•1. A particular 12 V car battery can send a total charge of 84 A•h (ampere-hours) through a circuit, from one terminal to the other. (a) How many coulombs of charge does this represent? (Hint: See Eq. 21-3.) (b) If this entire charge undergoes a change in electric potential of 12 V, how much energy is involved?

A: (a) We're looking for an amount of charge, in coulombs, and the problem already gives us the charge anyway (albeit in different units). The first part of this problem is then just a simple unit conversion problem - just like in chapter 1! Just remember that 1 A = 1 C/s and 1 h = 3600 s.

84 A•h = (84 C•h/s)(3600 s/h) = 3.0 x 10⁵ C.

(b) A change in electric potential (ΔV) leads to a change in potential energy (ΔU). The variables are related by:

ΔU = qΔV = (3.0 x 10⁵ C)(12 V) = 3.6 x 10⁶ J

FINAL ANSWERS: (a) 3.0 x 10⁵ C
(b) 3.6 x 10⁶ J

•2. The electric potential difference between the ground and a cloud in a particular thunderstorm is 1.2 x 10⁹ V. In the unit electron-volts, what is the magnitude of the change in the electric potential energy of an electron that moves between the ground and the cloud?

A: A change in potential energy due to a change in electric potential is given by ΔU = qΔV, where q is the charge of the particle moving between each point. In this case, the particle is an electron, so q = e, the elementary charge. However, since we want an answer in electron-volts anyway, we end up not actually having to do much math:

ΔU = eΔV = 1.2 x 10⁹ eV = 1.2 GeV

FINAL ANSWER: 1.2 GeV