Review the Force on the-10nc Charge Is as Shown in the Figure (Figure 1)

18.4 Electric Potential

Resources ID: ohuql3dr@half dozen

Grade Range: PreK - 12

Learning Objectives

Learning Objectives

By the end of this section, you will be able to do the following:

• Explain the similarities and differences between electric potential energy and gravitational potential energy
• Calculate the electric potential difference betwixt two indicate charges and in a compatible electric field

Section Primal Terms

electrical potential

electric potential energy

Every bit you learned in studying gravity, a mass in a gravitational field has potential energy, which means it has the potential to advance and thereby increase its kinetic energy. This kinetic energy can be used to practise work. For case, imagine you want to use a stone to pound a nail into a piece of wood. Y'all first lift the stone high higher up the nail, which increases the potential energy of the stone-Earth system—because Earth is so big, it does not move, and then nosotros usually shorten this by maxim simply that the potential energy of the rock increases. When y'all drib the rock, gravity converts the potential free energy into kinetic energy. When the stone hits the nail, it does work by pounding the blast into the wood. The gravitational potential energy is the piece of work that a mass can potentially do past virtue of its position in a gravitational field. Potential energy is a very useful concept, because it can be used with conservation of energy to summate the motion of masses in a gravitational field.

Electric potential energy works much the same fashion, but it is based on the electric field instead of the gravitational field. By virtue of its position in an electric field, a accuse has an electric potential energy. If the charge is free to move, the force due to the electric field causes it to accelerate, so its potential free energy is converted to kinetic energy, just like a mass that falls in a gravitational field. This kinetic energy tin be used to practise work. The electric potential energy is the piece of work that a charge tin can do by virtue of its position in an electric field.

The illustration betwixt gravitational potential free energy and electrical potential energy is depicted in Figure 18.23. On the left, the ball-Globe arrangement gains gravitational potential energy when the brawl is higher in Earth's gravitational field. On the right, the two-charge system gains electrical potential energy when the positive charge is farther from the negative accuse.

Effigy eighteen.23 On the left, the gravitational field points toward Earth. The higher the brawl is in the gravitational field, the higher the potential energy is of the World-brawl system. On the correct, the electric field points toward a negative charge. The further the positive accuse is from the negative charge, the college the potential energy is of the two-charge organisation.

Let's utilise the symbol $U_{\mathrm{Thou}}$ to denote gravitational potential energy. When a mass falls in a gravitational field, its gravitational potential free energy decreases. Conservation of free energy tells u.s. that the work done past the gravitational field to make the mass accelerate must equal the loss of potential energy of the mass. If nosotros use the symbol $W_{\text{done}\text{past}\text{gravity}}$ to denote this piece of work, then

18.xviii

$$-\Delta U_{\text{One thousand}}={\mathrm{Westward}}_{\text{done}\text{by}\text{gravity,}}$$

where the minus sign reflects the fact that the potential energy of the ball decreases.

The piece of work done by gravity on the mass is

18.19

$${\mathrm{Due\; west}}_{\text{done}\text{by}\text{gravity}}=-F(y_{\text{f}}-y_{\text{i}}),$$

where F is the force due to gravity, and $y_{\text{i}}$ and $y_{\text{f}}$ are the initial and final positions of the brawl, respectively. The negative sign is because gravity points down, which we consider to be the negative direction. For the abiding gravitational field near Earth'due south surface, $F=\mathrm{chiliad}\mathrm{one\; thousand}$. The alter in gravitational potential energy of the mass is

18.xx

$$-\Delta U_{\text{1000}}=W_{\text{done}\text{by}\text{gravity}}=-F\left(y_{\text{f}}-y_{\text{i}}\right)=-\mathrm{one\; thousand}g\left(y_{\text{f}}-y_{\text{i}}\right)\text{, or}\Delta U_{\text{G}}=m\mathrm{thousand}\left(y_{\text{f}}-y_{\text{i}}\right)\text{.}$$

Notation that $y_{\text{f}}-y_{\text{i}}$ is just the negative of the meridian h from which the mass falls, and then we usually only write $\Delta U_{\text{M}}=-mmh$.

We at present use the same reasoning to a charge in an electric field to notice the electric potential energy. The change $\Delta U_{\text{East}}$ in electric potential free energy is the work done by the electric field to move a charge q from an initial position ${\mathrm{ten}}_{\text{i}}$ to a concluding position $x_{\text{f}}$ ($-\Delta U_{\text{East}}={\mathrm{West}}_{\text{washed}\text{by}\text{E-field}}$). The definition of work does not change, except that now the work is washed by the electric field: $W_{\text{done}\text{by}\text{E-field}}=F({\mathrm{ten}}_{\text{f}}-{\mathrm{ten}}_{\text{i}})$. For a charge that falls through a constant electric field Eastward, the force practical to the charge by the electric field is $F=qE$. The alter in electric potential free energy of the charge is thus

18.21

$$-\Delta U_{\text{E}}=W_{\text{washed}\text{by}\text{Due east-field}}=Fd=qE\left({\mathrm{10}}_{\text{f}}-{\mathrm{ten}}_{\text{i}}\right)$$

or

eighteen.22

$$\Delta U_{\text{E}}=-qE\left({\mathrm{ten}}_{\text{f}}-x_{\text{i}}\right)\text{.}$$

This equation gives the change in electric potential energy of a charge q when information technology moves from position $x_{\text{i}}$ to position ${\mathrm{10}}_{\text{f}}$ in a constant electrical field East.

Figure 18.24 shows how this analogy would work if we were close to World's surface, where gravity is constant. The top image shows a charge accelerating due to a abiding electric field. Besides, the round mass in the bottom paradigm accelerates due to a constant gravitation field. In both cases, the potential energy of the particle decreases, and its kinetic free energy increases.

Figure 18.24 In the top motion-picture show, a mass accelerates due to a constant electric field. In the lesser picture, the mass accelerates due to a constant gravitational field.

Watch Physics

Illustration between Gravity and Electricity

This video discusses the analogy betwixt gravitational potential energy and electric potential energy. It reviews the concepts of work and potential energy and shows the connection between a mass in a uniform gravitation field, such as on Earth'due south surface, and an electric accuse in a uniform electric field.

If the electric field is not constant, and so the equation $\Delta U_{\text{E}}=-qE\left({\mathrm{10}}_{\text{f}}-{\mathrm{10}}_{\text{i}}\right)$ is not valid, and deriving the electrical potential energy becomes more than involved. For example, consider the electric potential energy of an associates of two signal charges $q_{\mathrm{ane}}$ and $q_{\mathrm{ii}}$ of the same sign that are initially very far apart. We start by placing accuse $q_1$ at the origin of our coordinate system. This takes no electrical energy, considering there is no electric field at the origin (because charge $q_2$ is very far away). Nosotros then bring charge $q_2$ in from very far away to a distance r from the center of charge $q_{\mathrm{one}}$. This requires some attempt, because the electrical field of charge $q_1$ applies a repulsive strength on accuse $q_2$. The free energy it takes to gather these 2 charges can be recuperated if we permit them fly autonomously again. Thus, the charges accept potential energy when they are a distance r apart. It turns out that the electrical potential free energy of a pair of betoken charges $q_1$ and $q_2$ a distance r apart is

To epitomize, if charges $q_{\mathrm{one}}$ and $q_2$ are gratis to move, they tin accumulate kinetic energy by flight apart, and this kinetic energy can be used to exercise piece of work. The maximum amount of work the 2 charges can do (if they fly infinitely far from each other) is given by the equation above.

Notice that if the two charges accept opposite signs, then the potential energy is negative. This means that the charges have more than potential to practice work when they are far apart than when they are at a distance r apart. This makes sense: Opposite charges concenter, so the charges can gain more kinetic free energy if they attract each other from far away than if they start at only a short distance apart. Thus, they accept more potential to do piece of work when they are far apart. Figure eighteen.25 summarizes how the electric potential energy depends on charge and separation.

Effigy eighteen.25 The potential free energy depends on the sign of the charges and their separation. The arrows on the charges indicate the direction in which the charges would move if released. When charges with the aforementioned sign are far apart, their potential energy is depression, as shown in the top console for two positive charges. The situation is the reverse for charges of opposite signs, as shown in the lesser console.

Electrical Potential

Electric Potential

Remember that to find the forcefulness applied by a fixed accuse Q on any capricious test charge q, it was convenient to define the electric field, which is the force per unit charge practical by Q on any test charge that we place in its electric field. The same strategy is used here with electrical potential energy: Nosotros now define the electric potential V, which is the electric potential energy per unit charge.

Misconception Alert

Emphasize the difference between electric potential energy and electric potential. Although the latter seems to be shorthand for the former, the two terms accept different meanings.

Normally, the electric potential is simply called the potential or voltage . The units for the potential are J/C, which are given the proper noun volt (5) after the Italian physicist Alessandro Volta (1745–1827). From the equation $U_{\text{Due east}}=k{q}_{\mathrm{i}}{q}_{\mathrm{two}}/r$, the electric potential a distance r from a point charge $q_1$ is

18.25

$$V=\frac{U_{\text{E}}}{q_{\mathrm{ii}}}=\frac{\mathrm{yard}{q}_{\mathrm{i}}}{r}\text{.}$$

This equation gives the free energy required per unit of measurement charge to bring a charge $q_2$ from infinity to a altitude r from a point charge $q_1$. Mathematically, this is written every bit

18.26

$$\mathrm{Five}={\frac{U_{\mathrm{Due\; east}}}{q_2}|}_{R=r}-{\frac{U_{\mathrm{Eastward}}}{q_2}|}_{R=\infty }\text{.}$$

Notation that this equation really represents a deviation in electric potential. However, considering the 2d term is naught, information technology is normally non written, and we speak of the electric potential instead of the electric potential departure, or we but say the potential difference, or voltage). Below, when we consider the electrical potential energy per unit charge betwixt two points not infinitely far apart, we speak of electrical potential difference explicitly. Just remember that electrical potential and electric potential deviation are really the same thing; the sometime is used just when the electric potential energy is zero in either the initial or terminal charge configuration.

Coming dorsum now to the electric potential a distance r from a point charge $q_1$, note that $q_{\mathrm{ane}}$ can be any arbitrary point accuse, so we can driblet the subscripts and simply write

Now consider the electric potential almost a group of charges q ₁, q ₂, and q _iii, as drawn in Figure eighteen.26. The electric potential is derived by considering the electric field. Electric fields follow the principle of superposition and tin can exist merely added together, so the electrical potential from different charges also add together. Thus, the electric potential of a signal nearly a group of charges is

18.28

$$\mathrm{Five}=\frac{\mathrm{thou}{q}_1}{r_{\mathrm{one}}}+\frac{k{q}_2}{r_{\mathrm{two}}}+\frac{k{q}_3}{r_3}+\cdots .$$

where $r_1,r_{\mathrm{ii}},r_3,\dots ,$ are the distances from the heart of charges $q_{\mathrm{ane}},q_{\mathrm{two}},q_3,\dots$ to the point of interest, as shown in Figure 18.26.

Effigy 18.26 The potential at the cerise point is just the sum of the potentials due to each private accuse.

Now let's consider the electric potential in a uniform electric field. From the equation $\Delta U_{\text{E}}=-q\mathrm{East}\left(x_{\text{f}}-{\mathrm{ten}}_{\text{i}}\right)$, we see that the potential difference in going from ${\mathrm{10}}_{\text{i}}$ to $x_{\text{f}}$ in a uniform electrical field Eastward is

eighteen.29

$$\Delta 5=\frac{\Delta U_E}{q}=-E\left({\mathrm{ten}}_{\text{f}}-x_{\text{i}}\right)\text{.}$$

Tips For Success

Discover from the equation $\Delta V=-E\left(x_{\text{f}}-x_{\text{i}}\right)$ that the electrical field can be written as

which means that the electric field has units of V/one thousand. Thus, if you know the potential difference between two points, computing the electric field is very uncomplicated—y'all simply divide the potential difference past the distance!

Detect that a positive charge in a region with high potential will feel a force pushing information technology toward regions of lower potential. In this sense, potential is similar force per unit area for fluids. Imagine a pipe containing fluid, with the fluid at 1 stop of the pipe under high force per unit area and the fluid at the other end of the pipage under depression pressure. If nothing prevents the fluid from flowing, it will flow from the loftier-pressure end to the low-pressure end. Likewise, a positive charge that is costless to move will motion from a region with loftier potential to a region with lower potential.

Watch Physics

Voltage

This video starts from electrical potential energy and explains how this is related to electric potential (or voltage). The lecturer calculates the electric potential created by a uniform electric field.

Grasp Check

What is the voltage difference between the positions

$x_{\text{f}}=11\text{m}$

and

$x_{\text{i}}=\mathrm{v.0}\text{chiliad}$

in an electric field of

$\left(2.0\text{Due north/C}\right)\hat{x}$

?

1. $6\text{V}$
2. $12\text{5}$
3. $24\text{Five}$
4. $32\text{Five}$

Links To Physics

Electric Animals

Many animals generate and/or detect electric fields. This is useful for activities such every bit hunting, defence, navigation, communication, and mating. Because salt h2o is a relatively good conductor, electric fish have evolved in all the earth's oceans. These fish accept intrigued humans since the earliest times. In the nineteenth century, parties were even organized where the primary allure was getting a jolt from an electric fish! Scientists also studied electric fish to learn near electricity. Alessandro Volta based his research that led to batteries in 1799 on electric fish. He even referred to batteries as artificial electric organs, considering he saw them as imitations of the electrical organs of electric fish.

Animals that generate electricity are called electrogenic and those that detect electric fields are called electroreceptive. Most fish that are electrogenic are besides electroreceptive. One of the most well-known electric fish is the electrical eel (come across Figure eighteen.27), which is both electrogenic and electroreceptive. These fish accept three pairs of organs that produce the electric charge: the main organ, Hunter's organ, and Sach'southward organ. Together, these organs business relationship for more than 80percent of the fish's body.

Electric eels can produce electrical discharges of much greater voltage than what yous would get from a standard wall socket. These discharges can stun or even kill their prey. They too use low-intensity discharges to navigate. The electrical fields they generate reverberate off nearby obstacles or animals and are so detected by electroreceptors in the eel's peel. The three organs that produce electricity incorporate electrolytes, which are substances that ionize when dissolved in water (or other liquids). An ionized atom or molecule is 1 that has lost or gained at to the lowest degree one electron, so information technology carries a internet charge. Thus, a liquid solution containing an electrolyte conducts electricity, because the ions in the solution can move if an electric field is practical.

To produce large discharges, the primary organ is used. It contains approximately half-dozen,000 rows of electroplaques connected in a long chain. Connected this fashion, the voltage between electroplaques adds up, creating a large concluding voltage. Each electroplaque consists of a column of cells controlled past an excitor nerve. When triggered by the excitor nerve, the electroplaques allow ionized sodium to catamenia through them, creating a potential difference between electroplaques. These potentials add together up, and a large current tin can flow through the electrolyte.

This geometry is reflected in batteries, which also use stacks of plates to produce larger potential differences.

Figure 18.27 An electric eel in its natural environment. (credit: Steven Thou. Johnson)

Grasp Check

If an electrical eel produces 1,000 V, which voltage is produced by each electroplaque in the principal organ?

1. 0.17 mV
2. i.vii mV
3. 17 mV
4. 170 mV

Worked Case

Ten-ray Tube

Dentists use X-rays to prototype their patients' teeth and bones. The X-ray tubes that generate X-rays contain an electron source separated by about x cm from a metallic target. The electrons are accelerated from the source to the target by a uniform electric field with a magnitude of virtually 100 kN/C, as drawn in Effigy eighteen.28. When the electrons hit the target, X-rays are produced. (a) What is the potential difference between the electron source and the metallic target? (b) What is the kinetic energy of the electrons when they reach the target, assuming that the electrons start at remainder?

Figure 18.28 In an X-ray tube, a large current flows through the electron source, causing electrons to be ejected from the electron source. The ejected electrons are accelerated toward the target by the electric field. When they strike the target, X-rays are produced.

STRATEGY FOR (A)

Use the equation $\Delta \mathrm{Five}=-E\left(x_{\text{f}}-x_{\text{i}}\right)$ to find the potential difference given a abiding electrical field. Define the source position as $x_{\text{i}}=0$ and the target position as $x_{\text{f}}=10\text{cm}$. To accelerate the electrons in the positive x direction, the electric field must point in the negative x management. This way, the force $F=qE$ on the electrons volition point in the positive ten direction, because both q and E are negative. Thus, $\mathrm{Due\; east}=-100\times {\mathrm{ten}}^{\mathrm{iii}}\text{N/C}$.

Solution for (a)

Using ${\mathrm{ten}}_{\text{i}}=0$ and ${\mathrm{ten}}_{\text{f}}=10\text{cm}=\mathrm{0.x}\text{m}$, the equation $\Delta V=-\mathrm{Eastward}\left({\mathrm{ten}}_{\text{f}}-{\mathrm{ten}}_{\text{i}}\right)$ tells us that the potential difference between the electron source and the target is

eighteen.31

$$\Delta 5=-\mathrm{Due\; east}\left(x_{\text{f}}-x_{\text{i}}\right)=-\left(-100\times {10}^3\text{N/C}\right)\left(0.10\text{m}-0\right)=+\mathrm{x}\text{kV.}$$

Discussion for (a)

The potential difference is positive, so the free energy per unit of measurement positive charge is college at the target than at the source. This ways that gratis positive charges would fall from the target to the source. However, electrons are negative charges, and so they accelerate from the source toward the target, gaining kinetic energy every bit they go.

STRATEGY FOR (B)

Apply conservation of energy to notice the concluding kinetic energy of the electrons. In going from the source to the target, the change in electric potential energy plus the change in kinetic energy of the electrons must be zero, so $\Delta U_{\mathrm{East}}+\Delta \mathrm{Thou}=0\text{.}$ The change in electrical potential energy for moving through a constant electric field is given by the equation

$$\Delta U_{\text{East}}=-qE\left(x_{\text{f}}-x_{\text{i}}\right)\text{,}$$

where the electric field is $E=-100\times {\mathrm{x}}^3\text{N/C}$. Because the electrons start at remainder, their initial kinetic energy is cypher. Thus, the modify in kinetic free energy is just their final kinetic free energy, so $\Delta \mathrm{One\; thousand}=K_{\text{f}}$.

Solution for (b)

Again $x_{\text{i}}=0$ and $x_{\text{f}}=10\text{cm}=0.10\text{m}$. The charge of an electron is $q=-1.602\times {10}^{-\mathrm{nineteen}}\text{C}$. Conservation of free energy gives

18.32

$$\begin{array}{ccc}\hfill \Delta U_E+\Delta G& \hfill =\hfill & 0.\hfill \\ \hfill -qE\left(x_{\text{f}}-{\mathrm{10}}_{\text{i}}\right)+K_{\text{f}}& \hfill =\hfill & 0.\hfill \\ \hfill {\mathrm{Thousand}}_{\text{f}}& \hfill =\hfill & qE\left(x_{\text{f}}-x_{\text{i.}}\right)\text{.}\hfill \end{array}$$

Inserting the known values into the correct-manus side of this equation gives

18.33

$$\begin{array}{ccc}\hfill K_{\text{f}}& \hfill =\hfill & \left(-1.60\times {\mathrm{ten}}^{-19}\text{C}\right)\left(-100\times {10}^3\text{N/C}\right)\left(0.10\text{m}-0\right)\hfill \\ \hfill & \hfill =\hfill & \mathrm{i.6}\times {10}^{-\mathrm{fifteen}}\text{J.}\hfill \end{array}$$

Give-and-take for (b)

This is a very modest energy. Nonetheless, electrons are very modest, and then they are piece of cake to accelerate, and this free energy is enough to make an electron go extremely fast. You lot tin can find their speed by using the definition of kinetic energy, $K=\frac{1}{2}\mathrm{1000}{v}^{\mathrm{two}}$. The result is that the electrons are moving at more than than 100 million miles per hour!

Worked Example

Electric Potential Free energy of Doorknob and Dust Speck

Consider again the doorknob from the example in the previous department. The doorknob is treated every bit a spherical usher with a uniform static accuse $q_1=\mathrm{-1.v}\text{nC}$ on its surface. What is the electric potential energy between the doorknob and a speck of dust carrying a charge $q_{\mathrm{ii}}=\mathrm{0.twenty}\text{nC}$ at 1.0 cm from the front surface of the doorknob? The bore of the doorknob is 5.0 cm.

STRATEGY

Equally nosotros did in the previous department, we care for the charge as if it were concentrated at the middle of the doorknob. Again, as you will be able to validate in later physics classes, we tin can make this simplification, because the charge is uniformly distributed over the surface of the spherical object. Make a sketch of the state of affairs and define a coordinate organization, as shown in the image below. We use $+\mathrm{10}$ to indicate the outward direction perpendicular to the door, with $x=0$ at the center of the doorknob. If the bore of the doorknob is v.0 cm, its radius is 2.v cm. Thus, the speck of dust 1.0 cm from the surface of the doorknob is a distance $r=2.5\text{cm}+1.0\text{cm}=\mathrm{iii.5}\text{cm}$ from the middle of the doorknob. To solve this problem, apply the equation $U_{\text{E}}=k{q}_1{q}_2/r$.

Solution

The charge on the doorknob is $q_1=\mathrm{-one.five}\text{nC}=\mathrm{-ane.5}\times {10}^{\mathrm{-9}}\text{C,}$ and the accuse on the speck of dust is $q_2=\mathrm{0.xx}\text{nC}=2.0\times {10}^{\mathrm{-x}}\text{C}$. The distance $r=\mathrm{3.v}\text{cm}=0.035\text{m}$. Inserting these values into the equation $U_{\text{E}}=\mathrm{chiliad}{q}_1{q}_2/r$ gives

xviii.34

$$\begin{array}{ccc}\hfill U_{\mathrm{Due\; east}}\hfill & \hfill =\hfill & \frac{k{q}_1{q}_{\mathrm{two}}}{r}\hfill \\ \hfill \hfill & \hfill =\hfill & \frac{\left(\mathrm{viii.99}\times {10}^{\mathrm{ix}}\text{Northward}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(-\mathrm{one.five}\times {10}^{\mathrm{-ix}}\text{C}\right)\left(\mathrm{ii.0}\times {10}^{\mathrm{-10}}\text{C}\right)}{\left(0.035\text{m}\right)}\hfill \\ \hfill \hfill & \hfill =\hfill & \mathrm{-7.vii}\times {10}^{\mathrm{-8}}\text{J}\text{.}\hfill \end{array}$$

Word

The energy is negative, which means that the energy volition decrease that is, get even more negative equally the speck of grit approaches the doorknob. This helps explain why grit accumulates on objects that carry a static accuse. Yet, note that insulators unremarkably collect more static charge than conductors, considering any accuse that accumulates on insulators cannot move about on the insulator to find a fashion to escape. They must simply look to be removed by some passing moist speck of grit or other host.

Practice Issues

Practice Problems

What is the electrical potential x cm from a −10 nC accuse?

1. ix.0 × 10^two V
2. ix.0 × 10³ V
3. 9.0 × x^four V
4. 9.0 × 10^five V

An electron accelerates from 0 to ten × 10⁴ grand/due south in an electric field. Through what potential difference did the electron travel? The mass of an electron is 9.eleven × 10^–31 kg, and its accuse is −1.60 × 10^–19 C.

1. 29 mV
2. 290 mV
3. two,900 mV
4. 29 V

Check Your Understanding

Check Your Agreement

Exercise 11

Gravitational potential energy is the $(–\text{10 N/C)}\widehat{x}$ potential for two masses to practise work by virtue of their positions with respect to each other. What is the analogous definition of electric potential energy?

1. Electric potential energy is the potential for 2 charges to do work past virtue of their positions with respect to the origin bespeak.
2. Electric potential free energy is the potential for two charges to practice work by virtue of their positions with respect to infinity.
3. Electric potential energy is the potential for two charges to exercise work by virtue of their positions with respect to each other.
4. Electrical potential free energy is the potential for single charges to practice work by virtue of their positions with respect to their final positions.

Do 12

A negative charge is 10 k from a positive accuse. Where would y'all have to motion the negative charge to increase the potential energy of the arrangement?

1. The negative charge should be moved closer to the positive charge.
2. The negative charge should be moved farther away from the positive charge.
3. The negative charge should be moved to infinity.
4. The negative charge should be placed just next to the positive charge.

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