Two protons are placed at some separation in Vacuum. Find the ratio of electric and gravitational force acting between them.

Explanation

Let the distance two protons having charge +e and mass m is placed at a distance r from each other as shown in figure.

` " " F_e = (1)/(4piin_0) .(q_1q_2)/(r^(2)) " " `( Coulomb.s law )
` F_g = G (m_1m_2)/(r^(2)) " " `(Gravitation law )
` = G (mxxm)/(r^(2))`
Where `F_e, and F_g` are electric and gravitational force respectively ,on putting the values , we get
` (F_e)/(F_g) = ((1)/(4piin_0) (e)/(r^(2)))/((Gm^(2))/(r^(2)))= (1)/(4piin_0) (e^(2))/(m^(2)) .(1)/(G) `
` " " = ((9xx 10 ^(9) ) (1.6 xx 10 ^(-19))^(2))/((1.67 xx 10^(-27))^(2)) xx (1)/((6.67 xx 10 ^(-11)))`
` " " = 1.2 xx 10 ^(36) = 10 ^(36)`
` rArr " " F gt gt F_g`