what is the electric potential difference needed to accelaate a proton

Learning Objectives

Past the stop of this department you will be able to:

• Define electrical potential and electrical potential energy.
• Describe the relationship between potential divergence and electrical potential energy.
• Explain electron volt and its usage in submicroscopic processes.
• Decide electric potential free energy given potential departure and amount of accuse.

The data presented in this department supports the following AP® learning objectives and scientific discipline practices:

• 2.C.1.one The student is able to predict the direction and the magnitude of the force exerted on an object with an electric charge q placed in an electric field Due east using the mathematical model of the relation between an electrical force and an electric field: F = q Eastward; a vector relation. (S.P. six.4, 7.2)
• 2.C.i.2 The student is able to summate any ane of the variables—electrical forcefulness, electric accuse, and electrical field—at a indicate given the values and sign or direction of the other two quantities. (South.P. 2.2)
• v.B.2.ane The student is able to summate the expected behavior of a arrangement using the object model (i.e., by ignoring changes in internal construction) to clarify a state of affairs. And so, when the model fails, the educatee can justify the use of conservation of energy principles to summate the change in internal energy due to changes in internal construction because the object is actually a system. (S.P. 1.4, 2.1)
• 5.B.3.1 The pupil is able to depict and brand qualitative and/or quantitative predictions about everyday examples of systems with internal potential energy. (Southward.P. ii.2, six.4, 7.two)
• 5.B.3.2 The student is able to make quantitative calculations of the internal potential energy of a system from a description or diagram of that system. (S.P. 1.4, 2.ii)
• v.B.iii.three The student is able to apply mathematical reasoning to create a description of the internal potential free energy of a system from a description or diagram of the objects and interactions in that system. (South.P. one.four, 2.2)
• 5.B.4.1 The pupil is able to describe and make predictions about the internal energy of systems. (S.P. vi.4, 7.2)
• 5.B.four.2 The student is able to calculate changes in kinetic energy and potential free energy of a organization, using information from representations of that organization. (S.P. 1.four, 2.ane, 2.ii)

When a gratuitous positive charge $q$ is accelerated by an electric field, such equally shown in Figure 19.2, it is given kinetic energy. The process is coordinating to an object existence accelerated past a gravitational field. Information technology is every bit if the charge is going down an electrical hill where its electric potential energy is converted to kinetic energy. Let united states explore the work washed on a charge $q$ by the electric field in this process, so that we may develop a definition of electric potential energy.

Figure 19.two A charge accelerated by an electrical field is analogous to a mass going downwardly a loma. In both cases potential free energy is converted to another form. Piece of work is done by a forcefulness, but since this force is bourgeois, nosotros can write $W=\mathrm{–\Delta }\text{PE}$.

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

We use the messages PE to denote electric potential energy, which has units of joules (J). The alter in potential free energy, $\text{Δ}\text{PE}$ , is crucial, since the work washed by a bourgeois force is the negative of the change in potential energy; that is, $W=\text{–Δ}\text{PE}$ . For example, piece of work $W$ done to accelerate a positive accuse from rest is positive and results from a loss in PE, or a negative $\text{Δ}\text{PE}$ . At that place must be a minus sign in front of $\text{Δ}\text{PE}$ to make $W$ positive. PE tin can be found at whatsoever point by taking 1 point as a reference and computing the work needed to move a charge to the other point.

Potential Energy

$W=\text{–ΔPE}$ . For example, work $\mathrm{Due\; west}$ done to accelerate a positive charge from rest is positive and results from a loss in PE, or a negative $\text{ΔPE}.$ There must exist a minus sign in front end of $\text{ΔPE}$ to brand $\mathrm{West}$ positive. PE tin can be found at any signal by taking ane indicate as a reference and calculating the work needed to move a charge to the other indicate.

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

Calculating the piece of work directly is generally difficult, since $\mathrm{Westward}=\text{Fd}\text{cos}\theta$ and the direction and magnitude of $F$ tin be complex for multiple charges, for odd-shaped objects, and along arbitrary paths. Simply nosotros exercise know that, since $F=\text{qE}$ , the piece of work, and hence $\text{ΔPE}$ , is proportional to the test charge $\mathrm{q.}$ To have a concrete quantity that is independent of exam accuse, we ascertain electric potential $V$ (or simply potential, since electric is understood) to be the potential energy per unit charge:

$$\mathrm{Five}=\frac{\text{PE}}{q}\text{.}$$

nineteen.one

Electric Potential

This is the electric potential free energy per unit charge.

$$V=\frac{\text{PE}}{q}$$

19.2

Since PE is proportional to $q$ , the dependence on $q$ cancels. Thus $\mathrm{Five}$ does non depend on $q$ . The modify in potential free energy $\text{ΔPE}$ is crucial, so we are concerned with the difference in potential or potential difference $\mathrm{\Delta }V$ between 2 points, where

$$\mathrm{\Delta }5=V_{\text{B}}-V_{\text{A}}=\frac{\mathrm{\Delta }\text{PE}}{q}\text{.}$$

19.3

The potential difference betwixt points A and B, $V_{\text{B}}–V_{\text{A}}$ , is thus divers to exist 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 (5) after Alessandro Volta.

$$\text{i V = 1}\frac{\mathrm{J}}{\mathrm{C}}$$

19.4

Potential Divergence

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

$$\text{1 V = ane}\frac{\mathrm{J}}{\mathrm{C}}$$

19.5

The familiar term voltage is the mutual proper noun for potential difference. Continue in mind that whenever a voltage is quoted, it is understood to exist the potential difference betwixt two points. For example, every bombardment has two terminals, and its voltage is the potential difference betwixt them. More fundamentally, the point you cull to be zero volts is capricious. This is coordinating to the fact that gravitational potential free energy has an arbitrary zero, such as sea level or perchance a lecture hall floor.

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

$$\mathrm{\Delta }V=\frac{\text{ΔPE}}{q}\text{and ΔPE =}q\mathrm{\Delta }V\text{.}$$

19.6

Potential Departure and Electric Potential Energy

The relationship betwixt potential deviation (or voltage) and electrical potential energy is given by

$$\mathrm{\Delta }V=\frac{\text{ΔPE}}{q}\text{and ΔPE =}q\mathrm{\Delta }5\text{.}$$

xix.7

The second equation is equivalent to the first.

Real World Connections: Electric Potential in Electronic Devices

You lot probably use devices with stored electric potential daily. Do you own or employ any electronic devices that practise not have to exist fastened to a wall socket? What happens if you utilize these items long enough? Do they stop operation? What do you lot do in that case? Choose 1 of these types of electronic devices and decide how much electric potential (measured in volts) the item requires for proper performance. So estimate the amount of fourth dimension between replenishments of potential. Describe how the time betwixt replenishments of potential depends on use.

Answer

Ready examples include calculators and prison cell phones. The former will either be solar powered, or accept replaceable batteries, probably four 1.5 V for a total of 6 5. The latter will need to be recharged with a specialized charger, which probably puts out five Five. Times will be highly dependent on which item is used, but should exist less with more intense use.

Voltage is non the same as energy. Voltage is the energy per unit of measurement charge. Thus a motorcycle bombardment and a auto battery can both have the same voltage (more precisely, the same potential divergence between bombardment terminals), however one stores much more energy than the other since $\text{ΔPE =}q\mathrm{\Delta }5$ . The car bombardment can move more charge than the motorcycle battery, although both are 12 V batteries.

Example 19.1

Calculating Energy

Suppose you take a 12.0 V motorcycle battery that can motion 5000 C of accuse, and a 12.0 V car bombardment that tin motility threescore,000 C of charge. How much energy does each evangelize? (Assume that the numerical value of each charge is authentic to three significant figures.)

Strategy

To say nosotros have a 12.0 V battery means that its terminals accept a 12.0 V potential difference. When such a battery moves accuse, it puts the charge through a potential departure of 12.0 5, and the charge is given a modify in potential energy equal to $\text{ΔPE =}q\mathrm{\Delta }5$ .

Then to observe the free energy output, nosotros multiply the accuse moved by the potential difference.

Solution

For the motorcycle battery, $q=\text{5000 C}$ and $\mathrm{\Delta }\mathrm{Five}=\text{12.0 V}$ . The full energy delivered by the motorcycle bombardment is

$$\begin{array}{lll}{\text{ΔPE}}_{\text{wheel}}& =& \left(\text{5000 C}\right)\left(\text{12.0 Five}\right)\\ & =& \left(\text{5000 C}\right)\left(\text{12.0 J/C}\right)\\ & =& \mathrm{vi.00}\times {\text{ten}}^4\text{J.}\end{array}$$

19.viii

Similarly, for the automobile bombardment, $q=\text{60},\text{000}\text{C}$ and

$$\begin{array}{lll}{\text{ΔPE}}_{\text{car}}& =& \left(\text{lx,000 C}\right)\left(\text{12.0 V}\right)\\ & =& 7.20\times {\text{ten}}^5\text{J.}\end{array}$$

19.ix

Give-and-take

While voltage and energy are related, they are not the same affair. The voltages of the batteries are identical, but the energy supplied by each is quite different. Notation also that as a battery is discharged, some of its free energy is used internally and its terminal voltage drops, such as when headlights dim because of a low car battery. The energy supplied by the battery is however calculated as in this example, merely not all of the free energy is available for external use.

Note that the energies calculated in the previous example are accented values. The change in potential energy for the bombardment is negative, since it loses energy. These batteries, similar many electrical systems, really move negative charge—electrons in detail. The batteries repel electrons from their negative terminals (A) through whatsoever circuitry is involved and attract them to their positive terminals (B) as shown in Figure xix.3. The change in potential is $\mathrm{\Delta }\mathrm{Five}=V_{\text{B}}{\mathrm{–V}}_{\text{A}}=\text{+12 5}$ and the charge $q$ is negative, so that $\text{ΔPE}=q\mathrm{\Delta }V$ is negative, meaning the potential free energy of the battery has decreased when $q$ has moved from A to B.

Figure 19.iii A battery moves negative charge from its negative terminal through a headlight to its positive terminal. Appropriate combinations of chemicals in the battery carve up charges so that the negative terminal has an excess of negative accuse, which is repelled by it and attracted to the excess positive accuse 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.

Making Connections: Potential Energy in a Battery

The previous instance stated that the potential energy of a battery decreased with each electron it pushed out. However, shouldn't this reduced internal energy reduce the potential, as well? Yes, it should. And so why don't we notice this?

Part of the answer is that the corporeality of energy taken by any i electron is extremely small, and therefore information technology doesn't reduce the potential much. But the main reason is that the energy is stored in the bombardment as a chemical reaction waiting to happen, not as electric potential. This reaction just runs when a load is attached to both terminals of the battery. Any one set of chemical reactants has a sure maximum potential that it tin can provide; this is why larger batteries consist of cells attached in series, so that the overall potential increases additively. As these reactants get used up, each jail cell gives less potential to the electrons it is moving; eventually this potential falls below a useful threshold. Then the battery either needs to be charged, which reverses the chemic reaction and reconstitutes the original reactants; or changed.

Case 19.2

How Many Electrons Move through a Headlight Each Second?

When a 12.0 5 car battery runs a single thirty.0 W headlight, how many electrons pass through information technology each second?

Strategy

To notice the number of electrons, we must first find the charge that moved in one.00 s. The accuse moved is related to voltage and free energy through the equation $\text{ΔPE}=q\mathrm{\Delta }5$ . A xxx.0 Due west lamp uses 30.0 joules per 2d. Since the bombardment loses energy, nosotros have $\text{ΔPE}=\text{–30.0 J}$ and, since the electrons are going from the negative terminal to the positive, nosotros run into that $\mathrm{\Delta }V=\text{+12.0 V}$ .

Solution

To discover the accuse $q$ moved, nosotros solve the equation $\text{ΔPE}=q\mathrm{\Delta }5$ :

$$q=\frac{\text{ΔPE}}{\mathrm{\Delta }V}\text{.}$$

19.10

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

$$q=\frac{\text{–thirty.0 J}}{\text{+12.0 V}}=\frac{\text{–30.0 J}}{\text{+12.0 J/C}}=\mathrm{–2.50\; C.}$$

xix.eleven

The number of electrons ${\text{due north}}_{\text{eastward}}$ is the total charge divided past the charge per electron. That is,

$${\text{n}}_{\text{due east}}=\frac{\mathrm{–2.50\; C}}{\mathrm{–1.lx}\times {\text{10}}^{\text{–xix}}{\text{C/eastward}}^{–}}=1.56\times {\text{10}}^{\text{19}}\text{electrons.}$$

19.12

Discussion

This is a very large number. It is no wonder that we practise not ordinarily observe individual electrons with so many beingness present in ordinary systems. In fact, electricity had been in use for many decades earlier it was determined that the moving charges in many circumstances were negative. Positive accuse moving in the contrary direction of negative accuse oft produces identical furnishings; this makes information technology difficult to determine which is moving or whether both are moving.

Applying the Science Practices: Work and Potential Free energy in Signal Charges

Consider a system consisting of two positive point charges, each two.0 µC, placed one.0 one thousand away from each other. We tin calculate the potential (i.e. internal) energy of this configuration by computing the potential due to one of the charges, so calculating the potential free energy of the second charge in the potential of the first. Applying Equations (19.38) and (nineteen.ii) gives us a potential energy of three.half-dozen×10^-ii J. If we move the charges closer to each other, say, to 0.50 g autonomously, the potential energy doubles. Annotation that, to create this second instance, some outside force would have had to exercise piece of work on this system to change the configuration, and hence it was non a closed system. However, because the electric force is conservative, we can use the work-energy theorem to state that, since in that location was no change in kinetic free energy, all of the work done went into increasing the internal energy of the system. Also notation that if the point charges had different signs they would be attracted to each other, so they would be capable of doing piece of work on an exterior system when the distance between them decreased. As work is done on the outside organization, the internal energy in the two-accuse system decreases.

Effigy 19.four Work is done by moving ii charges with the same sign closer to each other, increasing the internal energy of the 2-charge system.

The Electron Volt

The free energy per electron is very modest 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 case, even a tiny fraction of a joule can be neat enough for these particles to destroy organic molecules and harm living tissue. The particle may practice its damage past direct collision, or information technology may create harmful x rays, which can also inflict harm. It is useful to have an energy unit of measurement related to submicroscopic effects. Effigy 19.5 shows a situation related to the definition of such an energy unit. An electron is accelerated between two charged metal plates every bit it might be in an old-model television tube or oscilloscope. The electron is given kinetic energy that is subsequently 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 $\text{ΔPE}=q\mathrm{\Delta }5,$ we tin can call up of the joule as a coulomb-volt.

Figure 19.5 A typical electron gun accelerates electrons using a potential difference betwixt two metal plates. The energy of the electron in electron volts is numerically the aforementioned as the voltage between the plates. For example, a 5000 5 potential departure produces 5000 eV electrons.

On the submicroscopic calibration, it is more convenient to define an energy unit of measurement called the electron volt (eV), which is the energy given to a fundamental accuse accelerated through a potential departure of 1 5. In equation grade,

$$\begin{array}{lll}\text{ane eV}& =& \left(\mathrm{1.lx}\times {\text{10}}^{\text{–nineteen}}\text{C}\right)\left(\mathrm{1\; 5}\right)=\left(\mathrm{i.60}\times {\text{10}}^{\text{–19}}\text{C}\right)\left(\mathrm{1\; J/C}\right)\\ & =& \mathrm{i.lx}\times {\text{10}}^{\text{–19}}\text{J.}\end{array}$$

nineteen.13

Electron Volt

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

$$\begin{array}{lll}\text{1 eV}& =& \left(1.60\times {\text{10}}^{\text{–19}}\text{C}\right)\left(\mathrm{ane\; 5}\right)=\left(\mathrm{i.60}\times {\text{10}}^{\text{–19}}\text{C}\right)\left(\mathrm{1\; J/C}\right)\\ & =& \mathrm{1.threescore}\times {\text{10}}^{\text{–19}}\text{J.}\end{array}$$

19.14

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

Connections: Energy Units

The electron volt (eV) is the well-nigh common energy unit of measurement for submicroscopic processes. This will be particularly noticeable in the chapters on mod physics. Energy is and then of import to so many subjects that there is a tendency to define a special energy unit of measurement for each major topic. There are, for example, calories for nutrient energy, kilowatt-hours for electric free energy, and therms for natural gas free energy.

The electron volt is commonly employed in submicroscopic processes—chemic valence energies and molecular and nuclear binding energies are amidst the quantities often expressed in electron volts. For case, about 5 eV of energy is required to break up certain organic molecules. If a proton is accelerated from residue through a potential difference of thirty kV, it is given an energy of 30 keV (xxx,000 eV) and it can interruption upward as many every bit 6000 of these molecules ( $\text{30,000 eV}÷\text{5 eV per molecule}=\text{6000 molecules}$ ). Nuclear decay energies are on the guild of 1 MeV (one,000,000 eV) per outcome and can, thus, produce meaning biological damage.

Conservation of Free energy

The total free energy of a organisation is conserved if there is no internet addition (or subtraction) of work or heat transfer. For bourgeois forces, such as the electrostatic strength, conservation of free energy states that mechanical energy is a constant.

Mechanical free energy is the sum of the kinetic energy and potential free energy of a system; that is, $\text{KE}+\text{PE = constant}$ . A loss of PE of a charged particle becomes an increment in its KE. Here PE is the electric potential free energy. Conservation of energy is stated in equation course as

$$\text{KE}+\text{PE = constant}$$

19.xv

or

$${\text{KE}}_{\mathrm{i}}+{\text{PE}}_{\mathrm{i}}{\text{= KE}}_f+{\text{PE}}_f,$$

19.16

where i and f represent initial and final conditions. Equally we have institute many times before, considering energy can requite u.s. insights and facilitate problem solving.

Case 19.3

Electrical Potential Energy Converted to Kinetic Free energy

Summate the final speed of a complimentary electron accelerated from rest through a potential difference of 100 V. (Assume that this numerical value is accurate to iii significant figures.)

Strategy

Nosotros have a system with but conservative forces. Assuming the electron is accelerated in a vacuum, and neglecting the gravitational force (we will bank check on this assumption after), all of the electrical potential energy is converted into kinetic free energy. We tin can identify the initial and final forms of energy to be ${\text{KE}}_{\mathrm{i}}=\mathrm{0,}{\text{KE}}_{\mathrm{f}}=\mathrm{½}{\mathrm{mv}}^2,{\text{PE}}_{\mathrm{i}}=\mathrm{qV}{\text{, and PE}}_{\mathrm{f}}=0.$

Solution

Conservation of energy states that

$${\text{KE}}_{\mathrm{i}}+{\text{PE}}_{\mathrm{i}}{\text{= KE}}_{\mathrm{f}}+{\text{PE}}_{\mathrm{f}}\text{.}$$

19.17

Entering the forms identified above, we obtain

$$\text{qV}=\frac{{\text{mv}}^2}{\text{ii}}\text{.}$$

xix.18

We solve this for $v$ :

$$v=\sqrt{\frac{\mathrm{ii}\text{qV}}{m}}\text{.}$$

19.19

Entering values for $q,\mathrm{Five}\text{, and}\mathrm{one\; thousand}$ gives

$$\begin{array}{lll}v& =& \sqrt{\frac{2\left(\mathrm{–ane.60}\times {\text{10}}^{\text{–19}}\text{C}\right)\left(\text{–100 J/C}\right)}{\mathrm{nine.11}\times {\text{10}}^{\text{–31}}\text{kg}}}\\ & =& 5.93\times {\text{10}}^6\text{grand/s.}\end{array}$$

xix.20

Discussion

Note that both the charge and the initial voltage are negative, equally in Figure 19.5. From the discussions in Electric Charge and Electrical Field, we know that electrostatic forces on small particles are generally very big compared with the gravitational force. The large final speed confirms that the gravitational force is indeed negligible here. The big speed also indicates how easy it is to accelerate electrons with small voltages because of their very small mass. Voltages much higher than the 100 V in this problem are typically used in electron guns. Those college voltages produce electron speeds and so great that relativistic effects must exist taken into account. That is why a depression voltage is considered (accurately) in this example.

Making Connections: Kinetic and Potential Energy in Point Charges

Now consider some other system of two bespeak charges. Ane has a mass of g kg and a accuse of fifty.0 µC, and is initially stationary. The other has a mass of 1.00 kg, a charge of 10.0 µC, and is initially traveling directly at the first point charge at x.0 g/s from very far away. What will be the closest arroyo of these two objects to each other?

Figure nineteen.6 A system consisting of ii indicate charges initially has the smaller charge moving toward the larger accuse

Note that the internal energy of this 2-charge system will not change, due to an absence of external forces acting on the arrangement. Initially, the internal free energy is equal to the kinetic energy of the smaller charge, and the potential free energy is finer nada due to the enormous distance between the two objects. Conservation of energy tells the states that the internal energy of this system volition not change. Hence the distance of closest approach volition be when the internal energy is equal to the potential free energy between the two charges, and in that location is no kinetic free energy in this system.

The initial kinetic free energy may exist calculated as 50.0 J. Applying Equations (19.38) and (19.two), we find a altitude of ix.00 cm. Later on this, the mutual repulsion will ship the lighter object off to infinity again. Note that we did not include potential free energy due to gravity, every bit the masses concerned are and then small compared to the charges that the result volition never come close to showing up in significant digits. Furthermore, the first object is much more massive than the 2d. Equally a outcome, whatsoever motion induced in it will too be likewise small to show up in the significant digits.

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Source: https://openstax.org/books/college-physics-ap-courses/pages/19-1-electric-potential-energy-potential-difference

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