Question

Consider that earth is a positively charge object having charge equals to its mass (6×1024C)

and the moon is a negativity charge particle of charge (7×1022C)

a) Find the coulomb force between them if the distance is 384,403 km

b) if sun, moon and earth are collinear and the distance between sun and the center of moon and

earth is 149,597,887 km find Electric Field at sun?

Answer 1

a) The Coulomb's force between two charges is $F = \dfrac{kq_1q_2}{r^2}$ , where $q_1$ and $q_2$ are the charges and $r$ is the distance between charges. Here $k\approx 9\cdot10^9\,\mathrm{N\cdot m^2/C^2}$ .

In our case

$|F| = \dfrac{9\cdot10^9\,\mathrm{N\cdot m^2/C^2}\cdot 6\cdot10^{24}\,\mathrm{C}\cdot 7\cdot10^{22}\,\mathrm{C}}{384403000\,\mathrm{m}} = 9.8\cdot10^{48}\,\mathrm{N}.$

Due to the different signs of the charges the objects will attract.

b) The electric field made by the Earth will be

$E_e = \dfrac{kq_e}{R+r} = \dfrac{9\cdot10^9\,\mathrm{N\cdot m^2/C^2}\cdot 6\cdot10^{24}\,\mathrm{C}}{149597887000+384403000\,\mathrm{m}} = 3.60\cdot10^{23}\,\mathrm{N/C}$ .

The electric field made by the Moon will be

$E_m = \dfrac{kq_m}{R-r} = \dfrac{9\cdot10^9\,\mathrm{N\cdot m^2/C^2}\cdot 7\cdot10^{27}\,\mathrm{C}}{149597887000-384403000\,\mathrm{m}} =-4.22\cdot10^{21}\,\mathrm{N/C}$ .

The total field will be

$E_e+E_m = 3.60\cdot10^{23}\,\mathrm{N/C}-4.22\cdot10^{21}\,\mathrm{N/C} = 3.56\cdot10^{23}\,\mathrm{N/C}.$