Chapter 4. Electric Potential


 Henry Richardson
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1 Chapte 4 Electic Potential 4.1 Potential and Potential Enegy Electic Potential in a Unifom Field Electic Potential due to Point Chages Potential Enegy in a System of Chages Deiving Electic Field fom the Electic Potential Example 4.4.1: Calculating Electic Field fom Electic Potential Gadients and Equipotentials : Conductos and Equipotentials Continuous Symmetic Chage Distibutions Example 4.61: Electic Potential Due to a Spheical Shell Example Conducting Sphees Connected by a Wie Continuous NonSymmetic Chage Distibutions Example 4.7.1: Unifomly Chaged Rod Example 4.7.2: Unifomly Chaged Ring Example 4.7.3: Unifomly Chaged Disk Summay PoblemSolving Stategy: Calculating Electic Potential Solved Poblems Electic Potential Due to a System of Two Chages Electic Dipole Potential Electic Potential of an Annulus Chage Moving Nea a Chaged Wie Electic Potential of a Unifomly Chaged Sphee Conceptual Questions Additional Poblems Cube Thee Chages Wok Done on Chages Calculating E fom V
2 Electic Potential of a Rod Electic Potential Calculating Electic Field fom the Electic Potential Electic Potential and Electic Potential Enegy Electic Field, Potential and Enegy PN Junction Sphee with NonUnifom Chage Distibution Electic Potential Enegy of a Solid Sphee Calculating Electic Field fom Electical Potential
3 Electic Potential 4.1 Potential and Potential Enegy In the intoductoy mechanics couse, we have seen that foce on a paticle of mass m located at a distance fom Eath s cente due to the gavitational inteaction between the paticle and the Eath obeys an invesesquae law: Mm Fg = G 2, (4.1.1) whee G = N m 2 /kg 2 is the gavitational constant and is a unit vecto pointing adially outwad fom the Eath. The Eath is assumed to be a unifom sphee of mass M. The coesponding gavitation field g, defined as the gavitation foce pe unit mass, is given by F GM g = g = 2. (4.1.2) m Notice that g is a function of M, the mass that ceates the field, and, the distance fom the cente of the Eath. Figue Conside moving a paticle of mass m unde the influence of gavity (Figue 4.1.1). The wok done by gavity in moving m fom A to B is WG = FG d s = B A GMm GMm 2 d = B A 1 1 = GMm. B A (4.1.3) 43
4 The esult shows that W G is independent of the path taken; it depends only on the endpoints A and B. Nea Eath s suface, the gavitational field g is appoximately constant, with a magnitude g = GM / E 2 9.8m/s 2, whee E is the adius of Eath. The wok done by gavity in moving an object fom height y A to y B (Figue 4.1.2) is W g = F g d y B s = mg dy = mg( y B y A ). (4.1.4) y A Figue Moving an object fom A to B. The esult again is independent of the path, and is only a function of the change in vetical height y B y A. In the examples above, if the object etuns to its stating point, then the wok done by the gavitation foce on the object is zeo along this closed path. Any foce that satisfies this popety fo all closed paths is called a consevative foce: F d s = 0 (consevative foce). (4.1.5) all closed paths When dealing with a consevative foce, it is often convenient to intoduce the concept of change in potential enegy function, ΔU = U B U A between any two points in space, A and B, B ΔU = U B U A = F d s = W (4.1.6) whee W is the wok done by the foce on the object. In the case of gavity, W = W g and fom Eq. ((4.1.3)), the change in potential enegy can be witten as A 44
5 U G ( B ) U G ( A ) = GMm 1 1 B A (4.1.7) It is often convenient to choose a efeence point P whee U G ( P ) is equal to zeo. In the gavitational case, we choose infinity to be the efeence point, with U G ( P = ) = 0. Theefoe the change in potential enegy when two objects stat off an infinite distance apat and end up a distance apat is given by U G () U G ( ) = GMm 1 1 = GMm. (4.1.8) Thus we can define a potential enegy function U G () = GMm, U ( ) = 0. (4.1.9) When one object is much moe massive fo example the Eath and a satellite, then the scala quantity U G (), with units of enegy, coesponds to the negative of the wok done by the gavitation foce on the satellite as it moves fom an infinite distance away to a distance fom the cente of the Eath. The value of U G () depends on the choice that U G ( P = ) = 0. Howeve, the potential enegy diffeence U G ( B ) U G ( A ) between two points is independent of the choice of efeence point and by definition coesponds to a physical quantity, the negative of the wok done. Nea Eath s suface, whee the gavitation field g is appoximately constant, as an object moves fom the gound to a height h above the gound, the change in potential enegy is ΔU g = +mgh, and the wok done by gavity is W g = mgh. Let s again conside a gavitation field g. Let s define the change in potential enegy pe mass between points A and B by ΔV G V G ( B ) V G ( A ) = U G ( B ) U G ( A ) m ΔU G m (4.1.10) Accoding to ou definition, ΔV G = B A ( F G / m) d B s = g d s (4.1.11) A 45
6 ΔV G is called the gavitation potential diffeence. The teminology is unfotunate because it is vey easy to mixup potential diffeence with potential enegy diffeence. Fom Eq. (4.1.7), the gavitation potential diffeence between the points A and B is ΔV G = B A ( F G / m) d B s = g d s (4.1.12) Just like the gavitation field, the gavitation potential diffeence depends only on the M, the mass that ceates the field, and, the distance fom the cente of the Eath. Physically ΔV G epesents the negative of the wok done pe unit mass by gavity to move a paticle fom points A to B. Ou teatment of electostatics will be simila to gavitation because the electostatic foce Fe also obeys an invesesquae law. In addition, it is also consevative. In the pesence of an electic field E, in analogy to the gavitational field g, we define the electic potential diffeence between two points A and B as ΔV e = B A A ( F e / q t ) d B s = E d s, (4.1.13) whee q t is a test chage. The potential diffeence ΔV e, which we will now denote just by ΔV, epesents the negative of the wok done pe unit chage by the electostatic foce when a test chage q t moves fom points A to B. Again, electic potential diffeence should not be confused with electic potential enegy. The two quantities ae elated as follows. Suppose an object with chage q is moved acoss a potential diffeence ΔV, then the change in the potential enegy of the object is The SI unit of electic potential is volt [V] A ΔU = qδv. (4.1.14) 1volt = 1 joule/coulomb (1 V= 1 J/C). (4.1.15) When dealing with systems at the atomic o molecula scale, a joule [J]often tuns out to be too lage as an enegy unit. A moe useful scale is electon volt [ev], which is defined as the enegy an electon acquies (o loses) when moving though a potential diffeence of one volt: 1eV = ( C)(1V) = J. (4.1.16) 46
7 4.2 Electic Potential in a Unifom Field Conside a chage +q moving in the diection of a unifom electic field E = E( ĵ), as shown in Figue 4.2.1(a). (a) Figue (a) A chage q moving in the diection of a constant electic field E. (b) A mass m moving in the diection of a constant gavitation field g. Because the path taken is paallel to E, the electic potential diffeence between points A and B is given by B ΔV = V B V A = E d B s = E ds = Ed < 0. (4.2.1) A Theefoe point B is at a lowe potential compaed to point A. In fact, electic field lines always point fom highe potential to lowe. The change in potential enegy is ΔU = U B U A = qed. Because q > 0, fo this motion ΔU < 0, the potential enegy of a positive chage deceases as it moves along the diection of the electic field. The coesponding gavity analogy, depicted in Figue 4.2.1(b), is that a mass m loses potential enegy ( ΔU = mgd ) as it moves in the diection of the gavitation field g. A (b) Figue Potential diffeence in a unifom electic field 47
8 What happens if the path fom A to B is not paallel to E, but instead at an angle θ, as shown in Figue 4.2.2? In that case, the potential diffeence becomes ΔV = V B V A = B A E d s = E s = Escosθ = Ey. (4.2.2) Note that y inceases downwad in Figue Hee we see once moe that moving along the diection of the electic field E leads to a lowe electic potential. What would the change in potential be if the path wee A C B? In this case, the potential diffeence consists of two contibutions, one fo each segment of the path: ΔV = ΔV CA + ΔV BC. (4.2.3) When moving fom A to C, the change in potential is ΔV CA = Ey. When moving fom C to B, ΔV BC = 0 because the path is pependicula to the diection of E. Thus, the same esult is obtained iespective of the path taken, consistent with the fact that E is a consevative vecto field. Fo the path A C B, wok is done by the field only along the segment AC that is paallel to the field lines. Points B and C ae at the same electic potential, i.e., V B = V C. Because ΔU = qδv, this means that no wok is equied when moving the chage fom B to C. In fact, all points along the staight line connecting B and C ae on the same equipotential line. A moe complete discussion of equipotential will be given in Section Electic Potential due to Point Chages Next, let s compute the potential diffeence between two points A and B due to a chage +Q. The electic field poduced by Q is E = (Q / 2 )ˆ, whee ˆ is a unit vecto pointing adially away fom the location of the chage. Figue Potential diffeence between two points due to a point chage Q. 48
9 Fom Figue 4.3.1, we see that ˆ d s = dscosθ = d, which gives ΔV = V B V A = B A Q 2 ˆ d B Q s = d = Q 1 1 A 2 4πε 0 B A. (4.3.1) Once again, the potential diffeence ΔV depends only on the endpoints, independent of the choice of path taken. As in the case of gavity, only the diffeence in electical potential is physically meaningful, and one may choose a efeence point and set the potential thee to be zeo. In pactice, it is often convenient to choose the efeence point to be at infinity, so that the electic potential at a point P becomes P V P = E d s, V ( ) = 0. (4.3.2) With this choice of zeo potential, we intoduce an electic potential function, V (), whee is the distance fom the pointlike chaged object with chage Q: V () = 1 Q. (4.3.3) When moe than one point chage is pesent, by applying the supeposition pinciple, the electic potential is the sum of potentials due to individual chages: V () = 1 i q i i q = k i e (4.3.4) A summay of compaison between gavitation and electostatics is tabulated below: i i 49
10 Gavity Mass m Electostatics Chage q Gavitation foce F G = G Mm ˆ Electic foce Qq F 2 e = k e ˆ 2 Gavitation field g = F g / m Potential enegy change ΔU = B A F G d s Electic field E = F e / q Potential enegy change B ΔU = F e d s B Gavitational potential ΔV G = g d s Electic Potential ΔV = A A B A E d s Potential function, V G ( ) = 0 : V G = GM Q Potential function, V ( ) = 0 : V = k e ΔU g = mgd, (constant g ) ΔU = qed, (constant E) Potential Enegy in a System of Chages Suppose you lift a mass m though a height h. The wok done by the extenal agent (you), is positive, W ext = mgh > 0. The wok done by the gavitation field is negative, W g = mgh = W ext. The change in the potential enegy is theefoe equal to the wok that you do in lifting the mass, ΔU g = W g = +W ext = mgh. If an electostatic system of chages is assembled by an extenal agent, then ΔU = W = +W ext. That is, the change in potential enegy of the system is the wok that must be put in by an extenal agent to assemble the configuation. The chages ae bought in fom infinity and ae at est at the end of the pocess. Let s stat with just two chages q 1 and q 2 that ae infinitely fa apat with potential enegy U = 0. Let the potential due to q 1 at a point P be V 1 (Figue 4.3.2). Figue Two point chages sepaated by a distance
11 The wok W 2 done by an extenal agent in binging the second chage q 2 fom infinity to P is then W 2 = q 2 V 1. Because V 1 = q 1 / 12, whee 12 is the distance measued fom q 1 to P, we have that U 12 = W 2 = q 2 V 1 = 1 q 1 q (4.3.5) If q 1 and q 2 have the same sign, positive wok must be done to ovecome the electostatic epulsion and the change in the potential enegy of the system is positive, U 12 > 0. On the othe hand, if the signs ae opposite, then U 12 < 0 due to the attactive foce between the chages. To add a thid chage q 3 to the system (Figue 4.3.3), the wok equied is W 3 = q 3 ( V 1 + V 2 ) = q q q 2 4πε (4.3.6) 23 Figue A system of thee point chages. The potential enegy of this configuation is then U = W 2 + W 3 = 1 q 1 q 2 + q q q q 2 3 4πε = U + U + U. (4.3.7) The equation shows that the total potential enegy is simply the sum of the contibutions fom distinct pais. Genealizing to a system of N chages, we have U = 1 23 N N q i q j, (4.3.8) whee the constaint j > i is placed to avoid double counting each pai. Altenatively, one may count each pai twice and divide the esult by 2. This leads to i=1 j=1 j>i ij 411
12 U = 1 8πε 0 N N q i q j = 1 2 i=1 j=1 j i ij N q i i=1 1 4πε 0 N j=1 j i q j ij = 1 N q 2 i V ( i ). (4.3.9) i=1 whee V ( i ), the quantity in the paenthesis is the potential at i (location of q i ) due to all the othe chages. 4.4 Deiving Electic Field fom the Electic Potential In Eq. (4.3.2) we established the elation between E and V. If we conside two points that ae sepaated by a small distance d s, the following diffeential fom is obtained: dv = E d s. (4.4.1) In Catesian coodinates, E = E x î + E y ĵ + E z ˆk and d s = dx î + dyĵ + dz ˆk, and theefoe dv = (E x î + E y ĵ + E z ˆk )(dx î + dyĵ + dz ˆk) = E x dx + E y dy + E z dz. (4.4.2) We define diectional deivatives V / x, V / y, and V / z such that Theefoe dv = V x dx + V y E x = V x, E y = V y, dy + V z dz. (4.4.3) E z = V z. (4.4.4) By intoducing a diffeential quantity called the del (gadient) opeato î + ĵ + x y z ˆk (4.4.5) the electic field can be witten as E V = E x î + E y ĵ + E z ˆk = x î + V y ĵ + V z ˆk = V. (4.4.6) The diffeential opeato,, opeates on a scala quantity (electic potential) and esults in a vecto quantity (electic field). Mathematically, we can think of E as the negative of the gadient of the electic potential V. Physically, the negative sign implies that if V inceases as a positive chage moves along some diection, say x, with V / x > 0, then 412
13 thee is a nonvanishing component of E in the opposite diection E x = V / x < 0. In the case of gavity, if the gavitational potential inceases when a mass is lifted a distance h, the gavitational foce must be downwad. If the chage distibution possesses spheical symmety, then the esulting electic field is a function of the adial distance, i.e., E = E ˆ. In this case, dv = E d. If V () is known, then E may be obtained as E = E ˆ = dv d ˆ (4.4.7) Fo example, the electic potential due to a point chage q is V () = q /. Using the above fomula, the electic field is simply E = (q / 2 )ˆ. Example 4.4.1: Calculating Electic Field fom Electic Potential Suppose the electic potential due to a cetain chage distibution can be witten in Catesian Coodinates as V (x, y,z) = Ax 2 y 2 + Bxyz whee A, B and C ae constants. What is the associated electic field? Solution: The electic field can be found by using Eq. (4.4.4) E x = V x = 2Axy2 Byz E y = V y = 2Ax2 y Bxz E z = V z = Bxy Theefoe, the electic field is E = ( 2Axy 2 Byz) î (2Ax2 y + Bxz) ĵ Bxy ˆk. 4.5 Gadients and Equipotentials Suppose a system in two dimensions has an electic potential V (x, y). The cuves chaacteized by constant V (x, y) ae called equipotential cuves. Examples of equipotential cuves ae depicted in Figue below. 413
14 Figue Equipotential cuves In thee dimensions, sufaces such that V (x, y,z) = constant ae called equipotential sufaces. Because E = V, we can show that the diection of E at a point is always pependicula to the equipotential though that point. We shall show this in two dimensions. Genealization to thee dimensions is staightfowad. Refeing to Figue 4.5.2, let the potential at a point P(x, y) be V (x, y). What is the potential diffeence dv between P(x, y) and a neighboing point P(x + dx, y + dy)? Wite the diffeence as dv = V (x + dx, y + dy) V (x, y) = V (x, y) + V x dx + V y dy + V (x, y) V x dx + V y dy. (4.5.1) Figue Change in V when moving fom one equipotential cuve to anothe The displacement vecto connecting the points is given by d s = dxî + dyĵ. We can ewite dv as dv = V î + V ĵ x y ( dxî + dyĵ) = ( V ) ds = E d s. (4.5.2) 414
15 If the displacement d s is along the tangent to the equipotential cuve that passes though the point P with coodinates (x, y), then dv = 0 because V is constant eveywhee on that cuve. This implies that E d s along the equipotential cuve. That is, E is pependicula to the equipotential. In Figue we illustate some examples of equipotential cuves. In thee dimensions they become equipotential sufaces. Fom Eq. (4.5.8), we also see that the change in potential dv attains a maximum when the gadient V is paallel to d s : dv max = V. ds (4.5.3) Physically, this means that V always points in the diection of maximum ate of change of V with espect to the displacement d s. Figue Equipotential cuves and electic field lines fo (a) a constant E field, (b) a point chage, and (c) an electic dipole. The popeties of equipotential sufaces can be summaized as follows: (i) The electic field lines ae pependicula to the equipotentials and point fom highe to lowe potentials. (ii) By symmety, the equipotential sufaces poduced by a point chage fom a family of concentic sphees, and fo constant electic field, a family of planes pependicula to the field lines. (iii) The tangential component of the electic field along the equipotential suface is zeo, othewise nonvanishing wok would be done to move a chage fom one point on the suface to the othe. (iv) No wok is equied to move a paticle along an equipotential suface. A useful analogy fo equipotential cuves is a topogaphic map (Figue 4.5.4). Each contou line on the map epesents a fixed elevation above sea level. Mathematically it is 415
16 expessed as z = f (x, y) = constant. Since the gavitational potential nea the suface of Eath is Vg = gz, these cuves coespond to gavitational equipotentials. Figue A topogaphic map 4.5.1: Conductos and Equipotentials We aleady studies the basic popeties of a conducto in Chapte 3 which we now summaized: (1) the electic field inside a conducto is zeo; (2) any net chage must eside on the suface of the conducto; (3) the tangential component of the electic field on the suface is zeo; (4) just outside the conducto, the electic field is nomal to the suface; (5) the discontinuity in the nomal component of the electic field acoss the suface of a conducto is popotional to the suface chage density Because the tangential component of the electic field on the suface of a conducto vanishes, this implies that the suface of a conducto in electostatic equilibium is an equipotential suface. To veify this claim, conside two points A and B on the suface of a conducto. Since the tangential component Et = 0, the potential diffeence is B VB V A = E d s = 0 A 416
17 because E is pependicula to d s. Thus, points A and B ae at the same potential with V A = V B. 4.6 Continuous Symmetic Chage Distibutions We shall now calculate the electic potential diffeence between two points in space associated with a continuous symmetic distibution of chage in which we can fist use Gauss s Law to detemine the electic field eveywhee is space. Example 4.61: Electic Potential Due to a Spheical Shell Conside a metallic spheical shell of adius a and chage Q, as shown in Figue Figue A spheical shell of adius a and chage Q. (a) Find the electic potential eveywhee. (b) Calculate the potential enegy of the system. Solution: (a) In Example 3.3, we showed that the electic field fo a spheical shell of is given by E = Q ˆ, > a 2 0, < a. The electic potential may be calculated by using Eq. (4.1.13), Fo > a, we have B V B V A = E d s. A 417
18 Q V () V ( ) = d = 1 Q 2 = k e Q, (4.6.1) whee we have chosen V ( ) = 0 as ou efeence point. On the othe hand, fo < a, the potential becomes a V () V ( ) = Ed 0d a a Q = d = 1 Q 2 a = k e Q (4.6.2) a. A plot of the electic potential is shown in Figue Note that the potential V is constant inside a conducto. Figue Electic potential as a function of fo a spheical conducting shell (b) The potential enegy U can be thought of as the wok that needs to be done to build up the system. To chage up the sphee, an extenal agent must bing chage fom infinity and deposit it onto the suface of the sphee. Suppose the chage accumulated on the sphee at some instant is q. The potential at the suface of the sphee is then V = q / a. The amount of wok that must be done by an extenal agent to bing chage dq fom infinity and deposit it on the sphee is q dw ext = Vdq = a dq. (4.6.3) Theefoe, the total amount of wok needed to chage the sphee to Q is W ext = 0 Q dq q a = Q2 8πε 0 a. (4.6.4) Because V = Q / a and W ext = U, the above expession simplified to U = (1 / 2)QV. (4.6.5) 418
19 The esult can be contasted with the case of a point chage. The wok equied to bing a point chage Q fom infinity to a point whee the electic potential due to othe chages is V is W ext = QV. Theefoe, fo a point chage Q, the potential enegy is U = QV. Example Conducting Sphees Connected by a Wie Why does lightning stike the tip of a lightning od? Let s ty to answe that question. Suppose two metal sphees with adii 1 and 2 ae connected by a thin conducting wie, as shown in Figue Figue Two conducting sphees connected by a wie. Chage will continue to flow until equilibium is established such that both sphees ae at the same potential V 1 = V 2 = V. Suppose the chages on the sphees at equilibium ae q 1 and q 2. Neglecting the effect of the wie that connects the two sphees, the equipotential condition implies V = 1 q 1 = 1 q Theefoe q 1 = q 2, (4.6.6) 1 2 povided that the two sphees ae vey fa apat so that the chage distibutions on the sufaces of the conductos ae unifom. The electic fields can be expessed as E 1 = 1 q = σ 1 ε 0, E 2 = 1 q = σ 2 ε 0, (4.6.7) whee σ 1 and σ 2 ae the suface chage densities on sphees 1 and 2, espectively. Divided the magnitudes of the electic fields yields E 1 E 2 = σ 1 σ 2 = 2 1. (4.6.8) 419
20 With the suface chage density being invesely popotional to the adius, we conclude that the egions with the smallest adii of cuvatue have the geatest σ. Thus, the electic field stength on the suface of a conducto is geatest at the shapest point. The design of a lightning od is based on this pinciple. Lighting stikes the tip. 4.7 Continuous NonSymmetic Chage Distibutions If the chage distibution is continuous, the potential at a point P can be found by summing ove the contibutions fom individual diffeential elements of chage dq. Figue Continuous chage distibution Conside the chage distibution shown in Figue Taking infinity as ou efeence point with zeo potential, the electic potential at P due to dq is dv = 1 dq. (4.7.1) Summing ove contibutions fom all the diffeential elements, we have that V = 1 Example 4.7.1: Unifomly Chaged Rod dq. (4.7.2) Conside a nonconducting od of length having a unifom chage density λ. Find the electic potential at P, a pependicula distance y above the midpoint of the od. Solution: Conside a diffeential element of length d x that caies a chage dq = λ d x, as shown in Figue The souce element is located at ( x,0), while the field point P is located on the yaxis at (0,y). The distance fom d x to P is = ( x 2 + y 2 ) 1/ 2. Its contibution to the potential is given by 420
21 dv = 1 dq = 1 λ d x ( x 2 + y 2 ). 1/ 2 Figue A nonconducting od of length and unifom chage density λ. Taking V to be zeo at infinity, the total potential due to the entie od is V = λ / 2 d x = λ ln x + x 2 + y 2 / 2 x 2 + y 2 = λ ( / 2) + ( / 2) 2 + y 2 ln ( / 2) + ( / 2) 2 + y 2, / 2 / 2 (4.7.3) whee we have used the integation fomula d x = ln( x + x 2 + y 2 ). x 2 + y 2 A plot of V ( y) / V 0, whee V 0 = λ /, as a function of y / is shown in Figue
22 Figue Electic potential along the axis that passes though the midpoint of a nonconducting od. In the limit y, the potential becomes V = λ ( / 2) + / 2 1+ (2y / ) 2 ln ( / 2) + / 2 1+ (2y / ) 2 λ 2 ln 2y 2 / 2 = λ ln 2 4πε 0 y 2 = λ ln 2πε 0 y. = λ (2y / ) 2 ln (2y / ) 2 (4.7.4) The coesponding electic field can be obtained as E y = V y = λ 2πε 0 y / 2 ( / 2) 2 + y 2, in ageement with the esult obtained in Chapte 2, Eq. (2.10.9). Example 4.7.2: Unifomly Chaged Ring Conside a unifomly chaged ing of adius R and chage density λ (Figue 4.7.4). What is the electic potential at a distance z fom the cental axis? 422
23 Figue A nonconducting ing of adius R with unifom chage density λ. Solution: Conside a small diffeential element d = R d φ on the ing. The element caies a chage dq = λ d = λr d φ, and its contibution to the electic potential at P is dv = 1 dq = 1 λr d φ 4πε. 0 R 2 + z 2 The electic potential at P due to the entie ing is V = dv = 1 λr d φ = 1 R 2 + z 2 2πλR R 2 + z 2 = 1 Q R 2 + z 2, (4.7.5) whee we have substituted Q = 2π Rλ fo the total chage on the ing. In the limit z >> R, the potential appoaches its pointchage limit: V 1 Q z. Fom Eq. (4.4.4) the zcomponent of the electic field may be obtained as E z = V z = z 1 Q R 2 + z 2 = 1 Qz. (4.7.6) (R 2 + z 2 3/ 2 ) in ageement with Eq. ( ). Example 4.7.3: Unifomly Chaged Disk Conside a unifomly chaged disk of adius R and chage densityσ lying in the xyplane (Figue 4.7.5). What is the electic potential at a distance z fom the cental axis? 423
24 Figue A nonconducting disk of adius R and unifom chage density σ. Solution: Conside a cicula ing of adius and width d. The chage on the ing is d q = σ d A = σ(2π d ). The field point P is located along the z axis a distance z fom the plane of the disk. Fom the figue, we also see that the distance fom a point on the ing to P is = ( 2 + z 2 ) 1/ 2. Theefoe, the contibution to the electic potential at P is dv = 1 dq = 1 σ(2π d ). 2 + z 2 By summing ove all the ings that make up the disk, we have V = σ R 2π d z 2 = σ 2 + z 2 2ε 0 R 0 = σ R 2 + z 2 z 2ε 0. (4.7.7) In the limit z >> R, R 2 + z 2 = z 1+ R2 z 2 1/ 2 = z 1+ R2 2z + 2, and the potential simplifies to the pointchage limit: V σ 2ε 0 R 2 2 z = 1 σ(π R 2 ) = 1 z Q z. As expected, at lage distance, the potential due to a nonconducting chaged disk is the same as that of a point chage Q. A compaison of the electic potentials of the disk and a point chage is shown in Figue
25 Figue Compaison of the electic potentials of a nonconducting disk and a point chage. The electic potential is measued in tems of V 0 = Q / R. Note that the electic potential at the cente of the disk, z = 0, is finite, and its value is V c = σ R = Q 2ε 0 π R R = 1 2Q 2 2ε 0 R = 2V. (4.7.8) 0 This is the amount of wok that needs to be done to bing a unit chage fom infinity and place it at the cente of the disk. The coesponding electic field at P can be obtained as: E z = V z = σ 2ε 0 z z z R 2 + z 2, (4.7.9) which agees with Eq. ( ). In the limit R >> z, the above equation becomes E z = σ / 2ε 0, which is the electic field fo an infinitely lage nonconducting sheet. 425
26 4.8 Summay A foce F is consevative if the line integal of the foce aound a closed loop vanishes: F d s = 0. The change in potential enegy associated with a consevative foce F object as it moves fom A to B is B ΔU = U B U A = F d s. A acting on an The electic potential diffeence ΔV between points A and B in an electic field E is given by ΔV = V B V A = ΔU B = E d s q. A t The quantity epesents the amount of wok done pe unit chage to move a test chage q t fom point A to B, without changing its kinetic enegy. The electic potential due to a point chage Q at a distance away fom the chage is V = 1 Q. Fo a collection of chages, using the supeposition pinciple, the electic potential is V = 1 i Q i. The potential enegy associated with two point chages q 1 and q 2 sepaated by a distance 12 is i U = 1 q 1 q Fom the electic potential V, the electic field may be obtained by taking the gadient of V, E = V. In Catesian coodinates, the components may be witten as 426
27 E x = V x, E y = V y, E z = V z. The electic potential due to a continuous chage distibution is V = 1 dq. 4.9 PoblemSolving Stategy: Calculating Electic Potential In this chapte, we showed how electic potential could be calculated fo both the discete and continuous chage distibutions. Unlike electic field, electic potential is a scala quantity. Fo the discete distibution, we apply the supeposition pinciple and sum ove individual contibutions: q i V = k e. Fo the continuous distibution, we must evaluate the integal i i dq V = k e. In analogy to the case of computing the electic field, we use the following steps to complete the integation: (1) Stat with dv = k e dq. (2) Rewite the chage element dq as λ dl dq = σ da ρ dv (length) (aea) (volume) depending on whethe the chage is distibuted ove a length, an aea, o a volume. (3) Substitute dq into the expession fo dv. (4) Specify an appopiate coodinate system and expess the diffeential element ( dl, da o dv ) and in tems of the coodinates (see Table 2.1.) 427
28 (5) Rewite dv in tems of the integation vaiable. (6) Complete the integation to obtain V. Using the esult obtained fo V, one may calculate the electic field by E = V. Futhemoe, choosing a point P that lies sufficiently fa away fom the chage distibution can eadily check the accuacy of the esult. In this limit, if the chage distibution is of finite extent, the field should behave as if the distibution wee a point chage, and falls off as 1/ 2. Below we illustate how the above methodologies can be employed to compute the electic potential fo a line of chage, a ing of chage and a unifomly chaged disk. 428
29 Chaged Rod Chaged Ring Chaged disk dq = λ dx dq = λ dl dq = σ da Figue (2) Expess dq in tems of chage density (3) Substitute dq into expession fo dv (4) Rewite and the diffeential element in tems of the appopiate coodinates λ dx dv = ke dv = ke = x 2 + y 2 dv = ke = λ d x ( x + y 2 )1/ 2 Deive E fom V Pointchage limit fo E V= λ d x / 2 / 2 x 2 + y 2 ( / 2) + ( / 2)2 + y 2 λ = ln ( / 2) + ( / 2)2 + y 2 V y λ /2 = 2πε 0 y ( / 2)2 + y 2 Ey = Ey keq y 2 y Rλ (R + z 2 )1/ 2 = ke dφ (2π Rλ ) R2 + z 2 Q = ke R +z 2 Ez = Ez = 2 + z 2 λ R dφ (R 2 + z 2 )1/ keqz V = z (R 2 + z 2 )3/ 2 keq z2 σ da da = 2π d R2 + z 2 dv = ke 2 V = ke (6) Integate to get V dv = ke dl = R dφ dx (5) Rewite dv λ dl z R dv = ke 2πσ d ( 2 + z 2 )1/ 2 V = ke 2πσ = 2keπσ = 2keQ R Ez = 2 ( ( R d ( + z 2 )1/ ) z 2 + R2 z V 2keQ z = z R2 z Ez keq z2 ) z 2 + R2 z z +R z 2 z R
30 4.10 Solved Poblems Electic Potential Due to a System of Two Chages Conside a system of two chages shown in Figue Figue Electic dipole Find the electic potential at an abitay point on the xaxis and make a plot. Solution: The electic potential can be found by the supeposition pinciple. At a point on the xaxis, we have V (x) = 1 q x a + 1 The above expession may be ewitten as ( q) x + a = q 1 x a 1 x + a. whee V 0 = q / a. V (x) 1 = V 0 x / a 1 1 x / a + 1, Figue
31 The plot of the dimensionless electic potential as a function of x/a. is depicted in Figue As can be seen fom the gaph, V (x) diveges at x / a = ±1, whee the chages ae located Electic Dipole Potential Conside an electic dipole along the yaxis, as shown in the Figue Find the electic potential V at a point P in the xy plane, and use V to deive the coesponding electic field. Figue By supeposition pinciple, the potential at P is given by 1 V = V i = i 4πε 0 q q +, whee ± 2 = 2 + a 2 2acosθ. If we take the limit whee >> a, then 1 = 1 ± 1+ (a / 1/ 2 1 )2 2(a / )cosθ = (a / )2 ± (a / )cosθ +. The dipole potential can be appoximated as q V = (a / )2 + (a / )cosθ (a / )2 + (a / )cosθ + q 2acosθ = pcosθ = p ˆ 2,
32 whee p = 2aq ĵ is the electic dipole moment. In spheical pola coodinates, the gadient opeato is = ˆ + 1 θ ˆθ + 1 sinθ φ ˆφ Because the potential is now a function of both and θ, the electic field will have components along the ˆ  and ˆθ diections. Using E = V, we have E = V = pcosθ 2πε 0 3, E θ = 1 V θ = psinθ 3, E φ = Electic Potential of an Annulus Conside an annulus of unifom chage density σ, as shown in Figue Find the electic potential at a point P along the symmetic axis. Figue An annulus of unifom chage density. Solution: Conside a small diffeential element da at a distance away fom point P. The amount of chage contained in da is given by dq = σ da = σ( 'dθ)d '. Its contibution to the electic potential at P is dv = 1 dq = 1 σ 'd 'dθ ' 2 + z 2. Integating ove the entie annulus, we obtain V = σ b a 2π 0 'd 'dθ ' 2 + z 2 = 2πσ b a 'ds ' 2 + z 2 = σ b 2 + z 2 a 2 + z 2 2ε 0, 432
33 whee we have made used of the integal dss = s 2 + z 2. s 2 + z 2 Notice that in the limit a 0 and b R, the potential becomes V = σ R 2 + z 2 z 2ε 0, which agees with the esult of a nonconducting disk of adius R shown in Eq. (4.7.7) Chage Moving Nea a Chaged Wie A thin od extends along the z axis fom z = d to z = d. The od caies a positive chage Q unifomly distibuted along its length 2d with chage density λ = Q / 2d. (a) Calculate the electic potential at a point z > d along the z axis. (b) What is the change in potential enegy if an electon moves fom z = 4d to z = 3d? (c) If the electon stated out at est at the point z = 4d, what is its velocity at z = 3d? Solutions: (a) Fo simplicity, let s set the potential to be zeo at infinity, V ( ) = 0. Conside an infinitesimal chage element dq = λ d z located at a distance z ' along the zaxis. Its contibution to the electic potential at a point z > d is dv = λ dz ' z z '. Integating ove the entie length of the od, we obtain V (z) = λ z d dz' = λ ln z + d z + d z z' z d. (b) Using the esult deived in (a), the electical potential at z = 4d is V (z = 4d) = λ ln 4d + d 4d d = λ ln
34 Similaly, the electical potential at z = 3d is V (z = 3d) = λ ln 3d + d 3d d = λ ln 2. The electic potential diffeence between the two points is ΔV = V (z = 3d) V (z = 4d) = λ ln 6 5 > 0. Using the fact that the electic potential diffeence ΔV is equal to the change in potential enegy pe unit chage, we have whee q = e is the chage of the electon. ΔU = qδv = e λ ln 6 5 < 0, (c) If the electon stats out at est at z = 4d then the change in kinetic enegy is ΔK = 1 2 mv 2. f By consevation of enegy, the change in kinetic enegy is ΔK = ΔU = e λ ln 6 5 > 0. Thus, the magnitude of the velocity at z = 3d is v f = 2 e λ m ln Electic Potential of a Unifomly Chaged Sphee An insulated solid sphee of adius a has a unifom chage density ρ. Compute the electic potential eveywhee. Solution: Using Gauss s law, we showed in Example 3.4 that the electic field due to the chage distibution is 434
35 E = Q ˆ, > a 2 Q ˆ, < a. 3 a (4.10.1) The electic potential at P 1 (indicated in Figue ) outside the sphee is Q V 1 () V ( ) = d = 1 Q 2 = k e Q. (4.10.2) Figue Figue Electic potential due to a unifomly chaged sphee as a function of. On the othe hand, the electic potential at P 2 inside the sphee is given by ( ) ( ) a a Q Q V 2 () V ( ) = de > a E < a = d a d 2 a a 3 ( ) = 1 = 1 Q a 1 Q 1 a a 2 Q = k e 2a 3 2 a 2. Q 8πε 0 a 3 2 a 2 (4.10.3) A plot of electic potential as a function of is given in Figue : 4.11 Conceptual Questions 1. What is the diffeence between electic potential and electic potential enegy? 435
36 2. A unifom electic field is paallel to the xaxis. In what diection can a chage be displaced in this field without any extenal wok being done on the chage? 3. Is it safe to stay in an automobile with a metal body duing sevee thundestom? Explain. 4. Why ae equipotential sufaces always pependicula to electic field lines? 5. The electic field inside a hollow, unifomly chaged sphee is zeo. Does this imply that the potential is zeo inside the sphee? 4.12 Additional Poblems Cube How much wok is done to assemble eight identical point chages, each of magnitude q, at the cones of a cube of side a? Thee Chages Thee pointlike objects with chages with q = C and q 1 = C ae placed on the xaxis, as shown in the Figue The distance between q and q 1 is a = m. Figue (a) What is the net foce exeted on q by the othe two chages q 1? (b) What is the electic field at the oigin due to the two chages q 1? (c) What is the electic potential at the oigin due to the two chages q 1? Wok Done on Chages Two chages q 1 = 3.0µC and q 2 = 4.0µC initially ae sepaated by a distance 0 = 2.0cm. An extenal agent moves the chages until they ae f = 5.0cm apat. 436
37 (a) How much wok is done by the electic field in moving the chages fom 0 to f? Is the wok positive o negative? (b) How much wok is done by the extenal agent in moving the chages fom 0 to f? Is the wok positive o negative? (c) What is the potential enegy of the initial state whee the chages ae 0 = 2.0cm apat? (d) What is the potential enegy of the final state whee the chages ae f = 5.0cm apat? (e) What is the change in potential enegy fom the initial state to the final state? Calculating E fom V Suppose in some egion of space the electic potential is given by V (x, y,z) = V 0 E 0 z + E 0 a 3 z (x 2 + y 2 + z 2 ) 3/ 2, whee a is a constant with dimensions of length. Find the x, y, and the zcomponents of the associated electic field Electic Potential of a Rod A od of length L lies along the xaxis with its left end at the oigin and has a nonunifom chage density λ = αx, whee α is a positive constant (Figue ). (a) What ae the dimensions of α? Figue
38 (b) Calculate the electic potential at A. (c) Calculate the electic potential at point B that lies along the pependicula bisecto of the od a distance b above the xaxis Electic Potential Suppose that the electic potential in some egion of space is given by V (x, y,z) = V 0 exp( k z )cos kx. Find the electic field eveywhee. Sketch the electic field lines in the xz plane Calculating Electic Field fom the Electic Potential Suppose that the electic potential vaies along the xaxis as shown in Figue below. Figue The potential does not vay in the y o zdiection. Of the intevals shown (ignoe the behavio at the end points of the intevals), detemine the intevals in which E x has (a) its geatest absolute value. [Ans. 25 V/m in the inteval ab.] (b) its least absolute value. [Ans. (b) 0 V/m in the inteval cd.] (c) Plot E x as a function of x. (d) What sot of chage distibutions would poduce these kinds of changes in the potential? Whee ae they located? [Ans. sheets of chage extending in the yzdiection 438
39 located at points b, c, d, etc. along the xaxis. Note again that a sheet of chage with chage pe unit aea σ will always poduce a jump in the nomal component of the electic field of magnitudeσ / ε 0 ] Electic Potential and Electic Potential Enegy A ight isosceles tiangle of side a has chages q, +2q and q aanged on its vetices, as shown in Figue Figue (a) What is the electic potential at point P, midway between the line connecting the +q and q chages, assuming that V = 0 at infinity? [Ans. q / 2πε 0 a.] (b) What is the potential enegy U of this configuation of thee chages? What is the significance of the sign of you answe? [Ans. q 2 / 4 2πε 0 a, the negative sign means that wok was done on the agent who assembled these chages in moving them in fom infinity.] (c) A fouth chage with chage +3q is slowly moved in fom infinity to point P. How much wok must be done in this pocess? What is the significance of the sign of you answe? [Ans. +3q 2 / 2πε 0 a, the positive sign means that wok was done by the agent who moved this chage in fom infinity.] Electic Field, Potential and Enegy Thee chages, +5Q, 5Q, and +3Q ae located on the yaxis at y = +4a, y = 0, and y = 4a, espectively. The point P is on the xaxis at x = 3a. (a) How much enegy did it take to assemble these chages? (b) What ae the x, y, and z components of the electic field E at P? (c) What is the electic potential V at point P, taking V = 0 at infinity? 439
40 (d) A fouth chage of +Q is bought to P fom infinity. What ae the x, y, and z components of the foce F that is exeted on it by the othe thee chages? (e) How much wok was done (by the extenal agent) in moving the fouth chage +Q fom infinity to P? PN Junction When two slabs of Ntype and Ptype semiconductos ae put in contact, the elative affinities of the mateials cause electons to migate out of the Ntype mateial acoss the junction to the Ptype mateial. This leaves behind a volume in the Ntype mateial that is positively chaged and ceates a negatively chaged volume in the Ptype mateial. Let us model this as two infinite slabs of chage, both of thickness a with the junction lying on the plane z = 0. The Ntype mateial lies in the ange 0 < z < a and has unifom chage density +ρ 0. The adjacent Ptype mateial lies in the ange a < z < 0 and has unifom chage density ρ 0. Thus: (a) Find the electic field eveywhee. +ρ 0 0 < z < a ρ(x, y,z) = ρ(z) = ρ 0 a< z < 0 0 z > a. (b) Find the potential diffeence between the points P 1 and P 2.. The point P 1. is located on a plane paallel to the slab a distance z 1 > a fom the cente of the slab. The point P 2. is located on plane paallel to the slab a distance z 2 < a fom the cente of the slab Sphee with NonUnifom Chage Distibution A sphee made of insulating mateial of adius R has a chage density ρ = a whee a is a constant. Let be the distance fom the cente of the sphee. (a) Find the electic field eveywhee, both inside and outside the sphee. (b) Find the electic potential eveywhee, both inside and outside the sphee. Be sue to indicate whee you have chosen you zeo potential. (c) How much enegy does it take to assemble this configuation of chage? 440
41 (d) What is the electic potential diffeence between the cente of the cylinde and a distance inside the cylinde? Be sue to indicate whee you have chosen you zeo potential Electic Potential Enegy of a Solid Sphee Calculate the electic potential enegy of a solid sphee of adius R filled with chage of unifom density ρ. Expess you answe in tems of Q, the total chage on the sphee Calculating Electic Field fom Electical Potential Figue shows the vaiation of an electic potential V with distance z. The potential V does not depend on x o y. The potential V in the egion 1m < z < 1m is given in Volts by the expession V (z) = 15 5z 2. Outside of this egion, the electic potential vaies linealy with z, as indicated in the gaph. Figue (a) Find an equation fo the zcomponent of the electic field, E z, in the egion 1m < z < 1m. (b) What is E z in the egion z > 1 m? Be caeful to indicate the sign of E z? (c) What is E z in the egion z < 1 m? Be caeful to indicate the sign of E z? (d) This potential is due a slab of chage with constant chage pe unit volume ρ 0. Whee is this slab of chage located (give the zcoodinates that bound the slab)? What is the chage density ρ 0 of the slab in C/m 3? Be sue to give clealy both the sign and magnitude of ρ
Physics 202, Lecture 4. Gauss s Law: Review
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