How Do The Changes In Electric Potential Energy Compare?

Learning Objectives

By the end of this department, yous will be able to:

• Define electric potential, voltage, and potential difference
• Define the electron-volt
• Summate electric potential and potential difference from potential energy and electric field
• Depict systems in which the electron-volt is a useful unit
• Employ conservation of energy to electric systems

Recall that earlier we defined electric field to be a quantity contained of the exam charge in a given system, which would all the same allow us to calculate the strength that would effect on an capricious test charge. (The default assumption in the absenteeism of other information is that the test charge is positive.) We briefly defined a field for gravity, just gravity is always bonny, whereas the electric force tin exist either attractive or repulsive. Therefore, although potential energy is perfectly acceptable in a gravitational arrangement, it is convenient to ascertain a quantity that allows us to calculate the work on a charge independent of the magnitude of the accuse. Calculating the work directly may exist difficult, since $W=\overrightarrow{\text{F}}·\overrightarrow{\text{d}}$ and the management and magnitude of $\overrightarrow{\text{F}}$ can exist complex for multiple charges, for odd-shaped objects, and along arbitrary paths. But we practise know that because $\overrightarrow{\text{F}}=q\overrightarrow{\text{E}}$ , the work, and hence $\text{Δ}U,$ is proportional to the test charge q. To have a concrete quantity that is independent of test charge, we define electrical potential V (or simply potential, since electric is understood) to be the potential energy per unit of measurement charge:

Electric Potential

The electric potential free energy per unit of measurement charge is

Since U is proportional to q, the dependence on q cancels. Thus, V does non depend on q. The alter in potential energy $\text{Δ}U$ is crucial, and then nosotros are concerned with the divergence in potential or potential difference $\text{Δ}V$ between two points, where

$$\text{Δ}\mathrm{Five}={\mathrm{Five}}_B-5_A=\frac{\text{Δ}U}{q}.$$

Electric Potential Difference

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

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

The familiar term voltage is the mutual name for electric potential difference. Keep in mind that whenever a voltage is quoted, it is understood to be the potential difference between ii points. For case, every battery has ii terminals, and its voltage is the potential difference between them. More than fundamentally, the signal you lot choose to exist zero volts is arbitrary. This is coordinating to the fact that gravitational potential energy has an arbitrary zero, such equally sea level or maybe a lecture hall floor. Information technology is worthwhile to emphasize the stardom betwixt potential difference and electrical potential energy.

Potential Departure and Electrical Potential Energy

The human relationship between potential difference (or voltage) and electrical potential free energy is given by

$$\text{Δ}V=\frac{\text{Δ}U}{q}\text{or}\text{Δ}U=q\text{Δ}V.$$

vii.v

Voltage is not the same as free energy. Voltage is the energy per unit charge. Thus, a motorcycle battery and a machine bombardment can both take the aforementioned voltage (more precisely, the aforementioned potential deviation betwixt battery terminals), yet i stores much more energy than the other considering $\text{Δ}U=q\text{Δ}V.$ The car battery tin can move more charge than the motorcycle battery, although both are 12-5 batteries.

Instance vii.4

Calculating Energy

You have a 12.0-V motorcycle battery that can motility 5000 C of charge, and a 12.0-V car battery that tin can move lx,000 C of charge. How much energy does each evangelize? (Assume that the numerical value of each charge is accurate to 3 meaning figures.)

Strategy

To say we have a 12.0-V bombardment means that its terminals have a 12.0-V potential departure. 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 notice the energy output, we multiply the charge moved past the potential departure.

Solution

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

$$\text{Δ}{U}_{\text{cycle}}=(5000\text{C})(12.0\text{Five})=(5000\text{C})(12.0\text{J/C})=6.00\times {10}^{\mathrm{four}}\text{J}\text{.}$$

Similarly, for the motorcar battery, $q=\mathrm{60,000}\text{C}$ and

$$\text{Δ}{U}_{\text{car}}=(\mathrm{lx,000}\text{C})(12.0\text{V})=\mathrm{7.twenty}\times {10}^{\mathrm{v}}\text{J}\text{.}$$

Significance

Voltage and energy are related, only they are not the same thing. The voltages of the batteries are identical, but the energy supplied by each is quite different. A machine bombardment has a much larger engine to start than a motorcycle. Note too that equally 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 free energy supplied by the battery is even so calculated every bit in this example, but not all of the free energy is available for external employ.

Check Your Understanding vii.4

Check Your Agreement How much free energy does a one.5-5 AAA battery have that can movement 100 C?

Annotation that the energies calculated in the previous case are accented values. The change in potential free energy for the battery is negative, since it loses free energy. These batteries, like many electrical systems, actually move negative charge—electrons in detail. The batteries repel electrons from their negative terminals (A) through any circuitry is involved and attract them to their positive terminals (B), equally shown in Effigy vii.12. The change in potential is $\text{Δ}V=V_B-V_A=+12\text{5}$ and the charge q is negative, and so that $\text{Δ}U=q\text{Δ}V$ is negative, meaning the potential energy of the bombardment has decreased when q has moved from A to B.

Figure 7.12 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 final has an backlog of negative charge, which is repelled by information technology 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 last. Within the battery, both positive and negative charges move.

Instance vii.five

How Many Electrons Move through a Headlight Each Second?

When a 12.0-V auto battery powers a single 30.0-Due west headlight, how many electrons laissez passer through it each second?

Strategy

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

Solution

To observe the accuse 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 $\text{Δ}V$ , nosotros get

$$q=\frac{\mathrm{-xxx.0}\text{J}}{+12.0\text{V}}=\frac{\mathrm{-thirty.0}\text{J}}{+12.0\text{J/C}}=\mathrm{-two.50}\text{C}\text{.}$$

The number of electrons ${\mathrm{north}}_{\mathrm{due\; east}}$ is the total accuse divided by the charge per electron. That is,

$${\mathrm{due\; north}}_{\mathrm{due\; east}}=\frac{\mathrm{-two.50}\text{C}}{\mathrm{-i.60}\times {10}^{\mathrm{-xix}}{\text{C/e}}^{-}}=1.56\times {10}^{\mathrm{nineteen}}\text{electrons}\text{.}$$

Significance

This is a very large number. It is no wonder that we practice non ordinarily observe individual electrons with so many being present in ordinary systems. In fact, electricity had been in utilize for many decades before it was determined that the moving charges in many circumstances were negative. Positive accuse moving in the opposite management of negative charge ofttimes produces identical effects; this makes it hard to make up one's mind which is moving or whether both are moving.

Check Your Understanding 7.5

Cheque Your Understanding How many electrons would go through a 24.0-Due west lamp each 2nd from a 12-volt machine battery?

The Electron-Volt

The free energy per electron is very small in macroscopic situations similar that in the previous example—a tiny fraction of a joule. But on a submicroscopic calibration, such free energy per particle (electron, proton, or ion) can exist of great importance. For example, even a tiny fraction of a joule tin be great enough for these particles to destroy organic molecules and harm living tissue. The particle may practise its harm by straight standoff, or it may create harmful X-rays, which can also inflict damage. It is useful to have an energy unit of measurement related to submicroscopic effects.

Effigy 7.thirteen shows a situation related to the definition of such an energy unit of measurement. An electron is accelerated betwixt 2 charged metal plates, every bit it might be in an old-model television set tube or oscilloscope. The electron gains kinetic energy that is afterwards converted into some other course—light in the idiot box tube, for example. (Note that in terms of energy, "downhill" for the electron is "uphill" for a positive charge.) Since energy is related to voltage by $\text{Δ}U=q\text{Δ}V$ , we can retrieve of the joule equally a coulomb-volt.

Figure 7.13 A typical electron gun accelerates electrons using a potential difference betwixt two separated metal plates. By conservation of energy, the kinetic free energy has to equal the modify in potential energy, so $K\mathrm{East}=qV$. The energy of the electron in electron-volts is numerically the same equally the voltage between the plates. For instance, a 5000-V potential departure produces 5000-eV electrons. The conceptual construct, namely two parallel plates with a hole in one, is shown in (a), while a existent electron gun is shown in (b).

Electron-Volt

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 divergence of ane V. In equation form,

$$1\text{eV}=(1.60\times {\mathrm{ten}}^{\mathrm{-19}}\text{C})(1\text{V})=(1.60\times {\mathrm{x}}^{\mathrm{-19}}\text{C})(1\text{J/C})=1.60\times {\mathrm{x}}^{\mathrm{-19}}\text{J}\text{.}$$

An electron accelerated through a potential difference of 1 V is given an energy of 1 eV. Information technology follows that an electron accelerated through 50 V gains 50 eV. A potential difference of 100,000 V (100 kV) gives an electron an free energy of 100,000 eV (100 keV), and so on. Similarly, an ion with a double positive charge accelerated through 100 Five gains 200 eV of energy. These unproblematic relationships betwixt accelerating voltage and particle charges brand the electron-volt a unproblematic and user-friendly 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 oftentimes expressed in electron-volts. For case, almost 5 eV of energy is required to intermission up sure organic molecules. If a proton is accelerated from rest through a potential difference of 30 kV, information technology acquires an free energy of xxx keV (30,000 eV) and can break up as many every bit 6000 of these molecules $(\mathrm{30,000}\text{eV}÷5\text{eV}\text{per molecule}=6000\text{molecules).}$ Nuclear decay energies are on the gild of 1 MeV (1,000,000 eV) per event and can thus produce significant biological impairment.

Conservation of Energy

The full energy of a system is conserved if in that location is no net add-on (or subtraction) due to work or rut 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 organization; that is, $M+U=\text{abiding}\text{.}$ A loss of U for a charged particle becomes an increment in its K. Conservation of energy is stated in equation form as

$$M+U=\text{constant}$$

or

$$K_{\text{i}}+U_{\text{i}}=K_{\text{f}}+U_{\text{f}}$$

where i and f stand for initial and final conditions. As we accept found many times earlier, considering energy can give united states insights and facilitate problem solving.

Example 7.6

Electric Potential Free energy Converted into Kinetic Free energy

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

Strategy

Nosotros take a organization with but bourgeois forces. Bold the electron is accelerated in a vacuum, and neglecting the gravitational force (we will check on this assumption later), all of the electric potential energy is converted into kinetic energy. Nosotros can identify the initial and concluding forms of energy to be $K_{\text{i}}=0,G_{\text{f}}=\frac{1}{2}\mathrm{thou}{\mathrm{five}}^{\mathrm{ii}},U_{\text{i}}=qV,U_{\text{f}}=0.$

Solution

Conservation of energy states that

$${\mathrm{Yard}}_{\text{i}}+U_{\text{i}}=G_{\text{f}}+U_{\text{f}}.$$

Inbound the forms identified higher up, nosotros obtain

$$q\mathrm{Five}=\frac{\mathrm{1000}{v}^2}{2}.$$

We solve this for v:

$$\mathrm{five}=\sqrt{\frac{\mathrm{two}qV}{\mathrm{one\; thousand}}}.$$

Entering values for q, V, and grand gives

$$\mathrm{five}=\sqrt{\frac{\mathrm{ii}(\mathrm{-i.60}\times {10}^{\mathrm{-19}}\text{C})(-100\text{J/C})}{9.11\times {10}^{\mathrm{-31}}\text{kg}}}=5.93\times {\mathrm{x}}^{\mathrm{six}}\text{one thousand/south}\text{.}$$

Significance

Note that both the charge and the initial voltage are negative, every bit in Figure 7.xiii. From the give-and-take of electrical charge and electric field, we know that electrostatic forces on small particles are generally very large compared with the gravitational force. The large final speed confirms that the gravitational force is indeed negligible here. The large speed also indicates how piece of cake it is to accelerate electrons with small voltages considering of their very minor mass. Voltages much higher than the 100 5 in this problem are typically used in electron guns. These college voltages produce electron speeds so great that effects from special relativity must be taken into account and hence are reserved for a later chapter (Relativity). That is why nosotros consider a low voltage (accurately) in this example.

Check Your Understanding seven.half-dozen

Check Your Understanding How would this example change with a positron? A positron is identical to an electron except the charge is positive.

Voltage and Electric Field

So far, we have explored the relationship between voltage and energy. At present we desire to explore the relationship between voltage and electrical field. We volition commencement with the general case for a non-compatible $\overrightarrow{\text{E}}$ field. Call up that our general formula for the potential free energy of a test charge q at bespeak P relative to reference point R is

$$U_P=\text{−}{\displaystyle {\int }_R^P\overrightarrow{\text{F}}·d\overrightarrow{\text{l}}}.$$

When nosotros substitute in the definition of electric field $(\overrightarrow{\text{Due east}}=\overrightarrow{\text{F}}\text{/}q),$ this becomes

$$U_P=\text{−}q{\displaystyle {\int }_R^P\overrightarrow{\text{E}}·d\overrightarrow{\text{50}}}.$$

Applying our definition of potential $(V=U\text{/}q)$ to this potential energy, we find that, in general,

$$V_P=\text{−}{\displaystyle {\int }_R^P\overrightarrow{\text{East}}·d\overrightarrow{\text{l}}}.$$

7.half-dozen

From our previous discussion of the potential energy of a charge in an electric field, the result is independent of the path chosen, and hence we tin can pick the integral path that is most convenient.

Consider the special case of a positive signal charge q at the origin. To calculate the potential acquired by q at a altitude r from the origin relative to a reference of 0 at infinity (retrieve that we did the same for potential free energy), permit $P=r$ and $R=\infty ,$ with $d\overrightarrow{\text{l}}=d\overrightarrow{\text{r}}=\widehat{\text{r}}dr$ and apply $\overrightarrow{\text{Eastward}}=\frac{\mathrm{thousand}q}{r^{\mathrm{two}}}\widehat{\text{r}}.$ When we evaluate the integral

$$V_P=\text{−}{\displaystyle {\int }_R^P\overrightarrow{\text{E}}·d\overrightarrow{\text{l}}}$$

for this organisation, nosotros accept

$$V_r=\text{−}{\displaystyle {\int }_{\infty }^r\frac{\mathrm{one\; thousand}q}{r^2}\widehat{\text{r}}·\widehat{\text{r}}dr},$$

which simplifies to

$${\mathrm{Five}}_r=\text{−}{\displaystyle {\int }_{\infty }^r\frac{kq}{r^{\mathrm{two}}}dr}=\frac{kq}{r}-\frac{kq}{\infty }=\frac{gq}{r}.$$

This result,

$$5_r=\frac{kq}{r}$$

is the standard form of the potential of a point charge. This will be explored further in the next section.

To examine another interesting special case, suppose a uniform electrical field $\overrightarrow{\text{Eastward}}$ is produced past placing a potential difference (or voltage) $\text{Δ}V$ across two parallel metal plates, labeled A and B (Figure 7.14). Examining this situation volition tell us what voltage is needed to produce a certain electric field force. It will also reveal a more fundamental relationship between electric potential and electrical field.

Effigy seven.14 The relationship between V and East for parallel conducting plates is $\mathrm{Eastward}=V/d$. (Note that $\text{Δ}V=V_{AB}$ in magnitude. For a accuse that is moved from plate A at college potential to plate B at lower potential, a minus sign needs to exist included as follows: $\text{−}\text{Δ}V=V_A-V_B=V_{AB}.$)

From a physicist's point of view, either $\text{Δ}V$ or $\overrightarrow{\text{Due east}}$ can be used to draw whatever interaction between charges. However, $\text{Δ}5$ is a scalar quantity and has no management, whereas $\overrightarrow{\text{E}}$ is a vector quantity, having both magnitude and direction. (Notation that the magnitude of the electric field, a scalar quantity, is represented by E.) The relationship between $\text{Δ}V$ and $\overrightarrow{\text{E}}$ is revealed past calculating the work done by the electric force in moving a charge from bespeak A to point B. But, as noted before, capricious charge distributions require calculus. Nosotros therefore look at a uniform electrical field as an interesting special case.

The work done by the electric field in Figure 7.14 to motion a positive charge q from A, the positive plate, higher potential, to B, the negative plate, lower potential, is

$$\mathrm{Westward}=\text{−}\text{Δ}U=\text{−}q\text{Δ}V.$$

The potential difference betwixt points A and B is

$$\text{−}\text{Δ}5=\text{−}(V_B-V_A)={\mathrm{Five}}_A-V_B=V_{AB}.$$

Entering this into the expression for work yields

$$W=q{V}_{AB}.$$

Work is $W=\overrightarrow{\text{F}}·\overrightarrow{\text{d}}=Fd\text{cos}\theta$ ; here $\text{cos}\theta =1$ , since the path is parallel to the field. Thus, $W=Fd$ . Since $F=qE$ , we see that $\mathrm{West}=q\mathrm{East}d$ .

Substituting this expression for work into the previous equation gives

$$qEd=q{V}_{AB}.$$

The accuse cancels, so we obtain for the voltage betwixt points A and B

$$\begin{array}{c}{\mathrm{Five}}_{AB}=Ed\hfill \\ E=\frac{5_{AB}}{d}\hfill \end{array}\}(\text{uniform}E\text{-field but})$$

where d is the distance from A to B, or the altitude betwixt the plates in Effigy 7.14. Note that this equation implies that the units for electric field are volts per meter. We already know the units for electric field are newtons per coulomb; thus, the following relation among units is valid:

$$1\text{N}\text{/}\text{C}=1\text{V}\text{/}\text{m}.$$

Furthermore, we may extend this to the integral form. Substituting Equation seven.5 into our definition for the potential difference between points A and B, we obtain

$$5_{BA}=V_B-V_A=\text{−}{\displaystyle {\int }_R^B\overrightarrow{\text{E}}·d\overrightarrow{\text{fifty}}}+{\displaystyle {\int }_R^A\overrightarrow{\text{Due east}}·d\overrightarrow{\text{l}}}$$

which simplifies to

$$V_B-{\mathrm{Five}}_A=\text{−}{\displaystyle {\int }_A^B\overrightarrow{\text{Eastward}}·d\overrightarrow{\text{l}}}.$$

Equally a sit-in, from this we may calculate the potential difference betwixt two points (A and B) equidistant from a point charge q at the origin, as shown in Figure 7.15.

Effigy seven.15 The arc for calculating the potential departure betwixt ii points that are equidistant from a point charge at the origin.

To exercise this, we integrate effectually an arc of the circle of constant radius r between A and B, which means we let $d\overrightarrow{\text{50}}=r\widehat{\phi }d\phi ,$ while using $\overrightarrow{\text{Eastward}}=\frac{\mathrm{thou}q}{r^2}\widehat{\text{r}}$ . Thus,

$$\text{Δ}{V}_{BA}=V_B-V_A=\text{−}{\displaystyle {\int }_A^B\overrightarrow{\text{East}}·d\overrightarrow{\text{50}}}$$

vii.seven

for this organization becomes

$$V_B-V_A=\text{−}{\displaystyle {\int }_A^B\frac{\mathrm{thousand}q}{r^{\mathrm{two}}}}\widehat{\text{r}}·r\widehat{\phi }d\phi .$$

However, $\widehat{\text{r}}·\widehat{\phi }=0$ and therefore

$$V_B-V_A=0.$$

This outcome, that there is no difference in potential forth a constant radius from a point accuse, will come in handy when we map potentials.

Instance 7.7

What Is the Highest Voltage Possible between Two Plates?

Dry air can back up a maximum electric field strength of about $\mathrm{iii.0}\times {10}^6\text{V/m}\text{.}$ Above that value, the field creates enough ionization in the air to make the air a conductor. This allows a discharge or spark that reduces the field. What, then, is the maximum voltage betwixt ii parallel conducting plates separated by two.five cm of dry air?

Strategy

We are given the maximum electric field E between the plates and the distance d between them. We can utilize the equation $5_{AB}=Ed$ to summate the maximum voltage.

Solution

The potential difference or voltage between the plates is

$$V_{AB}=Ed.$$

Entering the given values for E and d gives

$$V_{AB}=(3.0\times {10}^{\mathrm{vi}}\text{V/yard})(0.025\text{m})=7.5\times {10}^4\text{V}$$

or

$$V_{AB}=75\text{kV}\text{.}$$

(The answer is quoted to just two digits, since the maximum field force is approximate.)

Significance

One of the implications of this outcome is that it takes about 75 kV to make a spark jump across a 2.v-cm (1-in.) gap, or 150 kV for a 5-cm spark. This limits the voltages that can exist betwixt conductors, mayhap on a ability manual line. A smaller voltage can cause a spark if there are spines on the surface, since precipitous points take larger field strengths than smooth surfaces. Humid air breaks downward at a lower field strength, significant that a smaller voltage will brand a spark spring through boiling air. The largest voltages can exist built up with static electricity on dry out days (Effigy seven.sixteen).

Figure 7.16 A spark chamber is used to trace the paths of high-energy particles. Ionization created by the particles equally they pass through the gas betwixt the plates allows a spark to bound. The sparks are perpendicular to the plates, following electrical field lines between them. The potential difference between adjacent plates is not high enough to cause sparks without the ionization produced past particles from accelerator experiments (or catholic rays). This form of detector is now primitive and no longer in employ except for sit-in purposes. (credit b: modification of work by Jack Collins)

Instance 7.8

Field and Force inside an Electron Gun

An electron gun (Figure seven.xiii) has parallel plates separated past 4.00 cm and gives electrons 25.0 keV of free energy. (a) What is the electric field force betwixt the plates? (b) What force would this field exert on a slice of plastic with a $0.500\text{-}\mu \text{C}$ charge that gets between the plates?

Strategy

Since the voltage and plate separation are given, the electric field forcefulness can be calculated directly from the expression $E=\frac{V_{AB}}{d}$ . Once we know the electric field forcefulness, nosotros can find the forcefulness on a charge by using $\overrightarrow{\text{F}}=q\overrightarrow{\text{Due east}}.$ Since the electric field is in just one direction, we tin can write this equation in terms of the magnitudes, $F=qE$ .

Solution

1. The expression for the magnitude of the electric field between two uniform metal plates is

   $$\mathrm{Due\; east}=\frac{V_{AB}}{d}.$$

   Since the electron is a single charge and is given 25.0 keV of energy, the potential difference must exist 25.0 kV. Entering this value for $V_{AB}$ and the plate separation of 0.0400 thousand, we obtain

   $$\mathrm{East}=\frac{25.0\text{kV}}{0.0400\text{grand}}=\mathrm{half-dozen.25}\times {10}^5\text{V/thousand}\text{.}$$

2. The magnitude of the force on a accuse in an electric field is obtained from the equation

   $$F=qE.$$

   Substituting known values gives

   $$F=(0.500\times {10}^{\mathrm{-half\; dozen}}\text{C})(6.25\times {10}^5\text{V/m})=0.313\text{N}\text{.}$$

Significance

Notation that the units are newtons, since $1\text{V/m}=1\text{N/C}$ . Because the electric field is compatible between the plates, the force on the accuse is the aforementioned no thing where the accuse is located between the plates.

Case 7.9

Calculating Potential of a Point Charge

Given a betoken charge $q=+\mathrm{ii.0}\text{nC}$ at the origin, summate the potential departure between indicate $P_{\mathrm{ane}}$ a distance $a=4.0\text{cm}$ from q, and $P_2$ a distance $b=12.0\text{cm}$ from q, where the ii points accept an angle of $\phi =24\text{°}$ between them (Figure 7.17).

Figure 7.17 Find the deviation in potential between $P_1$ and $P_2$.

Strategy

Do this in two steps. The start step is to use $V_B-5_A=\text{−}{\displaystyle {\int }_A^B\overrightarrow{\text{East}}·d\overrightarrow{\text{50}}}$ and permit $A=a=4.0\text{cm}$ and $B=b=12.0\text{cm},$ with $d\overrightarrow{\text{50}}=d\overrightarrow{\text{r}}=\widehat{\text{r}}dr$ and $\overrightarrow{\text{E}}=\frac{kq}{r^2}\widehat{\text{r}}.$ Then perform the integral. The 2nd step is to integrate $V_B-{\mathrm{Five}}_A=\text{−}{\displaystyle {\int }_A^B\overrightarrow{\text{E}}·d\overrightarrow{\text{l}}}$ around an arc of constant radius r, which ways we let $d\overrightarrow{\text{l}}=r\widehat{\phi }d\phi$ with limits $0\le \phi \le 24\text{°},$ nevertheless using $\overrightarrow{\text{E}}=\frac{\mathrm{thou}q}{r^2}\widehat{\text{r}}$ . Then add the two results together.

Solution

For the first part, ${\mathrm{Five}}_B-V_A=\text{−}{\displaystyle {\int }_A^B\overrightarrow{\text{E}}·d\overrightarrow{\text{l}}}$ for this organization becomes $V_b-5_a=\text{−}{\displaystyle {\int }_a^b\frac{kq}{r^2}\widehat{\text{r}}·\widehat{\text{r}}dr}$ which computes to

$$\begin{array}{cc}\text{Δ}V& =\text{−}{\displaystyle {\int }_a^b\frac{\mathrm{one\; thousand}q}{r^{\mathrm{ii}}}dr}=kq\left[\frac{1}{a}-\frac{1}{b}\right]\hfill \\ & =\left(8.99\times {\mathrm{x}}^{\mathrm{nine}}{\text{Nm}}^2{\text{/C}}^{\mathrm{ii}}\right)\left(2.0\times {10}^{\mathrm{-9}}\text{C}\right)\left[\frac{\mathrm{one}}{0.040\text{m}}-\frac{1}{0.12\text{m}}\right]=300\text{5}\text{.}\hfill \end{array}$$

For the 2d step, $5_B-V_A=\text{−}{\displaystyle {\int }_A^B\overrightarrow{\text{Eastward}}·d\overrightarrow{\text{50}}}$ becomes $\text{Δ}5=\text{−}{\displaystyle {\int }_0^{24\text{°}}\frac{\mathrm{chiliad}q}{r^2}\widehat{\text{r}}·r\widehat{\phi }d\phi }$ , merely $\widehat{\text{r}}·\widehat{\phi }=0$ and therefore $\text{Δ}V=0.$ Adding the two parts together, nosotros get 300 V.

Significance

We have demonstrated the use of the integral form of the potential difference to obtain a numerical issue. Detect that, in this detail organisation, we could have also used the formula for the potential due to a point accuse at the two points and just taken the difference.

Check Your Understanding 7.7

Check Your Understanding From the examples, how does the energy of a lightning strike vary with the superlative of the clouds from the ground? Consider the deject-ground system to be two parallel plates.

Earlier presenting problems involving electrostatics, we suggest a trouble-solving strategy to follow for this topic.

Problem-Solving Strategy

Electrostatics

1. Examine the situation to determine if static electricity is involved; this may concern separated stationary charges, the forces amidst them, and the electrical fields they create.
2. Identify the system of interest. This includes noting the number, locations, and types of charges involved.
3. Place exactly what needs to exist determined in the problem (identify the unknowns). A written list is useful. Determine whether the Coulomb force is to exist considered straight—if so, it may exist useful to draw a free-body diagram, using electric field lines.
4. Make a list of what is given or can be inferred from the problem as stated (identify the knowns). It is important to distinguish the Coulomb force F from the electric field East, for instance.
5. Solve the appropriate equation for the quantity to be determined (the unknown) or draw the field lines as requested.
6. Examine the answer to encounter if it is reasonable: Does it make sense? Are units correct and the numbers involved reasonable?

How Do The Changes In Electric Potential Energy Compare?,

Source: https://openstax.org/books/university-physics-volume-2/pages/7-2-electric-potential-and-potential-difference

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