Question

Example 15.1 The Forces in a Hydrogen Atom Goal Contrast the magnitudes of an electric force and a gravitational force. Problem The electron and proton of a hydrogen atom are separated (on the average) by a distance of about 5.30 x 1011 m. Find the magnitudes of the electric force and the gravitational force that each particle exerts on the other, and the ratio of the electric force, Fe, to the gravitational force, Fo Strategy Solving this problem is just a matter of substituting known quantities into the two force laws and then finding the ratio. Solution Substitute 1911 = 1921 = e and the distance into Coulomb's law to find the electric force. le? (... JON.m (1.6 x 10-19 C)? * 10-8N Fe= F,= Gmimin Substitute the masses and distance into Newton's law of gravity to find the gravitational force. F. = (6.67 x 10"kg) N. m² (9.11 x 10-31 kg) (1.67 x 10-27 kg) (5.3 x 10-11 m) Fg = * 10-47 N Find the ratio of the two forces. x 1039 Remarks The gravitational force between the charged constituents of the atom is negligible compared with the electric force between them. The electric force is so strong, however, that any net charge on an object quickly attracts nearby opposite charges, neutralizing the object. As a result, gravity plays a greater role in the mechanics of moving objects in everyday life. Exercise 15.1 Hints: Getting Started | I'm Stuck Find the magnitude of the electric force between two protons separated by one femtometer (10-15 m), approximately the distance between two protons in the nucleus of a helium atom, The answer may not appear large, but if not for the strong nuclear force, the two protons would fly apart at an initial acceleration of nearly 7 x 1028 m/s2 Fe = N

Answer 1

Fe = k*e^2/r^2

= 9*10^9*(1.6*10^-19)^2/(5.3*10^-11)^2

= 8.20*10^-8 N <<<<<<-------------Answer

Fg = G*me*mp/r^2

= 6.67*10^-11*9.1*10^-31*1.67*10^-27/(5.3*10^-11)^2

= 3.61*10^-47 N <<<<<<-------------Answer

Fe/Fg = 8.20*10^-8/(3.61*10^-47)

= 2.27*10^39 <<<<<<-------------Answer

magnitude of electric force between two protons,
Fe = k*q^2/r^2

= 9*10^9*(1.6*10^-19)^2/(10^-15)^2

= 230.4 N <<<<<<-------------Answer