# 7.2 Electric potential and potential difference  (Page 2/12)

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## Calculating energy

You have a 12.0-V motorcycle battery that can move 5000 C of charge, and a 12.0-V car battery that can move 60,000 C of charge. How much energy does each deliver? (Assume that the numerical value of each charge is accurate to three significant figures.)

## Strategy

To say we have a 12.0-V battery means that its terminals have a 12.0-V potential difference. When such a battery moves charge, it puts the charge through a potential difference of 12.0 V, and the charge is given a change in potential energy equal to $\text{Δ}U=q\text{Δ}V.$ To find the energy output, we multiply the charge moved by the potential difference.

## Solution

For the motorcycle battery, $q=5000\phantom{\rule{0.2em}{0ex}}\text{C}$ and $\text{Δ}V=12.0\phantom{\rule{0.2em}{0ex}}\text{V}$ . The total energy delivered by the motorcycle battery is

$\text{Δ}{U}_{\text{cycle}}=\left(5000\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left(12.0\phantom{\rule{0.2em}{0ex}}\text{V}\right)=\left(5000\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left(12.0\phantom{\rule{0.2em}{0ex}}\text{J/C}\right)=\phantom{\rule{0.2em}{0ex}}6.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}$

Similarly, for the car battery, $q=60,000\phantom{\rule{0.2em}{0ex}}\text{C}$ and

$\text{Δ}{U}_{\text{car}}=\left(60,000\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left(12.0\phantom{\rule{0.2em}{0ex}}\text{V}\right)=7.20\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}$

## Significance

Voltage and energy are related, but they are not the same thing. The voltages of the batteries are identical, but the energy supplied by each is quite different. A car battery has a much larger engine to start than a motorcycle. Note also that as a battery is discharged, some of its energy is used internally and its terminal voltage drops, such as when headlights dim because of a depleted car battery. The energy supplied by the battery is still calculated as in this example, but not all of the energy is available for external use.

Check Your Understanding How much energy does a 1.5-V AAA battery have that can move 100 C?

$\text{Δ}U=q\text{Δ}V=\left(100\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left(1.5\phantom{\rule{0.2em}{0ex}}\text{V}\right)=150\phantom{\rule{0.2em}{0ex}}\text{J}$

Note that the energies calculated in the previous example are absolute values. The change in potential energy for the battery is negative, since it loses energy. These batteries, like many electrical systems, actually move negative charge—electrons in particular. The batteries repel electrons from their negative terminals ( A ) through whatever circuitry is involved and attract them to their positive terminals ( B ), as shown in [link] . The change in potential is $\text{Δ}V={V}_{B}-{V}_{A}=+12\phantom{\rule{0.2em}{0ex}}\text{V}$ and the charge q is negative, so that $\text{Δ}U=q\text{Δ}V$ is negative, meaning the potential energy of the battery has decreased when q has moved from A to B .

## How many electrons move through a headlight each second?

When a 12.0-V car battery powers a single 30.0-W headlight, how many electrons pass through it each second?

## Strategy

To find the number of electrons, we must first find the charge that moves in 1.00 s. The charge moved is related to voltage and energy through the equations $\text{Δ}U=q\text{Δ}V.$ A 30.0-W lamp uses 30.0 joules per second. Since the battery loses energy, we have $\text{Δ}U=-30\phantom{\rule{0.2em}{0ex}}\text{J}$ and, since the electrons are going from the negative terminal to the positive, we see that $\text{Δ}V=\text{+12.0}\phantom{\rule{0.2em}{0ex}}\text{V}\text{.}$

## Solution

To find the charge q moved, we solve the equation $\text{Δ}U=q\text{Δ}V:$

$q=\frac{\text{Δ}U}{\text{Δ}V}.$

Entering the values for $\text{Δ}U$ and $\text{Δ}V$ , we get

$q=\frac{-30.0\phantom{\rule{0.2em}{0ex}}\text{J}}{+12.0\phantom{\rule{0.2em}{0ex}}\text{V}}=\frac{-30.0\phantom{\rule{0.2em}{0ex}}\text{J}}{+12.0\phantom{\rule{0.2em}{0ex}}\text{J/C}}=-2.50\phantom{\rule{0.2em}{0ex}}\text{C}\text{.}$

The number of electrons ${n}_{e}$ is the total charge divided by the charge per electron. That is,

${n}_{e}=\frac{-2.50\phantom{\rule{0.2em}{0ex}}\text{C}}{-1.60\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{\phantom{\rule{0.2em}{0ex}}10}^{-19}\phantom{\rule{0.2em}{0ex}}{\text{C/e}}^{-}}=1.56\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{\phantom{\rule{0.2em}{0ex}}10}^{19}\phantom{\rule{0.2em}{0ex}}\text{electrons}\text{.}$

## Significance

This is a very large number. It is no wonder that we do not ordinarily observe individual electrons with so many being present in ordinary systems. In fact, electricity had been in use for many decades before it was determined that the moving charges in many circumstances were negative. Positive charge moving in the opposite direction of negative charge often produces identical effects; this makes it difficult to determine which is moving or whether both are moving.

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