Need help in electric field problem

[chinasome]

Part A
Suppose all the electrons in a quantity of carbon atoms with a mass of 25.0 g were located at the North Pole of the Earth and all the protons at the South Pole. What would be the total force of attraction exerted on each group of charges by the other? The atomic number of carbon is 6, and the atomic mass of carbon is 12.0 g/mol.

Use 8.85×10−12 C^2/N*m^2 for the permittivity of free space, 1.60×10−19 C for the magnitude of the charge on an electron, 6.02×1023 mol^-1 for Avagadro's number, and 6.38×106 m for radius of the earth.

Part B
What would be the magnitude of the force exerted by the charges in part (a) on a third charge that is equal to the charge at the South Pole, and located at a point on the surface of the Earth at the equator?
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[Nate810]

Part AThe total number of electrons in 25.0 g of carbon is given by: N_e= (25.0 g * (6.02*10^23 mol^-1))/(12.0 g/mol) = 2.51*10^24 electrons The total force of attraction on each group of charges is given by: F_attraction = (2.51*10^24 * 1.60*10^-19 C^2)/(8.85*10^-12 C^2/N*m^2 * (6.38*10^6 m)^2) = 2.65*10^13 N Part BThe magnitude of the force exerted by the charges on a third charge located at the equator is given by: F_equator = (2.51*10^24 * 1.60*10^-19 C^2)/(8.85*10^-12 C^2/N*m^2 * (4.00*10^7 m)^2) = 5.32*10^11 N

 

[Moetasim]

Part A: To solve this problem, we first need to calculate the number of carbon atoms in 25.0 g of carbon, using the atomic mass and Avogadro's number. This comes out to be 1.25×10^23 atoms.

Next, we need to calculate the total charge of the electrons and protons at the North and South poles, respectively. Since each carbon atom has 6 electrons and 6 protons, the total charge at each pole would be 7.50×10^23 C.

Using Coulomb's law, we can calculate the force of attraction between the two groups of charges:

F = k(q1q2)/r^2

Where k is the Coulomb constant (8.99×10^9 N*m^2/C^2), q1 and q2 are the charges of the two groups, and r is the distance between them.

Plugging in the values, we get:

F = (8.99×10^9)(7.50×10^23)(7.50×10^23)/(6.38×10^6)^2 = 1.99×10^13 N

Therefore, the total force of attraction between the two groups of charges would be 1.99×10^13 N.

Part B: To calculate the force exerted on a third charge at the equator, we can use the same formula as above, but with the distance between the third charge and the South Pole being the radius of the Earth (6.38×10^6 m).

F = (8.99×10^9)(7.50×10^23)(7.50×10^23)/(6.38×10^6)^2 = 1.99×10^13 N

Therefore, the magnitude of the force exerted on the third charge would also be 1.99×10^13 N. However, since this third charge is equal in magnitude to the charges at the South Pole, the direction of the force would be repulsive, as like charges repel each other.