Question

An $\alpha$-particle is the nucleus of a helium atom. It has a mass $m=6.64\times {10}^{-27}$kg and a charge $q=+2e=3.2\times {10}^{-19}$ C. Compare the force of the electric repulsion between two a-particles with the force of gravitational attraction between them.

• A. $\frac{F_e}{F_g}=3.1\times {10}^{32}$
• B. $\frac{F_e}{F_g}=3.1\times {10}^{35}$
• C. $\frac{F_e}{F_g}=3.1\times {10}^{38}$
• D. $\frac{F_e}{F_g}=1\times {10}^{32}$

Answer 1

Answer: (B)

$\frac{F_e}{F_g}=3.1\times {10}^{35}$

Electrostatic repulsion = $F_e=\frac{1}{4\pi ϵ}\frac{q^2}{r^2}=\frac{1}{4\pi ϵ}\frac{4e^2}{r^2}$

gravitational attraction = $F_g=G\frac{m^2}{r^2}$

$\frac{F_e}{F_g}=\frac{1}{4\pi ϵ}\frac{4e^2}{G{m}^2}=3.1\times {10}^{35}$