#E11.+8N12NFnet=4NHW 3#E12F=qE=(‐3X10‐6C) X(8.5X104N/C) =‐2.55X10‐1NThe magnitude = 2.55X10‐1NThe direction = toward the left(the minus sign means thatEastdueN/C,102.67C101.58)N(12qFEfieldElectric66‐net×=×−===‐Ethe direction of the force on a negativecharge is opposite to the direction of the field)#E13ΔPE =ΔV∙q=(Vfinal‐Vinitial)∙q = (60 – 10)V∙(0.25C) =12.5J#E14ΔPE =ΔV∙q= VV)∙q =(06 )V∙(4C) = 24q=(Vfinal‐initial‐(4C) =‐24J#E16a.ΔPE =ΔV∙q=(Vfinal‐Vinitial)∙q =(500‐100 )V∙(‐5X10‐4C) =‐0.2JSo, magnitude is 0.2Jb. potential energy decreases. (because of minus sign)#SP50V‐‐‐a.ΔPE =ΔV∙q=(Vfinal‐Vinitial)∙q =(0‐400 )V∙(3X10‐4C) =‐0.12Jb The terminal of higher potential is labeled + and the terminal of lower potential is400V0.12m++++b. The terminal of higher potential is labeled + and the terminal of lower potential is
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This note was uploaded on 10/20/2011 for the course PHY 309L taught by Professor Sai during the Spring '08 term at University of Texas at Austin.