Question

Coulomb’s Law

Two protons are separated by a distance of 3.00 x 10^-10 meters and fixed in place. a) What is the magnitude of the repulsive electric force between the protons? b) What is the magnitude of the attractive gravitational force between the protons? c) 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. d) 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)

In the diagram below, assume charges q1 and q2 are fixed in place. A charge q3 (-5.00 μC) is initially at point A and is later moved to point B. Find the magnitude and direction of the electric force on q3 when at location A.

q1 -12.0 μC 45.0 cm

Now find the electric force on charge q3 (-5.00 μC) when it is moved to location B.

will vote, please show all work if possible

Answer 1

a) The magnitude of the repulsive electric force Fez kezet ke r = 381510m = 9x109 x (116 Xl619]" = 2.56 X16 "M b) The magnitude of the gravitational force - - Gm m=1.6 fg = r2 r2 667X15" X (1:67x10-27) ? = 2.07 X1645 (3X1610) 2

- as compared c) d) There is no F = ke distance where Fe = fg as fy is very small So when d = a E. (Flaq