{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

1Bch23

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

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern