Chapter 23--ExamplesProblemFind distance from corners to PPotentials (Voltage) is a scalarSlide 5Slide 6First, use Gauss’s law to find the E-field inside and outside the sphereOutside is simplerInsideVoltage=Outside+InsidePart b) What is the potential difference between a point on the surface and the sphere’s center?Okay, that is if V=0 at infinity what if V=0 at the center of the sphere?Slide 13E=-grad(V)DirectionsSlide 16Draw ItPotential DifferencePotential EnergyPower=Work per unit time1Chapter 23--Examples2ProblemIn the figure below, point P is at the center of the rectangle. With V=0 at infinity, what is the net electric potential at P due to the six charged particles?-2q+3q+5q-2q -3q+5qPdddddd3Find distance from corners to P-2q+3q+5q-2q -3q+5qPddddddsdd/24Potentials (Voltage) is a scalar654321VVVVVVVVii222dqkVsqkV53-2q+3q+5q-2q -3q+5qPddddddsdd/21234 5 6sqkV31225dqkVsqkV56sqkV345654321VVVVVVVVii dkqVdkqVsqdqkVsqdqsqsqdqsqkVVVVVVVV94.08541024522352236543216ProblemA charge q is distributed uniformly throughout a spherical volume of radius, R. a) Setting V=0 at infinity shown that the potential at a distance r from the center, where r<R is given byb) What is the potential difference between a point on the surface and the sphere’s center? 302283RrRqV7First, use Gauss’s law to find the E-field inside and outside the sphere rrqEqrEqAdEoutsideoutsideenclosedoutsideˆ14420020   rrRqrrErrERqVrEqAdEinsideinsideinsideenclosedinsideˆ4ˆ33414341430030230208Outside is simplerRqVrqVVdrrqrdrEsdEVVrrqERRRRRfiifoutside14|1414ˆˆ140020209Inside RqrRqVVRrRqVVrRqVVrdrRqsdEVVrrRqrrERinrRinrrRRinrrRrRRinrinside0230223023030300888|84ˆ4ˆ310Voltage=Outside+Inside)3(88838814881422302302300230002300rRRqVrRqRRqVRqrRqRqVRqrRqRqVVVVRinrR11Part b) What is the potential difference between a point on the surface and the sphere’s center?RqRqRqVVVVRqVratrRRqVCenterRCenter000022308834830)3(812Okay, that is if V=0 at infinity what if V=0 at the center of the sphere?30203020030003003080844ˆ4RqrVVBu tRqrVrdrRqVVrdrRqVVrrRqrErrrrrinsideRqVRratRin08Same as previousSame as previous13ProblemThe electric potential at points in a space are given byV=2x2-3y2+5z3What is the magnitude and direction of the electric field at the point (3,2,-1)?14E=-grad(V)CNEzyxEzyxEzzVyyVxxVzyxVVE/6.22151212ˆ15ˆ12ˆ12)1,2,3(ˆ)1(15ˆ)2(6ˆ)3(4)1,2,3(15645322222232215Directions01222014.1386.2215cos6.22cos135)1(tan11212tanˆ15ˆ12ˆ12)1,2,3(zyxrwhererzxyzyxEDirection w.r.t +x axisDirection w.r.t +z axis16ProblemThree +0.12 C charges form an equilateral triangle, 1.7 m on a side. Using energy that is supplied at a rate of 0.83 kW, how many days would be required to move one of the charges to the midpoint of the line joining the other two charges?17Draw It0.12C0.12C0.12C1.7 m1.7 m1.7 mInitially, this charge is 1.7 m from the other two chargesFinally, this charge is 0.85 m from the other two charges0.85 m0.85 m18Potential DifferenceCqmLwhereLkqVLkqLkqVVVLkqLkqLkqVLkqLkqLkqVVVViffiif12.07.1224422219Potential EnergyCqmLwhereJxxULkqVqU12.07.1101527.1)12.0)(109(22629220Power=Work per unit timeP=W/tW=-USo 0.83 kW= 830 J/sAnd t= U/P=152x106/830t=183,699 s or 51 hours or 2.12