64 grams of Copper = 1 mole (6 times 10^23) of Copper Atoms that\'s about 64 cen

Question

64 grams of Copper = 1 mole (6 times 10^23) of Copper Atoms that's about 64 cents (of course, pennies aren't all Copper) Let's separate the protons from the electrons (29 each) in each Copper atom and put the protons and electrons in separate containers separated by 1000 meters. What would be the magnitude of the electrostatic force exerted by one container on the other? protons are 1836 times more massive than electrons Let's say we take one mole of Copper atoms (about one roll of pennies), and separate the protons and electrons from each atom (29 protons and 29 electrons per atom). We place all the protons and all the electrons at a one kilometer separation distance. What would be the force between these groups of charges at this distance? a. 1.9 times 10^-31 N b. 4.0 times 10^-9 N c. 8.4 times 10^13 N d. 2.5 times 10^10 N e. 7.0 times 10^16 N On the previous problem, when the protons and electrons are one kilometer apart, what is the magnitude of the electric field midway between these groups of charges? a. 1.0 times 10^11 N/C b. 2.0 times 10^11 N/C c. 5.0 times 10^10 N/C d. 1.0 times l0^14 N/C e. zero

Explanation / Answer

4) F = kq1q2/r^2, r = 1000 m

F = 9*10^9*29*29*(1.6*10^-19)^2/(1000^2)

F = 1.94*10^-31 N

Correct option is (a)

5) E = E1+E2

E = kq/r - kq/r

E =0

Correct option is (e)

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