Question

Two protons are separated by a distance of 3.00 x 10-10 meters and fixed in place. What is the magnitude of the repulsive electric force between the protons? What is the magnitude of the attractive gravitational force between the protons? Is there any value of the distance between the protons (besides infinity) for which the gravitational force is equal to the electric force? If yes, what is the distance. If no, explain why Assume the repulsive electric force between the protons is F. If the separation between the protons is reduced by a factor of 3, what is the new electric force in terms of F? (e.g. 0.200 F or 27.0 F or 0.00534 F) a) b) c) d)

Answer 1

(a) F = K_c q^2/r^2 = 9 times 10^9 times (1.6 times 10^-19)^2/(3 times 10^-10)^2 = 2.56 times 10^-9 N (b) F_G = G m_1^2/r^2 = 6.67 times 10^-11 times (1.67 times 10^-27)^2/(3 times 10^-10)^2 = 2 times 10^-45 N (c) F = F_G implies K_c q^2/r^2 = Gm^2/r^2 = 2.56 times 10^-9 r = (6.67 times 10^-11 times 1.67 times 10^-27/2.56 times 10^-9)^1/2 r = 6.6 times 10^-15 m (d) F = 2.56 times 10^-9 K_c q^2/r^12 = 2.56 times 10^-9 if r^' = r/3, F^' = 9F