What is the magnitude of a charge chosen so that the electric field at 45 cm from the charge has magnitude 2.0 N/C
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E =9×10^9 q/R2
q=ER2/q9×10^9
=2×45×45×10^-4-9\9
450×10^-13 C
q=ER2/q9×10^9
=2×45×45×10^-4-9\9
450×10^-13 C
arti44:
hi but answer
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