Question

SIMULATION Coulomb's Law 5.0× 10-7 Stationary ion +1 Mobile ion 2 Clear Graph Click and drag to reposition the blue "mobile" ion. -1 Force of Attraction= 7.3 x 10-9 N Distance 1.79 A Distance (A) Now examine the strength of the interaction between the ions. Set the mobile ion charge to -1, leaving the stationary ion charge at +1. The distance between the two should still be 1.79 Angstroms What is the strength of the force of attraction between the two? (4 of 9)

Answer 1

Concept: here we apply Coulomb's law of force between charges to find the solution, ***************************************************************************************************
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Concept: here we apply Coulomb's law of force between charges to find the solution, F = Vertical-bar q_1 Vertical-barVertical-bar q_2 Vertical-bar /4 pi epsilon_0 r^2 = Vertical-bar -e Vertical-barVertical-bar + e Vertical-bar /4 pi epsilon_0 r^2 = Vertical-bar -1.6 times 10^-19 C Vertical-barVertical-bar 1.6 times 10^-19 C Vertical-bar /4 pi epsilon_0 (1.79 times 10^-10 m)^2 = 7.2 times 10^-9 N