p0228 - 2.28. CHAPTER 2, PROBLEM 28 139 2.28 Chapter 2,...

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2.28. CHAPTER 2, PROBLEM 28 139 2.28 Chapter 2, Problem 28 The dimensional quantities and their dimensions are [ E ]= ML 2 T 2 , [ m o M, [ q Q, [ 6 o Q 2 T 2 3 , [ h 2 T where Q denotes dimensions of charge (coulombs). There are 5 dimensional quantities and 4 independent dimensions ( M,L,T,Q ) , so that the number of dimensionless groupings is 1. The appropriate dimensional equation is [ E ]=[ m o ] a 1 [ q ] a 2 [ 6 o ] a 3 [ h ] a 4 Substituting the dimensions for each quantity yields 2 T 2 = M a 1 Q a 2 Q 2 a 3 T 2 a 3 M a 3 L 3 a 3 M a 4 L 2 a 4 T a 4 = M a 1 a 3 + a 4 L 3 a 3 +2 a 4 T 2 a 3 a 4 Q a 2 +2 a 3 Thus, equating exponents, we arrive at the following four equations: 1= a 1 a 3 + a 4 2= 3 a 3 +2 a 4 2 a 3 a 4 0= a 2 a 3 We can solve the last two equations immediately for a 2 and a 3 ,v iz . , a 2 = 2 a 3 and a 3 = 1 2 a 4 1 Substituting into the second equation, we find 3 w 1 2 a 4 1 W a 4 =2 = a 4 = 2 Then, from our solution for a 2 and a 3 ,wehave a 2 =4 and a 3 = 2 Finally, substituting the values of a 3 and a 4 into the first equation, we have a 1 =1+ a 3 a 4 =1 2+2=1
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140 CHAPTER 2. DIMENSIONAL ANALYSIS Substituting back into the dimensional equation, we have [ E ]=[ m o ][ q ] 4 [ 6 o ] 2 [ h ] 2 = ^ m o q 4 6 2 o h 2 ± Therefore, we expect
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p0228 - 2.28. CHAPTER 2, PROBLEM 28 139 2.28 Chapter 2,...

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