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How many electrons have to be moved by a car battery containing $7.20 \times 10^{5} \textrm{ J}$ at 12 V to reduce the energy by 1%?
  1. $4.80 \times 10^{27}$
  2. $4.00 \times 10^{26}$
  3. $3.75 \times 10^{21}$
  4. $3.13 \times 10^{20}
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OpenStax College Physics for AP® Courses Solution, Chapter 19, Problem 5 (Test Prep for AP® Courses) (1:03)

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This is College Physics Answers with Shaun Dychko. Voltage is defined as the potential energy per charge. So, we can solve for the charge by multiplying both sides by <i>Q</i> over <i>V</i>. And then, we're left with the charge, it's the potential energy divided by the voltage. So, we're told that the battery is going to have its energy reduced by one percent, and we're given this initial amount of energy that it has, and so we multiply that by 0.01 to get one percent of that. And then, that's divided by the voltage of 12 volts. And then, we want to know the number of electrons, not the number of Coulombs, because all of this here will give us an answer in units of Coulombs, and so we're going to multiply that by one electron for every 1.6 times ten to the minus 19 Coulombs, and that will give us a number of electrons. This works out to 3.75 times ten to the 21 electrons, which is option C, as shown here.