1Refer to Interactive Solution 1925 to review one way in whi

1-Refer to Interactive Solution 19.25 to review one way in which this problem can be solved. Two protons are moving directly toward one another. When they are very far apart, their initial speeds are 1.79 106 m/s. What is the distance of closest approach?

Solution

A. When the protons are very far apart (rA = infinite), so that EPEa = 0 J. At the distance rB of closest approach, the speed of each proton is momentarily zero (vB = 0 m/s). With these substitutions, the conservation of energy equation reduces to

0.5*m*Va^2 + 0.5*m*Va^2 = k*e^2/rB

rB = ke^2/(m*Va^2) = 9*10^9*1.6*1.6*10^-38/(1.67*10^-27*1.79*1.79*10^12) = 4.30*10^-14 m

B. V =the potential energy per unit charge.

So the energy gained by a sodium ion is

W = Q*V = 0.074*1.6*10^-19 = 1.184*10^-20 Joule

F = W/d = QE

E= W/(d*Q) = 1.184*10^-20/(7.3*10^-9*1.6*10^-19) = 1.013*10^7 N/C