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Unformatted text preview: 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 xyplane. 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 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 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)....
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This note was uploaded on 01/21/2010 for the course PHYS 1B taught by Professor Kusenko during the Winter '07 term at UCLA.
 Winter '07
 Kusenko
 Charge, Magnetism

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