Electrostatics, coulomb force between 2 charges

AI Thread Summary
The discussion focuses on calculating the electric force between protons and electrons in a 12g mass of carbon, emphasizing the total charge and distance used in the calculations. The correct total charge for electrons and protons is established, leading to an initial force calculation using the Earth's diameter, which results in a significantly lower force than expected. A subsequent calculation using the Earth's radius yields a force that aligns closely with the book's answer. Participants note the importance of using scientific notation for clarity and precision in calculations. The conversation concludes with the acknowledgment that even textbooks can contain errors, reinforcing the need for careful verification.
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in a mass of 12g of carbon there are NA= 6.02215x1023 atoms, each atom has 6 electrons, 6 protons and 6 neutrons. each electron has a charge of -1.6x10-19C and each proton has a charge of 1.6x10-19C, 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 Earth's 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?
 
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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.
 
thanks
 
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