 1 += −+ + ++ = −+ i 4 j ki j k r r (b) ; 3 r r ˆ ˆ 3( 4 i 3 j 5 k ) 3 ( 7 4 4 )1 7 1 5 7 −=− + + + = −−− i j j k r r (c) ; r r Since and ˆˆ ˆˆ ˆˆ 1 ⋅=⋅=⋅= ii jj kk ˆ ˆ 0 ⋅=⋅ = ⋅= i jj kk i , the dot product is ( )( ) ( )( ) ( )( ) 47 34 54 3 6 ⋅= +− + = r r (d) ; × r r With ˆ ˆ ˆ ˆ ˆ , and , ×= × = i j k j j the cross product × r r is given by ˆ ˆ 43 5 3 2 1 93 7 ×= − = − + + ij k i j k r r (e) What is the angle between and B A r r ? The dot product of and B is A r r cos θ AB AB r r r r where is the angle between the two vectors. With: () 2 (4) 3 (5) 50 5 2 A == + −+ = = A r 222 (7) (4) (4) 81 9 B + + = = B r , and using the result from part (c), we obtain 30 cos 0.6 52.5 . 529 θθ = = r ° r r r
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Problem Set 4 Solutions p. 2 of 7 Fall 2009 Problem 1: Two Vectors continued (f) Find two unit vectors that are perpendicular to A r and B r . The cross product (or × AB r r × BA r r ) is perpendicular to both A r and B r . Therefore, from the result of part (d), the unit vectors may be obtained as ( ) () 222 ˆˆ ˆ 32 19 37 1 ˆ ˆ 32 19 37 2754 ( 32) (19) (37) −++ × + + × −+ + ijk ni j k r r r r 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 3.0 μ C while B & C both have negative charge -1.0 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 ˆ CA CA d = ri r . The displacement from B to A is ˆ BA d = − rj r . Using Coulomb’s law, the force exerted on A is 33 0 1 4 CAC A BAB A AC A CA BA qq rr πε ⎛⎞ =+= + ⎜⎟ ⎝⎠ FF F r r r where ,, A BC qqq are the charges on A, B and C respectively. Since 1.0 B C q C μ =≡ = this becomes: 3 2 00 1.5 44 1.5 BC A CA BA BC A A CA BA d =+ = Fi + j r 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: ( ) 66 92 - 2 23 2 1.0 10 C 3.0 10 C 1.5 ˆ ˆ 8.99 10 N m C 1.5 3.3 N 1.5 9.0 10 m A −− −× × = + × j i j r 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 A F r 22 2 2 ( 1.5N) (3.3N) 3.6N xy F = + = and the angle with respect to the + x axis is 11 3.3N tan tan 114º 1.5N y x F F ϕ == = CA F r BA F r A F r
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Problem Set 4 Solutions p. 3 of 7 Fall 2009 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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This note was uploaded on 05/06/2010 for the course 8 8.02 taught by Professor Hudson during the Fall '07 term at MIT.

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ps04sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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