Protons in an atomic nucleus are typically 10 to the negative 15th metres apart. What is the electrostatic force of repulsion between nuclear protons? The charge of a proton is 1.6021 times 10 to the negative 19th coulombs.
We can begin by calling the distance typically separating nuclear protons of 10 to the negative 15th metres 𝑑. We’re told that a proton’s charge is 1.6021 times 10 to the negative 19th coulombs, what we’ll call 𝑞. We want to solve for the electrostatic force of repulsion between nuclear protons. We’ll call this value 𝐹.
To begin on our solution, we can recall Coulomb’s law. Coulomb’s law tells us that the electrostatic force between two charges 𝑞 one and 𝑞 two is equal to the product of those charges times 𝑘, coulomb’s constant, divided by the distance between those two charges squared. We’ll treat Coulomb’s constant 𝑘 as exactly 8.99 times 10 to the ninth newton metres squared per coulomb squared.
Applying Coulomb’s law to our scenario, we use Coulomb’s constant and multiply it by the charge of a proton squared all divided by the distance between protons also squared. When we plug in for these values and then enter them on our calculator, we find that 𝐹 to two significant figures is equal to 230 newtons. That’s roughly the force that two atomic nuclei exert on one another to push one another apart.