Question

What is the emf of a battery that increases the electric potential energy of 0. 060 c of charge by 0. 80 j as it moves it from the negative to the positive terminal?

Answer 1

E(energy) = V * Q (joules /coulomb * coulombs) = joues)

V = E / Q = .80 joules / .060 coulombs) = 13.3 J/C