What is the repulsive force between two protons that are located  .120 nm apart ?    ( Ans: + 1.6 X 10 -8 N)

Coulomb’s law tells us that the electrostatic force between two charged objects can be calculated by: ?= ??1?2/?^2Where k is a constant with a value of 8.99x109 Nm2/C2; ?1 and ?2 are the charges of the two particles (1.602x10-19 C), and r is the distance between the two particles. The force is attractive if the charges of the two particles is opposite (one’s positive and one’s negative) and repulsive if the charges have the same charge (both positive or both negative). Calculate the electrostatic force on proton #1 from the other two protons.Will the gravitational force between the protons be attractive or repulsive? What about the electrostatic force? Which has the larger magnitude, the electrostatic force or the gravitational force? Can the electrostatic and gravitational forces alone create a stable Lithium nucleus?

An alpha particle (charge 2e) is aimed directly at a gold nucleus(charge 79e). What minimum initial kinetic energy must the alphaparticle have to approach within 7.0 * 10-14 m of the center of thegold nucleus before reversing direction? Assume that the gold nucleus,which has about 40 times the mass of an alpha particle, remains at rest.

What charge is required to produce a repulsive force of 10 microNewtons on a charge q= 4.5 nanoC that is 5.0 m away?