This expression multiplied by kqq gives the change in

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Unformatted text preview: al, choose U = 0 to be at infinity: Case A q2 1 UA = 4πε 0 d d Case B q1q2 1 U (r ) = 4πε 0 r 2d q2 1 UB = 4πε 0 2d U(r) UA > UB 0 0 23 r r U(d) U(2d) Preflight 2 Preflight 2 60 BB 50 40 30 20 10 0 A B C D E Change in potential energy equals the final potential energy minus the initial potential energy. To represent this, the expression((1/r2)-(1/r1)) is needed. This expression multiplied by kQq gives the change in potential energy. Since it is a change in potential energy, it must contain both r1 and r2, and since r1 is smaller than than r2, we subtract 1/r2 from 1/r1 34 U1 = 1 Qq 4πε 0 r1 ∆U ≡ U 2 − U 1 = U2 = 1 Qq 4πε 0 r2 Qq 1 1 − 4πε 0 r2 r1 Potential Energy of Many Charges Two charges are separated by a distance d. What is the change in potential energy when a third charge q is brought from far away to a distance d from the original two charges? Q2 d qQ1 1 qQ2 1 ∆U = + 4πε 0 d 4πε 0 d (superposition) d q Q1 d 25 Potential Energy of Many Charges What is the total energy required to bring in three identical charges, from infinitely far away to the points on an equilateral triangle shown. A) 0 B) Q1 4πε 0 d C) Q2 1 ∆U = 2 4πε 0 d D) Q2 1 ∆U = 3 4πε 0 d E) Q2 1 ∆U = 6 4πε 0 d ∆U = Q 2 d Q Work (by E) to bring in third charge : 27 3 Q2 W = ∑ Wi = − 4πε 0 d d d Work (by E field) to bring in first charge: W1 = 0 Work (by E field) to bring in second charge : BB Q 3 Q2 ∆U = + 4πε 0 d Q2 W2 = − 4π 0 d ε 1 Q2 1 Q2 2 Q2 W3 = − − =− 4πε0 d 4πε0 d 4πε0 d 1 Potential Energy of Many Charges Suppose one of the charges is negative. Now what is the total energy required to brin...
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This document was uploaded on 02/13/2014.

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