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Lesson 3 Electric Potential You have no doubt noticed that TV sets, light bulbs, and other electric appliances operate on 115 volts, but electric ovens and clothes dryers usually need 220 volts. Batteries may be rated at a harmless 1.5, 6, 9, or 12 volts, but a high-tension electric transmission line may provide electric power at 400,000 volts. Now just what physical quantity is measured by all these volts? How do volts relate force, energy, and power, about which you have learned in earlier lessons? The answer is that volts measure electric potential difference (sometimes called “voltage”), which is derived from the potential energy acquired by electrically charged objects as a result of the electric forces they experience. Even though your familiarity with volts probably stems from electric power supplied to your household, your introduction to the concept of electric potential in this lesson will be in the context of the interaction of stationary (static) electric charges. 3-1: Definitions and Relations OBJECTIVES: Relate electric potential to (a) work done on a displaced charge, (b) the electric field, and (c) electric potential energy. Use the electron volt to express energy and solve simple problems applying energy conservation. State and interpret the conservative nature of the electrostatic field. PREREQUISITES: Evaluating work integrals and applying the work-energy theorem to solve problems Calculating potential energy and identifying conservative forces Calculating electrostatic forces and fields Reading Assignment Study in your textbook Chapter 23. Commentary The reading in this chapter includes a fairly thorough review of the work done by a conservative force. This is an essential concept for understanding electric potential, and since it has been some time since we have applied it, the review is well worthwhile. In particular, make sure that you understand the sign convention for negative and positive work done on a particle in a force field: positive work must be done by an external agent to move a positive test charge from a region of lower potential to a region of higher potential. Another way of remembering this convention is to note that electric flux line point in the direction of decreasing potential. As you study the reading assignment notice also that the focus of the discussion sometimes shifts from potential difference between two points ( ) to the potential at a point ( ). To accomplish this transition, a certain point at infinity, at the coordinate origin, or at some other point having symmetry in relation to the charges and fields is arbitrarily chosen as the reference point in order to give potential as a function of position in relation to that point (“absolute potential”). Then the electric potential is simply a (scalar) function of the coordinates of any given point with the reference point at zero potential by definition. The convention of using infinite separation as a reference point for electric potential is consistent with the definition of electric potential energy as the work required to assemble a system of charges by bringing them together from an infinite distance. You may find it useful to explore the analogy between electrostatic forces and gravitation by asking yourself such questions as “What would be the unit of gravitational potential?” and “How does the variation in this potential as one approaches a mass differ from the variation of electric potential as one approaches a charge?”. EXAMPLE 3-1: Consider a region of constant electric field in the field having magnitude the . The four points coordinates , and direction, with the and have , . , There is no gravitational field. (a) An external agent slowly moves a body with charge from to . How much work is done on the body by the electric field, and how much is done on the body by the external agent? sketch to illustrate. Use a (b) Answer the questions in part (a) for displacement of the same body (i) from to , (ii) from to , and (iii) from to . (c) A bead of mass and charge long a frictionless wire between points is permitted to slide a and . At which of the two points should it be released from rest in order that it will get to the other one, and with what speed will it arrive? (d) Find the differences in electric potential between points points and , and points and and ; compare them with your answers to parts (a) and (b). Solution: (a) The situation is depicted at right. From the definition of work, . For a constant field, as in this case, done by field. The external agent exerts a force (b) (i) and hence does ; . (ii) . (iii) , of work. (c) The negative charge in this part experiences a force to the left, hence it must be released at and will acquire speed sliding toward . The kinetic energy gained is equal to the work done on it by the electric field, which is . Hence , and . (d) . . . . The electron volt ( . . ), introduced in Section 23.2, is a unit of energy commonly used in atomic physics. It has the advantage that when an electron moves in an electric field across a potential difference of, say, , its kinetic energy changes by . In other words, the potential difference gives the energy directly in electron volts if you are dealing with a proton, electron, or other singly charged atomic particle. For most other purposes, the electron volt is impractically small. The system of International Units (SI) asks that the electron volt only be used in atomic physics. Otherwise is should be converted to joules ( ). Remember to convert to SI units before using masses in kilograms and velocities in meters per second with an energy originally given in electron volts. EXAMPLE 3-2: In the figure at right, electrons acquire an increased speed by moving in an electric field between two plates with slits through which the electrons can pass. The electric potential difference between the plates is . (a) By how much does the electric potential energy of an electron change as it passes through the two-plate system? (Give answer in electron volts.) (b) By how much does the kinetic energy of an electron changes as it passes through the two-plate system? (Give answer in electron volts.) (c) With what speed does an electron leave the space between the plates if it enters with a speed of ? Solution: (a) It decreases by (b) It increases by . . (c) . . . . One of your objectives in this section is to be able to verify that electrostatic forces are conservative. You know that a force field is conservative (permits the definition of a potential energy function) if and only if (a) the work done by the force on a particle moved from point to point is independent of the path taken from to ; or (b) the work done by the force on a particle moving through a closed path back to its starting point is zero. Since the electrostatic force at any point is directly proportional to the electric field at that point, , (1) The two equivalent conditions on the work stated above lead to two equivalent conditions on the field: (a) is independent of path from to (2) is independent of path from to . (3) implies that (b) by any closed path (4) by any closed path. (5) implies that (Note: the line integral in Eqs. (4) and (5) along a closed path is sometimes indicated by a small circle on the integral sign: . Do not confuse this symbol with the same symbol used in many texts to represent a surface integral over a closed surface, as in Gauss’s law. You can distinguish between the two uses of this symbol by looking carefully at the infinitesimal element under the integral sign to see if it refers to a surface or to a displacement.) We shall now show that Coulomb field caused by a single point charge is conservative by evaluating the integral in Eq. (3) and observing that its value depends only on the end points, not on the path. First, we note that the magnitude of the electric field caused by the point charge depends only on the distance (not on the entire vector ) from the charge and is given by , where and (6) is the distance from Figure 3-1 illustrates the geometric relationships. . The direction of the field is radial. Figure 3-1. The value of the line integral element is found in the limit as the path approaches zero. In Figure 3-1(a), you can see points and arrows representing four electric field values We must now calculate , a particular path we have chosen, and and four infinitesimal line elements . , which we shall do with the help of the enlarged diagram in Figure 3-1(b). This figure shows the angle between the path and the radial direction and also the radial increment . Using the definition of the dot product, we find . (7) In other words, because the field is radially directed, only the radial increments and not the angular increments of the path contribute to the line integral. With the integrand in Eq. (7), we can calculate the integral over the radius variable , which varies from to : . (8) Evidently, this result depends only on the end points and not on the path. Our proof is hereby completed. Our conclusion of path independence can be extended easily to an electric field that is caused by several point charges. Since such a field is obtained by adding the fields from the individual charges, the line integral can be expressed as a sum of line integrals like that in Eq. (8), for each of which path independence has been established. It is not so easy to extend the proof to the fields caused by continuous charge distributions. As a matter of fact, more advanced treatments of the theory of static electric fields always begin with the postulate that the fields are conservative. Practice Exercise 3-1 Write your solutions to the following problems in your notebook. 1. An upward-directed, uniform electric field of exists in a certain region. (a) What is the potential difference between a certain point origin and the point (i) diagonally downward to the right above? (ii) that we shall call the to the left? (iii) ? (b) How much work does an external agent to when it moves a charge of very slowly from the origin to the point (i) in (a-ii)? (ii) in (a-ii)? (iii) in (a-iii)? (Hint: Draw a diagram and remember the definitions.) 2. An electron is placed at a distance of to move freely until it is only from a proton. It is then allowed from the proton. (a) What is the potential difference between the two points in the proton’s electric field? (b) How much kinetic energy does the electron gain during its motion? Express the result in electron volts. (c) With what speed does the electron arrive at the second point, assuming that it starts from rest? (Hint: Draw a diagram and remember the definitions.) 3. Given the two conditions that establish the conservative nature of the electric field: is independent of path from to (3) by any closed path (5) Show that these two conditions are equivalent. Check your answers to Practice Exercise 3-1 with the Module 1 Answer Key and review Section 3-1 as necessary before you begin working on the next section. 3-2: Finding Potentials from Charges OBJECTIVES: Use the definition of electric potential and/or the superposition principle to find the potential caused by (a) one or more given point charges, and (b) continuous charge distributions with planar, cylindrical, or spherical symmetry. PREREQUISITES: Finding the electric field using Gauss’s law and describing the electric field near conductors Finding the electric field using Coulomb’s law and using field lines to describe the electric field Commentary After the derivation of the potential due to a point charge in Section 23.2, the discussion in the reading assignment is all presented in terms of absolute potential by setting . With this approach, we can apply the superposition principle and find the potential at any point that results from an arrangement of charges by simply taking the (scalar) sum of the potentials due to the individual charges. Another technique that is sometimes useful for analyzing the potential due to an arrangement of point charges is to graph the potential as a function of position along a single axis. This method is applied in the following example. EXAMPLE 3-3: A point charge of magnitude magnitude ( is located at ( ). is located at ( ) and a point of ), as in the figure below. (a) Find a mathematical expression for the electric potential on the axis, and sketch it on a graph. (Draw the sketch from about six to eight well-placed points.) (b) Find a mathematical expression for the electric potential along the -axis, and sketch it on a graph. (Draw the sketch from about six to eight well-placed points.) How would you expect the potential to vary along the -axis? Explain. (c) Find the locus of points where the electric potential caused by the two charges is equal to zero. Solution: The electric potential caused by the two charges is , Where charges and to the and point are the distances from the locations of the ( ): and . (a) Since along the -axis, and . (The absolute values appear because distances are always to be taken as positive.) Hence, the potential on the -axis is given by . Rather than plotting though, we will plot the dimensionless quantity ). We first make a data table for values of this quantity at several points on the -axis: We further note that , because the charges as and and as are located at these points. Using the eight values in the table and asymptotes through and , we can sketch the graph. (b) Along the -axis, and , so that . Our data table is thus and the -axis potential is sketched below. The potential along the -axis should be identical to that along the -axis since the and axes are symmetric with respect to the -axis on which the charges are located. In fact, the potential is the same at all points of any circle lying in the -plane and centered at the origin. We note further that the potential is positive along the -axis between and , but that it is negative elsewhere on the -axis and on the other axes. This is to be expected, since the negative charge had the greater magnitude; hence the positive potential is confined to a region near the positive charge. At large distances from both charges, where the distance between them is negligible, the potential caused by the two-charge system resembles that caused by a charge at the origin. (c) We saw from the graphs and qualitative reasoning that the potential would be positive near the positive charge and negative elsewhere. Therefore there should be a finite surface on which the potential is zero. Setting , we find that or . In other words, . By combining the -dependent terms, we find . The other terms also combine and permit cancellation of a factor . The final equation is , which is a sphere of radius about the point passes through the origin and the point . The sphere , where we had already found the potential to be zero. Example 3-3 illustrates the kind of quantitative results that can be obtained by applying a bit of combined algebraic and geometric reasoning to a simple charge distribution. It also yields a generalizable result: the electric potential of two point charges of opposite sign is always zero on a spherical surface of certain radius and center that surrounds the smaller of the two charges. You can prove this result without much difficulty by using charges at the origin and at the point ( ), and proceeding as we did above. What do you expect for ? Finding the potential due to a continuous charge distribution (e.g., a charged conductor) is fundamentally an application of Gauss’s law, and thus has similar limitations in usefulness- to cases involving uniform sheets of charge and distributions with spherical or cylindrical symmetry. In these cases, the sum that gives the net potential by means of the superposition principle is replaced by an easily evaluated integral. EXAMPLE 3-4: A point charge of magnitude is located at the center of a hollow conducting spherical shell that is electrically neutral. The interior and exterior radii of the shell are and , respectively. Find the electric potential as a function of radius from the shell’s center, and draw a sketch of the dimensionless combination . Solution: Since the system of charges is spherically symmetric, the electric field outside the shell can be found from Gauss’s law. It is directed radially and has the magnitude , . In the shell itself, the electric field is zero because it is a conductor, (9) , . (10) Inside the hollow cavity of the shell, the field is again found from Gauss’s law (or memory): , . (11) To find the potential as a function of radius, relative to infinity as the reference point, we integrate the electric field: . (12) The important and new step (compared to evaluating this integral for a point charge) is to break up the integral into several parts since the electric field is given in different mathematical expressions for different intervals of the radius. We also use the fact that is radial so that . Then in the first interval ( ) we have one expression for , For the second interval ( for . for For the third interval ( (13) ) we break the integral into two parts, one using Eq. (9) and a second for for : using Eq. (10): ; . (14) ) we apply the result for integrate for the potential difference to from Eq. (14) and using Eq. (11): . In order to graph , we arbitrarily set the ratio (15) . Practice Exercise 3-2 Write your solutions to the following problems in your notebook. 1. Two point charges are placed on the -axis, one of magnitude one of magnitude at the point at the origin, and . (a) Find the mathematical expression for the electric potential on the -axis. Draw an approximate graph of your result. (b) Find the point(s) on the -axis where the potential is zero. (c) Find an approximate expression for the potential at large distances from the origin along the -axis and explain its physical significance. 2. A home electrostatic air cleaner has a wire of radius metal cylinder of radius . The wire is at a potential of at the center of a relative to the cylinder. (a) Find the electric charge per unit length needed on the wire to establish this potential difference. (b) Show that the electric field at the surface of the wire is greater than , a field at which air becomes ionized. (By the ionization of air, charged particles are produced that can attach themselves to the dust, thereby making the dusk subject to removal by the action of the electric forces.) Check your answers to Practice Exercise 3-2 with the Module 1 Answer Key and review section 3-2 as necessary. 3-3: Finding Fields from Potentials OBJECTIVES: Determine the electric field when given an electric potential that is a function of one position variable only. Use equipotential surfaces and field lines for describing the potential and field semiquantitatively near several given point charges and/or simply shaped metallic surfaces. Commentary Study Figure 23.24 in the text carefully now in light of your new understanding of the relationship between and . To find from graphically, we simply draw the flux lines everywhere perpendicular to the equipotential surfaces. The method for using partial derivatives of to find the components of in three dimensions is outlined in the text. We will focus our problem-solving efforts on one-dimensional cases- those that can be solved for by differentiating with respect to a single variable- but note that the analysis is essentially the same when depends on more than one variable. Study the following examples before doing the practice exercise. EXAMPLE 3-5: A certain spherically symmetric charge distribution generates the potential , where is the radius from the center. Find the electric field. Solution: Since depends only on the radius, the electric field is radially directed and has magnitude . If is positive, radially inward. is directed radially outward. If is negative, is directed EXAMPLE 3-6: They symmetry of the problem of the two point charges in Example 3-3 implies that on the -axis the electric field will be directed along the axis. Find the electric field on the -axis, , using the result for given in Example 3-3. Solution: We must calculate from the result . To obtain the correct sign for , we must eliminate the absolute-value signs in the expression for . This involves three separate cases corresponding to three intervals on the -axis. Case (i): , , . Case (ii): , , . Case (iii): , , . Note that since the squares in the denominators of the electric-field terms are always positive, you can identify the sign of a term form the sign in front of it. Thus we see that due to the charge at is positive in the interval and negative for , as it must be since this charge is positive and thus repels another positive charge. The field along the -axis due to the charge at positively for . is directed negatively for and EXAMPLE 3-7: Two equal positive charges are placed on the -axis at equal distances above and below the -axis. Draw the equipotential lines and field lines caused by this two-charge system in the intersections of the equipotential surfaces with the -plane (i.e., trace the -plane). Solution: For a problem like this, it is wise to exploit your knowledge of the potential of a single point charge. Thus (a) Near each charge, where the potential increases without limit, the equipotential surfaces will be virtually unaffected by the presence of the other charge. The surfaces will approximate spheres, and their intersection with the -plane will approximate circles centered around the charge. (b) Very far from all the charges, the whole system will act like a single point charge whose magnitude is the algebraic sum of all the charges in the system. Equipotential surfaces will again approximate spheres centered around the “center” of the charges. (c) Field lines are perpendicular to the equipotential surfaces and can terminate only at charged bodies. After constructing the near- and far-field lines with the help of (a) and (b), you must connect the two parts of the diagram. (d) At intermediate points, you may be able to use symmetry of the charges and clues from the need to connect field lines at small and large distances as described in part (c). You may also be able to locate special points where the electric field is zero; at such points the field lines have no welldefined direction, and it is possible for two equipotential surfaces to intersect. (Since these surfaces are perpendicular to the field lines, and a field line can have only one perpendicular surface at a point, two equipotential surfaces cannot ordinarily intersect.) These ideas have been used to construct Figure 23.24(c) in the text. Note the small, circular equipotentials near each charge, the large oval shapes that will become more nearly circular at larger distances, the total number of field lines for the system (twice that of each charge separately), and the field-free symmetry point half-way between the charges where equipotentials intersect. EXAMPLE 3-8: A point charge of magnitude charge of magnitude is located at is located at and a point , with both and positive. Draw approximate equipotential lines and electric-field lines in the -plane for this system of charges. Solution: We shall begin with an overall analysis, following the procedure outlined in Example 3-7. (a) Near each charge there will be equipotential circles, representing positive potentials near and negative potentials near . (b) Far from the origin (that is, for ) there will be equipotential circles representing the negative potentials of a charge located approximately at the origin. (c) Field lines radiate outward from charge at many field lines radiate inward to charge lines emanating from the charge at the negative charge at hemisphere at , and two times as at . All of the field go into the right hemisphere of , whereas the lines coming into the left come from infinity. The latter are uniformly distributed in angle at the large equipotential circles mentioned in (b). (d) Since the charges are unequal, there is no simple right-left symmetry, but there is up-down symmetry. Since the charges have opposite sign, the field between them will be very strong. However, to the right of the positive charge, which is weaker than the negative charge, there must be a reversal of the field because near the field lines point to the right, whereas farther away, where the negative charge dominates, the field lines point to the left. To find the zero-field position we need the root of (see Example 3-6). This equation is solved most easily by taking square roots of both sides and then collecting terms of the linear equation, with the result that . At this point equipotential lines may cross to make a transition from the behavior near the charges to the far behavior. We can now draw the figure below as the solution to this problem. Practice Exercise 3-3 Write your solutions to the following problems in your notebook. 1. A point charge of magnitude is placed at the point axis is given by . Find the . Its potential on the component of the electric field associated with this potential by differentiating and justify your procedure. (Hint: treat the absolute value carefully!) 2. A charge distribution at the coordinate origin gives rise to the potential , where is distance from the origin and is given constant vector called the electric dipole moment. (a) Find the (b) Find the component of the electric field for points on the -axis. component of the electric field for points on the -axis. (Hint: First state the potential on this axis.) 3. Draw approximate equipotentials and field lines in the system in problem 1 of Practice Exercise 3-2. -plane for the two-charge 4. The figure below shows a cross-section of equipotential surfaces in a region of electric field with cylindrical (not circular) symmetry. Five points have been marked by letters. (a) At which point is the field strongest? Explain. (b) At which point is the field weakest? Explain. Check your answers to Practice Exercise 3-3 with those given in the Module 1 Answer Key. Correct any errors in your solutions and review Section 3-3 as necessary before going on to the self-check test. To see if you have achieved the objectives in Lesson 3, try to solve the problems in Self-Check Test 3 without using any reference materials. Self-Check Test 3 Write your solutions to the following problems in your notebook. 1. A long, hollow, cylindrical conductor of radius has a charge of coulombs per meter of length. On one side it has a tiny hole through which electrons can emerge from the interior, but the hole has a negligible effect on the electric field due to the cylinder. Use the hole as the reference (zero) point for the electric potential. (a) Find the electric potential radius from the cylinder axis). both outside and inside the cylinder ( is the (b) How much kinetic energy is acquired from the field by an electron that emerges from the hole and hits a screen from the axis? Express the answer in electron volts. (c) With what speed does the electron hit the screen if it starts with negligible speed in the hole? (d) Suppose you had a frictionless guiding mechanism for the electron that made it spiral around the cylinder rather than traveling straight from the hole to the screen. How would that affect the answers to parts (b) and (c)? Explain briefly. 2. A spherically symmetric charge distribution gives rise to the potential , where is the radius from the center of the charge, and and are constants with dimensions of distance potential and distance, respectively. Find the electric field (magnitude and direction). 3. Two point charges of magnitudes and are placed with the separation . Make an approximate drawing of several equipotentials and field lines in a plane including the two charges. Check your answers with the key and review Lesson 3 as necessary. Assignment 3 When you have demonstrated mastery of the content of this lesson, log into the Mastering Physics website and work Homework Set 3. Module 1 Review Much of the rest of your work in this course will build upon what you have learned in this unit, so it is essential that you have the important concepts of electrostatics firmly in mind. You have considered these concepts from a number of different aspects; as you review the unit try to “tie things together” in order to strengthen your understanding. For example, lines of electric force (flux lines) can help you to visualize both equipotential surfaces and Gaussian surfaces. The analogy between the gravitational field and the electric field can also be useful; the only memory supplement you need for this analogy is that the field is directed inward toward a negative charge. Use the objectives in Lessons 1, 2, and 3 in this syllabus and the review-and-summary sections at the end of Chapters 21-23 in the text to guide your review. When you have demonstrated mastery of the content of this unit begin working on the next lesson.