Electrostatic forces between macroscopic objects.

Charge on each gram of copper is Q_{total} $Q_{total} = Q_{net}N$ $Q_{net} = 0.01 * 29 * 1.6 * 10^{-19} C = 4.64 * 10^{-20} C$ $Q_{total} = 4.64 * 10^{-20} C * 9.5 * 10^{21} = 440.8 C$ $F = \frac {k(Q_{total})^2} {r^{2}}$ $F = \frac {9 * 10^{9} N m^{2} C^{-2} * (440.8 C)^2} {(1m)^{2}}$ $F = 1.749 * 10^{15} N$

The net charge on one atom of copper will be $0.01Z|Q(electron)| = 0.01(29)(1.6 * 10^{-19}) = 4.64 * 10^{-20} C$ .

Since there are $N = 9.5 * 10^{21}$ atoms in one gram of copper, each ball has a net charge of $(9.5 * 10^{21})(4.64 * 10^{-20}) = 440.8 C$ .

Now Coulomb's law may be applied: $F = k q_{1}q_{2} / r^{2} = (9 * 10^{9})(440.8)(440.8) / (1^{2}) = 1.75 * 10^{15} N$ .

The number of atoms in $1$ $g$ of copper $= N$ and each copper atom has $Z$ electrons each with $0.01e$ extra charge, therefore, the total extra charge $Q$ of $1\space g$ of copper is given by:

$Q = 0.01eNZ = 0.01 \times 1.6\times 10^{-19}\times 9.5\times 10^{21}\times 29 = 440.8\space C$

By Coulomb's Law, the force $F$ between two $1$ $g$ of copper separated by a distance $r=1$ $m$ apart is:

$F = k\frac{Q^2}{r^2}=9\times 10^9\times \frac{440.8^2}{1}=\boxed {1.75\times 10^{15}}\space N$

First, we determine the net electric charge on a copper atom. Qnet= 0.01Z|Q(electron)|=0.01×29×1.6×10-19=46.4×10-21 coulombs. We multiply the number of charge with the number of atoms in a 1-gram copper ball. 9.5×1021×Qnet=440.8 coulombs. Then, we use F=kq1q2(r^-2). (note: q1=q2 since the balls are identical). F=(9E+9)(440.8)(440.8)(1^-2). Thus we get 1.75E+15 newtons.

1g = 9.5E21 atoms or electron

1 atom charge is 0.01ZQ = 0.01 x 29 x 1.6E-19 = 4.64E-20 = a

Total charge for 1g , T = 9.5E21 x a = 440.8

R = 1

F= (Q1 x Q2) / (4πε R^2 ) = (Q1 x Q2) / (KR^2)

The overall charge on 1 atom of copper is given by 0.01Z|Q(electron)|, which is 0.01 \times 29 \times 1.6 \times 10^{-19}. This is multiplied by the number of atoms, 9.5 \times 10^{21}, giving us the charge of 1 ball. Coulomb's law states that the electrostatic force experienced by 2 charged bodies is \frac {Q 1 \times Q 2}{4 \times \pi \times \varepsilon_0 \times r^2}. putting in the values gives us \frac {(9.5 \times 10^{21} \times 0.01(29) \times 1.6 \times 10^{-19})^2}{4 \times 3.1415 \times \8.854 \times 10^{-12} \times 1}=1.75 \times 10^{15}