hw6sol - HW 5 Physics 1B Spring 2010 C M Cooper Teaching...

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HW 5: Physics 1B Spring 2010 C. M. Cooper, Teaching Assitant W. Gekelman, Professor May 7, 2010 Abstract This hw covers all of chapter 23. This includes the electric potential energy U and electric potential V , as well as the relations between F, E, U, V . The emphasis is most on potential, V and includes two ways to determine V , specifically the formular for the voltage between two points V a - V b = - Z a b E · dl (1) with the generalized V ( r ) = V ( r ) - V ( ) = - Z r E · dl (2) In addition, the potential can be calulated by integrating over the charge as per Chapter 21: V ( r ) = k Z dQ ( r ) r (3) which is not a vector so is a little easier to calculate 1 Question 23.11: You’ve got Potential Energy This question requires you to realize that the potential energy is of a system , that is, between all charges. You wouldn’t ever try calculate the force of a many bodied problem because its not defined. You can find the force between two charges, or of multiple charges on a single one. However, you can find the potential energy between two charges, or of multiple charges on a single (the work required to put that lone charge there from infinity) and you can calulated the potential energy of the whole system, by adding the potential energy between all the charges with every other charge (the work require to put all the charges in their current locations). U > 0 implies that you had to put energy into the system by overcoming a force to do the work requied to move the charge from infinity. U < 0 implies that the charge is being pulled into its position from infinity. Let us calculate the total energy of the total system with the 3rd charge ( q a ) being the one we are placing with respect to the exisiting charges ( q o ) and set it equal to zero so that no work is required to set this up U tot = U 12 + U 23 + U 13 = kq 2 o d + kq o q a d + kq o q a d = 0 (4) 1
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k ’s and d ’s cancel out and solving for q a gives q a = - q o 2 (5) which makes sense as it should be negative to offset the positive U needed to put the first two charges together. 2 Question 23.24: Potential of a dipole 2.1 (a) an expression for the potential at all points If you think of the distance in vectors, it makes this problem a lot easier to set up and you get trivially that V = - kq | y - a | + kq | y + a | (6)
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