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PHYS 272 Lecture Notes - Lecture 13: Electric Potential Energy, Dot Product, Electric Field

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cyanwallaby953
29 Aug 2018
School
University of Maryland
Department
Physics
Course
PHYS 272
Professor
Orozco

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Document Summary

Therefore, electric field changes but flux remains constant: two non-identical charges q and q are separating by a distance d; at a point p between them, the potential v = 0. In the given example, image b is an example of an electrostatic field because the same number of field lines enter the region as exit it. 3/2 r or treat the sphere as a point charge at the origin): =(cid:3018)(cid:3116) use gauss"s law, =(cid:3018)(cid:3116) electric field is constant, 4(cid:2024)(cid:4672)(cid:2871)(cid:2870)(cid:1844)(cid:4673)(cid:2870)=(cid:3018)(cid:3116) integral of area = volume, calculating electric field within sphere radius. Apply gauss"s law: =(cid:4672) (cid:2869)(cid:2872)(cid:3095)(cid:3116)(cid:4673)(cid:4678) (cid:3018)(cid:4672)(cid:3119)(cid:3118)(cid:3019)(cid:4673)(cid:3118)(cid:4679)= (cid:2173)(cid:2786)(cid:2174, (cid:2025)=(cid:3018)= (cid:3018)(cid:3120)(cid:3119)(cid:3095)(cid:3019)(cid:3119) (cid:1843)(cid:3031)(cid:3032)=(cid:2025)(cid:1848)(cid:3031)(cid:3032)= (cid:4678) (cid:3018)(cid:3120)(cid:3119)(cid:3095)(cid:3019)(cid:3119)(cid:4679)((cid:2872)(cid:2871)(cid:2024)(cid:4672)(cid:3019)(cid:2870)(cid:4673)(cid:2870))=(cid:1843)/8 find charge within the given, =(cid:3018)(cid:3279)(cid:3280) (cid:3116, (cid:3031)(cid:3032)(4(cid:2024)(cid:4672)(cid:3019)(cid:2870)(cid:4673)(cid:2870))=(cid:3018)(cid:3279)(cid:3280) (cid:3116) =(cid:3018)8, (cid:3031)(cid:3032)= (cid:2869)(cid:2872)(cid:3095)(cid:3116)(cid:4678) (cid:3018)8(cid:4672)(cid:3118)(cid:4673)(cid:3118)(cid:4679)= (cid:2173)(cid:2785)(cid:2174, (cid:2025)=(cid:2025)(cid:2868)(cid:1870) (cid:2025)(cid:2868)= , (cid:1869)=(cid:2025)(cid:1848) (cid:1843)= (cid:2025)(cid:1848, (cid:1843)= (cid:2025)(cid:2868)(cid:1870)(cid:1848),(cid:1848)=4(cid:2024)(cid:1870)(cid:2870)(cid:1870) due to spherical symmetry. Q4 = q, find total electrostatic potential energy of the array, and find the potential at the center of the square assuming that at infinity it is zero.

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

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Related Questions

An infinite nonconducting sheet has a surface charge density s =5.40 pC/m2.
(a) How much work is done by the electric field due to the sheet ifa particle of charge q = +1.60 × 10-19 C is moved from the sheet toa point P at distance d = 3.14 cm from the sheet?
J

(b) If the electric potential V is defined to be zero on the sheet,what is V at P?
V

An infinite nonconducting sheet has a surface charge density s =4.40 pC/m2.

(a) How much work is done by the electric field due to the sheet ifa particle of charge q = +1.60 × 10-19 C is moved from the sheet toa point P at distance d = 3.68 cm from the sheet?


(b) If the electric potential V is defined to be zero on the sheet,what is V at P?


A charge q that moves between two points A and B, and has anelectric force acting on it ( and perhaps several other forces) haswork W done on it by the electric force. The electrostatic force isconservative and so we can define a change in electric potentialenergy, Delta U = -W. We defined electric field using the relationF^^\-> = q E^^\->, and in a similar way we define electricpotential commonly called voltage as
Delta V = V_B - V_A = -W/q=(Delta U)/q or equivalently q Delta V =q (V_B - V_A) = -W=Delta U

A ball has mass 3.0 µg and charge -6 nC moves from point A to pointB. It has the same speed at points A and B, and the work done bythe electric field during the move is -1.5 mJ.
(a) What is the change in electric potential energy of the ball?1
(b) What is the change in voltage between points A and B? 2
(c) At which point does the ball have a higher electric potentialenergy?
(d) Which point has the higher voltage?
(e) Has the kinetic energy of the ball changed?