Question

Two protons are 9.261 fm apart. (1 fm= 1 femtometer = 1 x 10^-15 m.) What is the ratio of the electric force to the gravitational force on one proton due to the other proton?

Answer 1

Answer:

$1.24\times 10^{36}$

Explanation:

$q$ = magnitude of charge on each proton = 1.6 x 10⁻¹⁹ C

$m$ = mass of each proton = 1.67 x 10⁻²⁷ kg

r = distance between the two protons = 1 x 10⁻¹⁵ m

Electric force between the two protons is given as

$F_{e} = \frac{kq^{2}}{r^{2}}$

$F_{e} = \frac{(9\times 10^{9})(1.6\times 10^{-19})^{2}}{(1\times 10^{-15})^{2}}$

$F_{e} = 230.4$ N

Gravitational force between the two protons is given as

$F_{g} = \frac{Gm^{2}}{r^{2}}$

$F_{g} = \frac{(6.67\times 10^{-11})(1.67\times 10^{-27})^{2}}{(1\times 10^{-15})^{2}}$

$F_{g} = 1.86\times 10^{-34}$ N

Ratio is given as

$Ratio =\frac{F_{e}}{F_{g}}$

$Ratio =\frac{230.4}{1.86\times 10^{-34}}$

$Ratio = 1.24\times 10^{36}$

Answer 2

Answer:

$F_e:F_g = 1:1.237 * 10^{36}$

Explanation:

Parameters given:

Distance between the protons, r = $1 * 10^{-15} m$

Electric charge of proton, Q = $1.602 * 10^{-19} C$

Mass of proton, m = $1.673 *10^{-27} kg$

The electric force on one proton due to the other given as:

$F_e = \frac{k*Q*Q}{r^2}$

where k = Coulomb's constant

$F_e = \frac{9 * 10^9 * 1.602 * 10^{-19} * 1.602 * 10^{-19} }{(1 * 10^{-15})^2} \\\\\\F_e = 230.98 N$

The gravitational force on one proton due to the other is given as:

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

where G = gravitational constant

$F_g = \frac{6.67408 * 10^{-11}*1.673 *10^{-27} * 1.673 *10^{-27}}{(1 * 10^{-15})^2} \\\\\\F_g = 1.868 * 10^{-34} N$

Therefore, the ratio of electric force, $F_e$, and gravitational force, $F_g$, is:

$\frac{F_e}{F_g} = \frac{230.98}{1.868 * 10^{-34}} = \frac{1}{1.237 * 10^{36}} \\\\\\F_e:F_g = 1:1.237 * 10^{36}$