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Solved: Figure 24-59 shows a ring of outer radius R = 13.0

Fundamentals of Physics Extended | 9th Edition | ISBN: 9780470469088 | Authors: David Halliday ISBN: 9780470469088 189

Solution for problem 80 Chapter 24

Fundamentals of Physics Extended | 9th Edition

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Fundamentals of Physics Extended | 9th Edition | ISBN: 9780470469088 | Authors: David Halliday

Fundamentals of Physics Extended | 9th Edition

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Problem 80

Figure 24-59 shows a ring of outer radius R = 13.0 cm, inner radius r = 0.200R, and uniform surface charge density if = 6.20 pC/m2. With V = 0 at infinity, find the electric potential at point P on the central axis of the ring, at distance z = 2.00R from the center of the ring.

Step-by-Step Solution:
Step 1 of 3

**Workischangeintotalenergyamount. **Kindsofenergycanchangeevenifnoworkisdone. • GravitationalPotentialEnergy o Lifting/loweringthings;“height” o U =Gmgh § U G=(mass)(gravity)(height) § Example:Massof60kgliftedfromheight3mà4m.Beforeand afterGPE,howmuchwork 2 U Gi(60kg)(9.8m/s )(3m)=1764J 2 U Gf(60kg)(9.8m/s )(4m)=2352J W=U GfU =Gi88J • KineticEnergy“motion” 2 o K=½mv Example:Howmuchenergydoesa0.25kgballhaveat5m/s K=½mv =½(0.25kg)(5m/s) 2

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Chapter 24, Problem 80 is Solved
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Textbook: Fundamentals of Physics Extended
Edition: 9
Author: David Halliday
ISBN: 9780470469088

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Solved: Figure 24-59 shows a ring of outer radius R = 13.0