qqepe−=0 0000010. then the actual difference would be −=×−16 1025..CAmplified by a factor of 29 ×3 ×1022as indicated in the problem, this amounts to a deviation from perfect neutrality of ∆q=×××=−29 3 1016 100142225chchCCin a copper penny. Two such pennies, at
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.