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Electrostatics, coulomb force between 2 charges 
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#1
Mar1109, 02:55 AM

P: 591

in a mass of 12g of carbon there are NA= 6.02215x10^{23} atoms, each atom has 6 electrons, 6 protons and 6 neutrons. each electron has a charge of 1.6x10^{19}C and each proton has a charge of 1.6x10^{19}C, each neutron has no charge. if all the electrons were at the north pole and all the protons at the south pole, what would be the electric force between them.
(the (correct??) answer is F=73439043N) but i get: in 12g there are 6*NA = 3.61328994x10^24 electrons and 3.61328994x10^24 protons the total electron charge of the 12g is 3.61328994x10^24 * 1.6x10^19 = (q)= 578126.39C the total proton charge of the 12g is 3.61328994x10^24 * 1.6x10^19 = (q)= 578126.39C the distance between them is the diameter of earth, ie 2*radius of earth =2*6400000m=12800000m F=K(q)(q)/R^2 = K(q/R)^2 =9x10^24*(578126.39/12800000)^2 = 18359808.99N F=18359808.99N which is not the answer in my book, but if i use earths radius instead of diameter, i get F=K(q)(q)/R^2 = K(q/R)^2 =9x10^24*(578126.39/6400000)^2 = 73439235.97 N which is close enough to the answer, surely the distance between them MUST be the diameter if they are on opposite poles? where have i gone wrong? 


#2
Mar1109, 09:50 AM

Sci Advisor
HW Helper
P: 2,002

You are right, you should use the diameter. Sometimes even books make mistakes.
You should really use scientific notation. The way you write is is nearly illegible. Also, if you use 6 decimal places everywhere and then use K=9x10^9 (one decimal place) makes your final answer reliable to only decimal place. 


#3
Mar1109, 10:21 AM

P: 591

thanks



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