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1Bch23 - Chapter 23 As before the diculty is listed...

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Chapter 23 As before, the difficulty is listed immediately after each problem, ranging from 1 (least difficult) to 10 (most difficult). Problem 1 (Difficulty: 5) Consider a collection of N point charges { Q 1 , Q 2 , .... , Q N } , where each charge Q j is located at a point ( x j , y j ) in the xy -plane. 1: What is the potential at a point ( x, y )? 2: What is the electrostatic energy of the configuration? 3: Let Q > 0. Consider the specific case of a point charge - 2 Q located at ( - a, 0) and a point charge - Q located at (2 a, 0). A third point charge Q is initially located at (0 , 0). How much work is required to move charge Q from (0 , 0) to (0 , 3 a )? 4: For the case considered in part 3, what is the electrostatic energy of the system when charge Q is located at (0 , 3 a ), with charge - 2 Q at ( - a, 0) and charge - Q at (2 a, 0)? 5: More generally, what is the electrostatic energy of the system when charge Q is located at (0 , y ), with charge - 2 Q at ( - a, 0) and charge - Q at (2 a, 0)? For what values of y is this total energy positive? For what values of y is it negative? Solution: We begin with part 1 . Let r j denote the distance from the point ( x, y ) to the location ( x j , y j ) of charge Q j . Then r j = q ( x - x j ) 2 + ( y - y j ) 2 The total potential at ( x, y ) is obtained by adding the contributions from the N point charges: V ( x, y ) = N X j =1 kQ j r j = N X j =1 kQ j p ( x - x j ) 2 + ( y - y j ) 2 (1) We now consider part 2 . We must first determine the distance r ij between two point charges Q i and Q j . This separation is given by r ij = q ( x i - x j ) 2 + ( y i - y j ) 2 The interaction energy between two charges Q i and Q j is U ij = kQ i Q j r ij The total electrostatic energy of the system is obtained by summing over the interaction energy of each distinct pair of point charges. 1
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U = N - 1 X i =1 N X j = i +1 U ij = N - 1 X i =1 N X j = i +1 kQ i Q j r ij = N - 1 X i =1 N X j = i +1 kQ i Q j p ( x i - x j ) 2 + ( y i - y j ) 2 (2) The above double sum insures that each pair { i, j } occurs only once, and we have i 6 = j . We now consider part 3 . We have a charge - 2 Q at ( - a, 0) and a charge - Q at (2 a, 0). We wish to find the potential V i , due to these two charges, at (0 , 0), and the potential V f at (0 , 3 a ). The work required to move charge Q from (0 , 0) to (0 , 3 a ) is then W = Q ( V f - V i ) From equation (1) for V ( x, y ), we have for V i : V i = V (0 , 0) = k ( - 2 Q ) a + k ( - Q ) 2 a = - 5 kQ 2 a We have for V f : V f = V (0 , 3 a ) = k ( - 2 Q ) p a 2 + (3 a ) 2 + k ( - Q ) p (2 a ) 2 + (3 a ) 2 = - 2 kQ a 10 - kQ a 13 = - kQ a 2 10 + 1 13 The work required to move charge Q from (0 , 0) to (0 , 3 a ) is therefore W = Q ( V f - V i ) = Q - kQ a 2 10 + 1 13 + 5 kQ 2 a = kQ 2 a 5 2 - 2 10 - 1 13 1 . 59 kQ 2 a 2
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We now consider part 4 . We have charge - 2 Q at ( - a, 0), charge - Q at (2 a, 0), and charge Q at (0 , 3 a ). There are three distinct pairs of charges, and therefore three terms in the expression for the electrostatic energy, which is given by equation (2): U = k ( - 2 Q ) Q p a 2 + (3 a ) 2 + k ( - Q ) Q p (2 a ) 2 + (3 a ) 2 + k ( - 2 Q )( - Q ) 3 a = - 2 kQ 2 a 10 - kQ 2 a 13 + 2 kQ 2 3 a = kQ 2 a 2 3 - 2 10 - 1 13 ≈ - 0 . 24 kQ 2 a We now consider part 5 . We have charge - 2 Q at ( - a, 0), charge - Q at (2 a, 0), and charge Q at (0 , y ). Let us make the definitions β = y a U 0 = kQ 2 a Then equation (2) for the energy now gives U = k ( - 2 Q ) Q p a 2 + y
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