Question

Strength of the electric force imagine two 1.0-g bags of protons, one at the earth's north pole and the other at the south pole.

Answer 1

(a) The mass of one proton is $m_p = 1.67 \cdot 10^{-27} kg$. Each bag has a mass of M=1.0 g=0.001 kg, so the number of protons in each bag is
$N= \frac{M}{m_p}= \frac{0.001 kg}{1.67 \cdot 10^{-27}kg}=6.0 \cdot 10^{23}$

(b) The distance between the two bags is twice the Earth's radius:
$r=2 \cdot 6.37 \cdot 10^6 m=1.27 \cdot 10^7 m$

The gravitational attraction between the two bags is given by
$F_g = G \frac{M_1 M_2}{r^2}=(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2}) \frac{(0.001 kg)^2}{(1.27 \cdot 10^7 m)^2}=4.1 \cdot 10^{-31}N$

To calculate the electrostatic repulsion, we should first calculate the total charge of each bag, which is the charge of one proton times the number of protons:
$Q=eN=(1.6 \cdot 10^{-19} C)(6.0 \cdot 10^{23})=9.6 \cdot 10^4 C$

And now we can calculate the electrostatic repulsion between the two bags:
$F_e = k_e \frac{Q_1 Q_2}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(9.6 \cdot 10^4 C)^2}{(1.27 \cdot 10^7 m)^2}=5.14 \cdot 10^5 N$

(c) we can easily see from point (b) that the gravitational attraction between the two bags is very tiny, so it cannot be felt. Instead, the electrostatic repulsion between the two bags is a very large number, so this can be felt.