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An atomic nucleus has a charge of +40e. What is the magnitude of the electric field...

An atomic nucleus has a charge of +40e. What is the magnitude of the electric field at a distance of 1.0 m from the center of the nucleus? (k=1/4πϵ0=8.99×109 N · m2/C2, e=1.60×10−19 C)

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Answer #1

1) 5.4536 x10

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Answer #2

charge Q = 40e = 40*1.6*10^-19 = 64*10^-19 C

Electric field E = kQ/r^2 = (9*10^9) * (64*10^-19)/1^2 = 5.76*10^-8 V/m

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