PHYS102: Introduction to Electromagnetism

When a free positive charge $q$ is accelerated by an electric field, such as shown in Figure 19.2, it is given kinetic energy. The process is analogous to an object being accelerated by a gravitational field. It is as if the charge is going down an electrical hill where its electric potential energy is converted to kinetic energy. Let us explore the work done on a charge $q$ by the electric field in this process, so that we may develop a definition of electric potential energy.

Figure 19.2 A charge accelerated by an electric field is analogous to a mass going down a hill. In both cases potential energy is converted to another form. Work is done by a force, but since this force is conservative, we can write $W=-\Delta PE$.

The electrostatic or Coulomb force is conservative, which means that the work done on $q$ is independent of the path taken. This is exactly analogous to the gravitational force in the absence of dissipative forces such as friction. When a force is conservative, it is possible to define a potential energy associated with the force, and it is usually easier to deal with the potential energy (because it depends only on position) than to calculate the work directly.

We use the letters PE to denote electric potential energy, which has units of joules (J). The change in potential energy, $\Delta PE$, is crucial, since the work done by a conservative force is the negative of the change in potential energy; that is, $W=-\Delta PE$. For example, work 𝑊 done to accelerate a positive charge from rest is positive and results from a loss in PE, or a negative $\Delta PE$. There must be a minus sign in front of $\Delta PE$ to make $W$. PE can be found at any point by taking one point as a reference and calculating the work needed to move a charge to the other point.

Gravitational potential energy and electric potential energy are quite analogous. Potential energy accounts for work done by a conservative force and gives added insight regarding energy and energy transformation without the necessity of dealing with the force directly. It is much more common, for example, to use the concept of voltage (related to electric potential energy) than to deal with the Coulomb force directly.

Calculating the work directly is generally difficult, since $W=fd\: cos\: \theta$ and the direction and magnitude of $F$ can be complex for multiple charges, for odd-shaped objects, and along arbitrary paths. But we do know that, since $F=qE$, the work, and hence $\Delta PE$, is proportional to the test charge $q$. To have a physical quantity that is independent of test charge, we define electric potential $V$ (or simply potential, since electric is understood) to be the potential energy per unit charge:

$V=\frac{PE}{q}$ [equation 19.1]

Since PE is proportional to $q$, the dependence on $q$ cancels. Thus $V$ does not depend on $q$. The change in potential energy $\Delta PE$ is crucial, and so we are concerned with the difference in potential or potential difference $\Delta V$ between two points, where

$\Delta V=V_{B}-V_{A}=\frac{\Delta PE}{q}$ [equation 19.3]

The potential difference between points A and B, $V_{B}-V_{A}$, is thus defined to be the change in potential energy of a charge $q$ moved from A to B, divided by the charge. Units of potential difference are joules per coulomb, given the name volt (V) after Alessandro Volta.

$1V=1\: \frac{J}{C}$ [equation 19.4]

The familiar term voltage is the common name for potential difference. Keep in mind that whenever a voltage is quoted, it is understood to be the potential difference between two points. For example, every battery has two terminals, and its voltage is the potential difference between them. More fundamentally, the point you choose to be zero volts is arbitrary. This is analogous to the fact that gravitational potential energy has an arbitrary zero, such as sea level or perhaps a lecture hall floor.

In summary, the relationship between potential difference (or voltage) and electrical potential energy is given by

$\Delta V=\frac{\Delta PE}{q}$ and $\Delta PE=q\Delta V$ [equation 19.6]

Voltage is not the same as energy. Voltage is the energy per unit charge. Thus a motorcycle battery and a car battery can both have the same voltage (more precisely, the same potential difference between battery terminals), yet one stores much more energy than the other since $\Delta PE=q\Delta V$. The car battery can move more charge than the motorcycle battery, although both are 12 V batteries.

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 Figure 19.3.

The change in potential is $\Delta V=V_{B}-V_{A}=+12\: V$ and the charge $q$ is negative, so that $\Delta PE=q\Delta V$ is negative, meaning the potential energy of the battery has decreased when $q$ has moved from A to B.

Figure 19.3 A battery moves negative charge from its negative terminal through a headlight to its positive terminal. Appropriate combinations of chemicals in the battery separate charges so that the negative terminal has an excess of negative charge, which is repelled by it and attracted to the excess positive charge on the other terminal. In terms of potential, the positive terminal is at a higher voltage than the negative. Inside the battery, both positive and negative charges move.

The Electron Volt

The energy per electron is very small in macroscopic situations like that in the previous example – a tiny fraction of a joule. But on a submicroscopic scale, such energy per particle (electron, proton, or ion) can be of great importance. For example, even a tiny fraction of a joule can be great enough for these particles to destroy organic molecules and harm living tissue. The particle may do its damage by direct collision, or it may create harmful x rays, which can also inflict damage. It is useful to have an energy unit related to submicroscopic effects.

Figure 19.4 shows a situation related to the definition of such an energy unit. An electron is accelerated between two charged metal plates as it might be in an old-model television tube or oscilloscope. The electron is given kinetic energy that is later converted to another form – light in the television tube, for example. (Note that downhill for the electron is uphill for a positive charge.) Since energy is related to voltage by $\Delta PE=q\Delta V$, we can think of the joule as a coulomb-volt.

Figure 19.4 A typical electron gun accelerates electrons using a potential difference between two metal plates. The energy of the electron in electron volts is numerically the same as the voltage between the plates. For example, a 5000 V potential difference produces 5000 eV electrons.

On the submicroscopic scale, it is more convenient to define an energy unit called the electron volt (eV), which is the energy given to a fundamental charge accelerated through a potential difference of 1 V. In equation form,

$1\: eV=(1.60\times 10^{-19}\: C)(1\: V)=(1.60\times 10^{-19}\: C)(1\: J/C)$ [equation 19.13]

$=1.60\times 10^{-19}\: J$

An electron accelerated through a potential difference of 1 V is given an energy of 1 eV. It follows that an electron accelerated through 50 V is given 50 eV. A potential difference of 100,000 V (100 kV) will give an electron an energy of 100,000 eV (100 keV), and so on. Similarly, an ion with a double positive charge accelerated through 100 V will be given 200 eV of energy. These simple relationships between accelerating voltage and particle charges make the electron volt a simple and convenient energy unit in such circumstances.

The electron volt is commonly employed in submicroscopic processes – chemical valence energies and molecular and nuclear binding energies are among the quantities often expressed in electron volts. For example, about 5 eV of energy is required to break up certain organic molecules. If a proton is accelerated from rest through a potential difference of 30 kV, it is given an energy of 30 keV (30,000 eV) and it can break up as many as 6000 of these molecules (30,000 eV÷5 eV per molecule=6000 molecules). Nuclear decay energies are on the order of 1 MeV (1,000,000 eV) per event and can, thus, produce significant biological damage.

Conservation of Energy

The total energy of a system is conserved if there is no net addition (or subtraction) of work or heat transfer. For conservative forces, such as the electrostatic force, conservation of energy states that mechanical energy is a constant.

Mechanical energy is the sum of the kinetic energy and potential energy of a system; that is, $KE+PE = constant$. A loss of PE of a charged particle becomes an increase in its KE. Here PE is the electric potential energy. Conservation of energy is stated in equation form as

$KE+PE = constant$ [equation 19.15]

or

$KE_{i}+PE_{i}=KE_{f}+PE_{f}$ [equation 19.16]

where i and f stand for initial and final conditions. As we have found many times before, considering energy can give us insights and facilitate problem solving.

Source: Rice University, https://openstax.org/books/college-physics/pages/19-1-electric-potential-energy-potential-difference
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