What Is The Change In Potential Energy Of A +2.00 Nc Test Charge

Learning Objectives

Past the end of this section, yous will exist able to:

• Define electric potential, voltage, and potential difference
• Ascertain the electron-volt
• Summate electrical potential and potential difference from potential energy and electric field
• Describe systems in which the electron-volt is a useful unit
• Apply conservation of free energy to electrical systems

Recall that before we divers electrical field to exist a quantity independent of the test charge in a given organization, which would nonetheless allow the states to calculate the force that would result on an arbitrary exam accuse. (The default supposition in the absence of other information is that the test charge is positive.) We briefly divers a field for gravity, but gravity is ever attractive, whereas the electrical forcefulness can be either bonny or repulsive. Therefore, although potential energy is perfectly acceptable in a gravitational organization, it is convenient to define a quantity that allows us to calculate the work on a charge contained of the magnitude of the charge. Calculating the work directly may be difficult, since $W=\overrightarrow{\text{F}}·\overrightarrow{\text{d}}$ and the direction and magnitude of $\overrightarrow{\text{F}}$ can be complex for multiple charges, for odd-shaped objects, and forth arbitrary paths. Only we practise know that because $\overrightarrow{\text{F}}=q\overrightarrow{\text{Eastward}}$ , 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 5 (or only potential, since electric is understood) to exist the potential free energy per unit 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 modify in potential free energy $\text{Δ}U$ is crucial, so nosotros are concerned with the difference in potential or potential difference $\text{Δ}\mathrm{Five}$ between two points, where

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

Electric Potential Difference

The electric potential difference betwixt points A and B, $V_B-V_A,$ is defined to be the alter in potential free energy of a charge q moved from A to B, divided by the accuse. Units of potential divergence are joules per coulomb, given the name volt (5) after Alessandro Volta.

$$1\text{Five}=1\text{J/C}$$

The familiar term voltage is the common name for electric potential difference. Go along in mind that whenever a voltage is quoted, it is understood to exist the potential difference betwixt two points. For example, every battery has two terminals, and its voltage is the potential difference betwixt them. More than 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 equally sea level or perhaps a lecture hall floor. It is worthwhile to emphasize the stardom between potential difference and electrical potential energy.

Potential Difference and Electric Potential Energy

The relationship between potential deviation (or voltage) and electrical potential energy is given past

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

7.v

Voltage is not the same every bit free energy. Voltage is the energy per unit charge. Thus, a motorcycle battery and a car battery can both have the same voltage (more than precisely, the same potential deviation betwixt battery terminals), still one stores much more energy than the other because $\text{Δ}U=q\text{Δ}V.$ The motorcar bombardment can move more charge than the motorcycle battery, although both are 12-5 batteries.

Example 7.4

Calculating Energy

You have a 12.0-5 motorcycle battery that can move 5000 C of accuse, and a 12.0-V car battery that tin move 60,000 C of charge. How much energy does each deliver? (Assume that the numerical value of each charge is accurate to three significant figures.)

Strategy

To say we have a 12.0-5 battery ways that its terminals have a 12.0-V potential difference. When such a battery moves charge, it puts the charge through a potential difference of 12.0 5, and the charge is given a change in potential energy equal to $\text{Δ}U=q\text{Δ}V.$ To find the free energy output, we multiply the charge moved past the potential difference.

Solution

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

$$\text{Δ}{U}_{\text{cycle}}=(5000\text{C})(12.0\text{V})=(5000\text{C})(12.0\text{J/C})=6.00\times {10}^4\text{J}\text{.}$$

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

$$\text{Δ}{U}_{\text{auto}}=(\mathrm{60,000}\text{C})(12.0\text{V})=7.20\times {10}^5\text{J}\text{.}$$

Significance

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

Check Your Understanding 7.4

Cheque Your Understanding How much energy does a 1.5-Five AAA battery have that can movement 100 C?

Note that the energies calculated in the previous example are absolute values. The modify in potential energy for the battery is negative, since information technology loses energy. These batteries, similar many electric 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), equally shown in Figure vii.12. The alter in potential is $\text{Δ}\mathrm{Five}=V_B-V_A=+12\text{V}$ and the charge q is negative, so that $\text{Δ}U=q\text{Δ}V$ is negative, meaning the potential free energy of the battery has decreased when q has moved from A to B.

Effigy 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 excess of negative charge, which is repelled by it and attracted to the excess positive accuse on the other final. In terms of potential, the positive terminal is at a higher voltage than the negative terminal. Inside the bombardment, both positive and negative charges move.

Instance vii.5

How Many Electrons Move through a Headlight Each Second?

When a 12.0-V car battery powers a unmarried 30.0-W headlight, how many electrons laissez passer through it each 2d?

Strategy

To discover 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-W lamp uses 30.0 joules per 2nd. Since the battery loses energy, we take $\text{Δ}U=\mathrm{-30}\text{J}$ and, since the electrons are going from the negative terminal to the positive, nosotros run across that $\text{Δ}V=\text{+12.0}\text{Five}\text{.}$

Solution

To notice the charge q moved, we solve the equation $\text{Δ}U=q\text{Δ}\mathrm{Five}:$

$$q=\frac{\text{Δ}U}{\text{Δ}V}.$$

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

$$q=\frac{\mathrm{-30.0}\text{J}}{+12.0\text{V}}=\frac{\mathrm{-xxx.0}\text{J}}{+12.0\text{J/C}}=\mathrm{-2.50}\text{C}\text{.}$$

The number of electrons ${\mathrm{northward}}_e$ is the total charge divided past the charge per electron. That is,

$${\mathrm{due\; north}}_e=\frac{\mathrm{-2.50}\text{C}}{\mathrm{-ane.threescore}\times {10}^{\mathrm{-19}}{\text{C/e}}^{-}}=1.56\times {10}^{19}\text{electrons}\text{.}$$

Significance

This is a very large number. Information technology is no wonder that nosotros practice not ordinarily discover individual electrons with and then many being nowadays in ordinary systems. In fact, electricity had been in use for many decades before it was adamant that the moving charges in many circumstances were negative. Positive charge moving in the opposite direction of negative accuse often produces identical furnishings; this makes it difficult to decide which is moving or whether both are moving.

Cheque Your Agreement 7.5

Check Your Understanding How many electrons would become through a 24.0-W lamp each second from a 12-volt car bombardment?

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. Merely on a submicroscopic calibration, such energy per particle (electron, proton, or ion) can exist of great importance. For example, fifty-fifty a tiny fraction of a joule can exist slap-up enough for these particles to destroy organic molecules and harm living tissue. The particle may practise its damage by direct standoff, or it may create harmful X-rays, which can also inflict damage. Information technology is useful to have an free energy unit of measurement related to submicroscopic effects.

Figure 7.13 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 quondam-model television tube or oscilloscope. The electron gains kinetic energy that is afterward converted into another course—light in the television tube, for instance. (Note that in terms of energy, "downhill" for the electron is "uphill" for a positive charge.) Since energy is related to voltage by $\text{Δ}U=q\text{Δ}5$ , nosotros tin recall of the joule as a coulomb-volt.

Effigy 7.13 A typical electron gun accelerates electrons using a potential difference between two separated metallic plates. By conservation of free energy, the kinetic free energy has to equal the alter in potential energy, so $K\mathrm{Eastward}=qV$. The energy of the electron in electron-volts is numerically the aforementioned as the voltage between the plates. For case, a 5000-V potential difference produces 5000-eV electrons. The conceptual construct, namely 2 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 free energy unit called the electron-volt (eV), which is the energy given to a key accuse accelerated through a potential departure of 1 Five. In equation form,

$$\mathrm{ane}\text{eV}=(\mathrm{one.60}\times {10}^{\mathrm{-xix}}\text{C})(1\text{V})=(\mathrm{one.threescore}\times {10}^{\mathrm{-xix}}\text{C})(\mathrm{one}\text{J/C})=1.60\times {10}^{\mathrm{-19}}\text{J}\text{.}$$

An electron accelerated through a potential difference of ane 5 is given an energy of 1 eV. It follows that an electron accelerated through 50 5 gains fifty eV. A potential difference of 100,000 V (100 kV) gives an electron an energy of 100,000 eV (100 keV), and and so on. Similarly, an ion with a double positive charge accelerated through 100 V gains 200 eV of energy. These simple relationships between accelerating voltage and particle charges make the electron-volt a simple and user-friendly energy unit in such circumstances.

The electron-volt is ordinarily employed in submicroscopic processes—chemical valence energies and molecular and nuclear binding energies are among the quantities ofttimes expressed in electron-volts. For example, nigh 5 eV of energy is required to break upward certain organic molecules. If a proton is accelerated from rest through a potential departure of 30 kV, it acquires an free energy of 30 keV (30,000 eV) and tin break up as many as 6000 of these molecules $(\mathrm{30,000}\text{eV}÷5\text{eV}\text{per molecule}=6000\text{molecules).}$ Nuclear decay energies are on the order of 1 MeV (1,000,000 eV) per event and can thus produce pregnant biological harm.

Conservation of Free energy

The total energy of a arrangement is conserved if in that location is no net addition (or subtraction) due to work or heat transfer. For conservative forces, such as the electrostatic force, conservation of energy states that mechanical energy is a abiding.

Mechanical free energy is the sum of the kinetic energy and potential energy of a system; that is, $G+U=\text{constant}\text{.}$ A loss of U for a charged particle becomes an increase in its K. Conservation of free energy is stated in equation form as

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

or

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

where i and f represent initial and terminal conditions. Every bit we accept found many times before, considering energy can give usa insights and facilitate problem solving.

Example vii.6

Electrical Potential Energy Converted into Kinetic Energy

Calculate the final speed of a free electron accelerated from residue through a potential difference of 100 V. (Presume that this numerical value is accurate to iii meaning figures.)

Strategy

We have a system with merely conservative forces. Bold the electron is accelerated in a vacuum, and neglecting the gravitational force (nosotros will cheque on this supposition after), all of the electrical potential energy is converted into kinetic free energy. We can place the initial and final forms of energy to be $K_{\text{i}}=0,K_{\text{f}}=\frac{1}{2}\mathrm{yard}{v}^2,U_{\text{i}}=qV,U_{\text{f}}=0.$

Solution

Conservation of energy states that

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

Inbound the forms identified in a higher place, we obtain

$$qV=\frac{\mathrm{thousand}{v}^{\mathrm{ii}}}{2}.$$

We solve this for v:

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

Entering values for q, Five, and grand gives

$$v=\sqrt{\frac{2(\mathrm{-ane.lx}\times {10}^{\mathrm{-nineteen}}\text{C})(-100\text{J/C})}{9.11\times {10}^{\mathrm{-31}}\text{kg}}}=\mathrm{v.93}\times {10}^6\text{m/s}\text{.}$$

Significance

Note that both the charge and the initial voltage are negative, as in Figure 7.13. From the discussion of electric charge and electric field, we know that electrostatic forces on small-scale particles are more often than not very big compared with the gravitational strength. The large last speed confirms that the gravitational force is indeed negligible here. The big speed also indicates how easy information technology is to accelerate electrons with minor voltages because of their very small mass. Voltages much higher than the 100 V in this problem are typically used in electron guns. These higher voltages produce electron speeds so bang-up that effects from special relativity must be taken into account and hence are reserved for a later affiliate (Relativity). That is why nosotros consider a low voltage (accurately) in this example.

Check Your Understanding 7.6

Check Your Understanding How would this example alter 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 free energy. Now we want to explore the relationship betwixt voltage and electrical field. We will start with the general case for a not-uniform $\overrightarrow{\text{E}}$ field. Recall that our full general formula for the potential energy of a exam accuse q at point P relative to reference point R is

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

When we substitute in the definition of electrical field $(\overrightarrow{\text{East}}=\overrightarrow{\text{F}}\text{/}q),$ this becomes

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

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

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

7.6

From our previous give-and-take of the potential free energy of a accuse in an electric field, the result is independent of the path called, and hence nosotros can pick the integral path that is most convenient.

Consider the special instance of a positive point charge q at the origin. To calculate the potential caused by q at a altitude r from the origin relative to a reference of 0 at infinity (recall that we did the same for potential energy), let $P=r$ and $R=\infty ,$ with $d\overrightarrow{\text{l}}=d\overrightarrow{\text{r}}=\widehat{\text{r}}dr$ and use $\overrightarrow{\text{E}}=\frac{kq}{r^2}\widehat{\text{r}}.$ When we evaluate the integral

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

for this system, we have

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

which simplifies to

$$5_r=\text{−}{\displaystyle {\int }_{\infty }^r\frac{gq}{r^2}dr}=\frac{kq}{r}-\frac{kq}{\infty }=\frac{\mathrm{chiliad}q}{r}.$$

This outcome,

$${\mathrm{Five}}_r=\frac{kq}{r}$$

is the standard grade of the potential of a point charge. This will be explored farther in the side by side section.

To examine some other interesting special instance, suppose a compatible electric field $\overrightarrow{\text{E}}$ is produced past placing a potential divergence (or voltage) $\text{Δ}V$ beyond two parallel metal plates, labeled A and B (Figure 7.fourteen). Examining this situation will tell us what voltage is needed to produce a certain electrical field strength. It volition also reveal a more than fundamental relationship between electrical potential and electric field.

Figure seven.14 The relationship betwixt V and E for parallel conducting plates is $\mathrm{East}=V/d$. (Note that $\text{Δ}V=5_{AB}$ in magnitude. For a accuse that is moved from plate A at higher 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 $\overrightarrow{\text{E}}$ can be used to describe any interaction between charges. However, $\text{Δ}V$ is a scalar quantity and has no management, whereas $\overrightarrow{\text{E}}$ is a vector quantity, having both magnitude and management. (Note that the magnitude of the electrical field, a scalar quantity, is represented by Due east.) The relationship between $\text{Δ}V$ and $\overrightarrow{\text{E}}$ is revealed by calculating the piece of work washed by the electric strength in moving a charge from betoken A to point B. But, equally noted earlier, capricious charge distributions require calculus. We therefore look at a compatible electric field as an interesting special case.

The work done past the electric field in Figure vii.14 to move a positive accuse q from A, the positive plate, higher potential, to B, the negative plate, lower potential, is

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

The potential deviation between points A and B is

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

Entering this into the expression for piece of work yields

$$W=q{5}_{AB}.$$

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

Substituting this expression for piece of work into the previous equation gives

$$q\mathrm{Eastward}d=q{\mathrm{Five}}_{AB}.$$

The charge cancels, so we obtain for the voltage between points A and B

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

where d is the distance from A to B, or the distance between the plates in Figure seven.14. Notation that this equation implies that the units for electrical field are volts per meter. We already know the units for electric field are newtons per coulomb; thus, the following relation amidst units is valid:

$$\mathrm{one}\text{North}\text{/}\text{C}=\mathrm{i}\text{V}\text{/}\text{m}.$$

Furthermore, we may extend this to the integral form. Substituting Equation 7.5 into our definition for the potential difference between points A and B, we obtain

$$V_{BA}=V_B-{\mathrm{Five}}_A=\text{−}{\displaystyle {\int }_R^B\overrightarrow{\text{E}}·d\overrightarrow{\text{fifty}}}+{\displaystyle {\int }_R^A\overrightarrow{\text{Due east}}·d\overrightarrow{\text{l}}}$$

which simplifies to

$$V_B-V_A=\text{−}{\displaystyle {\int }_A^B\overrightarrow{\text{E}}·d\overrightarrow{\text{l}}}.$$

As a demonstration, from this nosotros may calculate the potential departure between two points (A and B) equidistant from a point accuse q at the origin, as shown in Figure 7.15.

Figure seven.15 The arc for computing the potential difference betwixt ii points that are equidistant from a point charge at the origin.

To do this, we integrate around an arc of the circle of constant radius r between A and B, which means we let $d\overrightarrow{\text{l}}=r\widehat{\phi }d\phi ,$ while using $\overrightarrow{\text{Due east}}=\frac{kq}{r^{\mathrm{two}}}\widehat{\text{r}}$ . Thus,

$$\text{Δ}{5}_{BA}=V_B-V_A=\text{−}{\displaystyle {\int }_A^B\overrightarrow{\text{E}}·d\overrightarrow{\text{l}}}$$

7.seven

for this system becomes

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

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

$$V_B-V_A=0.$$

This outcome, that at that place is no deviation in potential along a abiding radius from a betoken accuse, will come up in handy when we map potentials.

Example 7.vii

What Is the Highest Voltage Possible between Ii Plates?

Dry air tin can support a maximum electric field forcefulness of most $3.0\times {10}^{\mathrm{half-dozen}}\text{Five/m}\text{.}$ Higher up that value, the field creates plenty 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 two parallel conducting plates separated by two.5 cm of dry air?

Strategy

We are given the maximum electric field E betwixt the plates and the distance d between them. Nosotros can apply the equation $V_{AB}=Ed$ to summate the maximum voltage.

Solution

The potential difference or voltage betwixt the plates is

$$5_{AB}=Ed.$$

Entering the given values for E and d gives

$$V_{AB}=(\mathrm{three.0}\times {10}^{\mathrm{half\; dozen}}\text{V/m})(0.025\text{m})=\mathrm{vii.5}\times {10}^{\mathrm{four}}\text{Five}$$

or

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

(The reply is quoted to simply 2 digits, since the maximum field strength is approximate.)

Significance

I of the implications of this result is that information technology takes about 75 kV to make a spark jump across a two.five-cm (1-in.) gap, or 150 kV for a 5-cm spark. This limits the voltages that can exist between conductors, perhaps on a ability transmission line. A smaller voltage can cause a spark if there are spines on the surface, since sharp points have larger field strengths than smooth surfaces. Boiling air breaks down at a lower field strength, meaning that a smaller voltage volition make a spark jump through humid air. The largest voltages tin can be congenital upwardly with static electricity on dry out days (Figure 7.xvi).

Figure 7.xvi A spark sleeping accommodation is used to trace the paths of loftier-energy particles. Ionization created past the particles as they laissez passer through the gas between the plates allows a spark to jump. The sparks are perpendicular to the plates, following electrical field lines between them. The potential difference betwixt adjacent plates is not loftier enough to cause sparks without the ionization produced by particles from accelerator experiments (or catholic rays). This form of detector is now primitive and no longer in utilize except for demonstration purposes. (credit b: modification of work past Jack Collins)

Case 7.viii

Field and Force within an Electron Gun

An electron gun (Effigy seven.13) has parallel plates separated past four.00 cm and gives electrons 25.0 keV of energy. (a) What is the electric field strength between the plates? (b) What force would this field exert on a piece of plastic with a $0.500\text{-}\mu \text{C}$ charge that gets betwixt the plates?

Strategy

Since the voltage and plate separation are given, the electrical field forcefulness can be calculated directly from the expression $E=\frac{5_{AB}}{d}$ . Once we know the electrical field strength, we tin can notice the force on a charge by using $\overrightarrow{\text{F}}=q\overrightarrow{\text{E}}.$ Since the electric field is in only one direction, nosotros tin write this equation in terms of the magnitudes, $F=qE$ .

Solution

1. The expression for the magnitude of the electric field betwixt two compatible metallic plates is

   $$E=\frac{V_{AB}}{d}.$$

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

   $$E=\frac{25.0\text{kV}}{0.0400\text{thousand}}=6.25\times {10}^5\text{V/m}\text{.}$$

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

   $$F=qE.$$

   Substituting known values gives

   $$F=(0.500\times {\mathrm{x}}^{\mathrm{-6}}\text{C})(6.25\times {10}^{\mathrm{five}}\text{V/m})=0.313\text{N}\text{.}$$

Significance

Note that the units are newtons, since $\mathrm{ane}\text{V/m}=1\text{N/C}$ . Considering the electric field is uniform betwixt the plates, the force on the charge is the same no matter where the charge is located between the plates.

Example 7.9

Calculating Potential of a Point Charge

Given a indicate charge $q=+2.0\text{nC}$ at the origin, summate the potential deviation between bespeak $P_{\mathrm{ane}}$ a distance $a=4.0\text{cm}$ from q, and $P_2$ a distance $b=12.0\text{cm}$ from q, where the two points have an bending of $\phi =24\text{°}$ between them (Figure 7.17).

Effigy 7.17 Detect the difference in potential betwixt $P_1$ and $P_{\mathrm{two}}$.

Strategy

Do this in two steps. The commencement step is to use $V_B-V_A=\text{−}{\displaystyle {\int }_A^B\overrightarrow{\text{E}}·d\overrightarrow{\text{fifty}}}$ and let $A=a=4.0\text{cm}$ and $B=b=12.0\text{cm},$ with $d\overrightarrow{\text{l}}=d\overrightarrow{\text{r}}=\widehat{\text{r}}dr$ and $\overrightarrow{\text{E}}=\frac{gq}{r^{\mathrm{ii}}}\widehat{\text{r}}.$ And so perform the integral. The 2d step is to integrate $5_B-V_A=\text{−}{\displaystyle {\int }_A^B\overrightarrow{\text{E}}·d\overrightarrow{\text{50}}}$ around an arc of abiding radius r, which means nosotros allow $d\overrightarrow{\text{50}}=r\widehat{\phi }d\phi$ with limits $0\le \phi \le 24\text{°},$ still using $\overrightarrow{\text{E}}=\frac{kq}{r^2}\widehat{\text{r}}$ . Then add the two results together.

Solution

For the starting time function, $V_B-5_A=\text{−}{\displaystyle {\int }_A^B\overrightarrow{\text{E}}·d\overrightarrow{\text{l}}}$ for this organisation becomes $V_b-{\mathrm{Five}}_a=\text{−}{\displaystyle {\int }_a^b\frac{\mathrm{yard}q}{r^{\mathrm{ii}}}\widehat{\text{r}}·\widehat{\text{r}}dr}$ which computes to

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

For the second pace, $5_B-V_A=\text{−}{\displaystyle {\int }_A^B\overrightarrow{\text{E}}·d\overrightarrow{\text{l}}}$ becomes $\text{Δ}5=\text{−}{\displaystyle {\int }_0^{24\text{°}}\frac{kq}{r^2}\widehat{\text{r}}·r\widehat{\phi }d\phi }$ , but $\widehat{\text{r}}·\widehat{\phi }=0$ and therefore $\text{Δ}\mathrm{Five}=0.$ Adding the 2 parts together, we go 300 V.

Significance

We have demonstrated the use of the integral form of the potential difference to obtain a numerical upshot. Observe that, in this item system, nosotros could have as well used the formula for the potential due to a point charge at the two points and simply taken the difference.

Bank check Your Understanding vii.7

Bank check Your Understanding From the examples, how does the energy of a lightning strike vary with the pinnacle of the clouds from the ground? Consider the cloud-footing organisation to be two parallel plates.

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

Trouble-Solving Strategy

Electrostatics

1. Examine the state of affairs to make up one's mind if static electricity is involved; this may concern separated stationary charges, the forces among them, and the electrical fields they create.
2. Place 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 trouble (identify the unknowns). A written list is useful. Determine whether the Coulomb strength is to be considered direct—if so, it may be useful to describe a free-trunk diagram, using electric field lines.
4. Brand a listing of what is given or can be inferred from the problem as stated (identify the knowns). Information technology is important to distinguish the Coulomb force F from the electric field Due east, for case.
5. Solve the appropriate equation for the quantity to be adamant (the unknown) or describe the field lines every bit requested.
6. Examine the answer to come across if it is reasonable: Does it make sense? Are units correct and the numbers involved reasonable?

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

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