Question

It is known that the gravitational force of attraction between two alpha particles is much weaker than the electrical repulsion. For two alpha particles 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 Gravitational/Electrical.

Answer 1

Answer:

The ratio of gravitational force to electrical force is 3.19 x 10^-36

Explanation:

mass of an alpha particle = 6.64 x $10^{-27}$ kg

charge on an alpha particle = +2e = +2(1.6 x $10^{-19}$ C) = 3.2 x $10^{-19}$ C

distance between particles = d

For gravitational attraction:

The force of gravitational attraction F = $\frac{Gm^{2} }{r^{2} }$

where G = gravitational constant = 6.67 x $10^{-11}$ m^3 kg^-1 s^-2

r = the distance between the particles = d

m = the mass of each particle

therefore, gravitational force = $\frac{6.67*10^{-11}*(6.64*10^{-27} )^{2} }{d^{2} }$ = $\frac{2.94*10^{-63} }{d^{2} }$  Newton

For electrical repulsion:

Electrical force between the particles = $\frac{-kQ^{2} }{r^{2} }$

where k is the Coulomb's constant = 9.0 x $10^{9}$ N•m^2/C^2

r = distance between the particles = d

Q = charge on each particle

therefore, electrical force = $\frac{-9*10^{9}*(3.2*10^{-19} )^{2} }{d^{2} }$ = $\frac{-9.216*10^{-28} }{d^{2} }$ Newton

the negative sign implies that there is a repulsion on the particles due to their like charges.

Ratio of the magnitude of gravitation to electrical force = $\frac{2.94*10^{-63} }{9.216*10^{-28} }$

==> 3.19 x 10^-36