f Facebook cowww.webassi \\ e solved: Consider the dia ¾ Chapter 25 Pre-Lecture

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f Facebook cowww.webassi e solved: Consider the dia ¾ Chapter 25 Pre-Lecture × G G calculator scientific. Gox × x / particle with charge c My Notes Ask Your A conducting sphere with aradius o0.25 m has a total charg a 5.d m.. A particle with a tharge ot -1.50 mC is initially 0.35 fram the sphere's center arnd is mawed to a tinal positian 0.44 n fram the spheres certer. (a) What is the difference in electnc potental between the part cle's final and initial positions. V- -K? (b) What is the change in the syste's eletric potetial energy? Need Help?Read IMastor It o1 points KPSLn 26P027 My Notes Ask Your A particle with charge a^--/ os uc is located at 0 3.25 cm and a second particle with charge qe-3.30 C s located at 0 2.75 cm what is the electric potential due to the two charges at the ollowing locabons? (a) at the onigin (b (2.70 un, 0) Need Help? Read ItMaster t Suomit Answer 5sve Practice Ancther version -1 points KarzPSE1 25.P0G4M Two identical metal balls of radi r 1.70 cm are at a center-to-center distance of d-2.60 m from each other as shown in the igure not drawn to scale). Each ball s charged so that a point at the surface of the first ball has an electric polential ofV1.70 x 103 V and a point at h surfae l he er ball has a elctripetetial e 1.70103 V. Whet is the tolal hrye on eauh bali? My Notes AskYour Type: here to search 920/2017 21

Explanation / Answer

Here,

q = 5.60 mC

the other charge is q1 = -1.50 mC

part a) due to sphere ,

change in potential difference = k * q * (1/r2 - 1/r1)

change in potential difference = 9 *10^9 * 5.6 *10^-3 * (1/0.44 - 1/.35)

change in potential difference = -2.95 *10^7 V

part b)

change in elctric potential energy = change in potential difference * charge

change in elctric potential energy = -2.95 *10^7 * (-1.5 *10^-3)

change in elctric potential energy = 44250 J

change in elctric potential energy is 44250 J

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