Problem: What is the change in potential energy if the distance separating two protons is increased to from 62 pm to 1.0 nm?

FREE Expert Solution

E → electrostatic potential energy

E=Efinal-Einitial


E=keQ1·Q2d


Initial distance of  62 pm:

1 pm = 10-12 m

d=62 pm×10-12 m1 pm

d = 6.2x10-11 m

E=(8.99×109N·m2C2)(1.6×10-19 C)·(1.6×10-19 C)(6.2×10-11 m)E=(8.99×109N·m2C2)(2.56×10-38 C2)(6.2×10-11 m)

Einitial = 3.712x10-18 N•m = 3.712x10-18 J


Final distance of 1.0 nm:

View Complete Written Solution
Problem Details

What is the change in potential energy if the distance separating two protons is increased to from 62 pm to 1.0 nm?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Coulomb's Law concept. If you need more Coulomb's Law practice, you can also practice Coulomb's Law practice problems.