# HW3 - HW3 #E11. 8N 12N + Electricfield=E = Fnet =4N Fnet...

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#E11. + 8N 12N F net =4N HW 3 #E12 F =q E=( 3X10 6 C) X ( 8.5X10 4 N/C) = 2.55X10 1 N The magnitude = 2.55X10 1 N The direction = toward the left(the minus sign means that East due N/C, 10 2.67 C 10 1.5 8)N (12 q F E field Electric 6 6 net × = × = = = E the direction of the force on a negative charge is opposite to the direction of the field) #E13 Δ PE = Δ V q=(V final V initial ) q = (60 – 10)V (0.25C) =12.5J #E14 Δ PE = Δ V q= V V ) q =(0 6 )V (4C) = 24 q=(V final initial (4C) = 24J #E16 a. Δ PE = Δ V q=(V final V initial ) q =(500 100 )V ( 5X10 4 C) = 0.2J So, magnitude is 0.2J b. potential energy decreases. (because of minus sign) #SP5 0V ‐‐‐ a. Δ PE = Δ V q=(V final V initial ) q =(0 400 )V (3X10 4 C) = 0.12J b The terminal of higher potential is labeled + and the terminal of lower potential is 400V 0.12m + + + + 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.

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