Question

It is known that the gravitational force of attraction between two protons is much weaker than the electrical repulsion. For two protons at a distance d apart, calculate the ratio of the size of the gravitational attraction to that of the electrical repulsion. Specifically, find the magnitude of Fgravitational/Felectrical. Use the following data: k = 8.99×109 Nm2/C2 charge on one of the protons = 1.60×10-19 C G = 6.67×10-11 Nm2/kg2 mass of one of the protons = 1.67×10-27 kg.

Answer 1

Answer:

$\frac{F_g}{F_e}=8.08\times 10^{-37}$

Explanation:

We are given that

$k=8.99\times 10^9 Nm^2/C^2$

Charge on proton=$q=1.6\times 10^{-19} C$

$G=6.67\times 10^{-11} Nm^2/Kg^2$

Mass of one proton=$m=1.67\times 10^{-27} Kg$

The distance between two protons=d

Gravitational attraction force=$F_g=\frac{Gm_1m_2}{d^2}$

Using the formula

Gravitational attraction force between two protons=$F_g=\frac{6.67\times 10^{-11}\times (1.67\times 10^{-27})^2}{d^2}$..(1)

Electric force =$F_e=\frac{kq_1q_2}{d^2}$

Electric repulsive force between two protons=$F_e=\frac{8.99\times 10^9\times (1.6\times 10^{-19})^2}{d^2}$

$\frac{F_g}{F_e}=\frac{\frac{6.67\times 10^{-11}\times (1.67\times 10^{-27})^2}{d^2}}{\frac{8.99\times 10^9\times (1.6\times 10^{-19})^2}{d^2}}$

$\frac{F_g}{F_e}=\frac{6.67\times (1.67\times 10^{-27})^2}{8.99\times 10^9\times (1.6\times 10^{-19})^2}$

$\frac{F_g}{F_e}=8.08\times 10^{-37}$