# Chapter 4. Electric Potential

 To view this video please enable JavaScript, and consider upgrading to a web browser that supports HTML5 video
Save this PDF as:

Size: px
Start display at page:

## Transcription

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 Non-Symmetic Chage Distibutions Example 4.7.1: Unifomly Chaged Rod Example 4.7.2: Unifomly Chaged Ring Example 4.7.3: Unifomly Chaged Disk Summay Poblem-Solving 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 P-N Junction Sphee with Non-Unifom 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 invese-squae 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) 4-3

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 4-4

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 4-5

6 ΔV G is called the gavitation potential diffeence. The teminology is unfotunate because it is vey easy to mix-up 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 invese-squae 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) 4-6

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 4-7

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. 4-8

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 point-like 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 4-9

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 4-11

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 4-12

13 thee is a non-vanishing 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. 4-13

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) 4-14

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 non-vanishing 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 4-15

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 4-16

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 4-17

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) 4-18

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) 4-19

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 Non-Symmetic 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 non-conducting 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 y-axis 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 4-20

21 dv = 1 dq = 1 λ d x ( x 2 + y 2 ). 1/ 2 Figue A non-conducting 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? 4-22

23 Figue A non-conducting 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 point-chage limit: V 1 Q z. Fom Eq. (4.4.4) the z-component 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? 4-23

24 Figue A non-conducting 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 point-chage 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 non-conducting 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 non-conducting 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 non-conducting sheet. 4-25

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 4-26

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 Poblem-Solving 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.) 4-27

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. 4-28

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 x-axis and make a plot. Solution: The electic potential can be found by the supeposition pinciple. At a point on the x-axis, 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 y-axis, as shown in the Figue Find the electic potential V at a point P in the x-y 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, 4-32

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 non-conducting 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 z-axis. 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 4-34

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? 4-35

36 2. A unifom electic field is paallel to the x-axis. 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 point-like objects with chages with q = C and q 1 = C ae placed on the x-axis, 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. 4-36

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 z-components of the associated electic field Electic Potential of a Rod A od of length L lies along the x-axis 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 x-axis 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 x-axis as shown in Figue below. Figue The potential does not vay in the y- o z-diection. 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 yz-diection 4-38

39 located at points b, c, d, etc. along the x-axis. 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 y-axis at y = +4a, y = 0, and y = 4a, espectively. The point P is on the x-axis 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? 4-39

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? P-N Junction When two slabs of N-type and P-type semiconductos ae put in contact, the elative affinities of the mateials cause electons to migate out of the N-type mateial acoss the junction to the P-type mateial. This leaves behind a volume in the N-type mateial that is positively chaged and ceates a negatively chaged volume in the P-type 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 N-type mateial lies in the ange 0 < z < a and has unifom chage density +ρ 0. The adjacent P-type 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 Non-Unifom 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? 4-40

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 z-component 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 z-coodinates 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

Physics 202, Lectue 4 Today s Topics Review: Gauss s Law Electic Potential (Ch. 25-Pat I) Electic Potential Enegy and Electic Potential Electic Potential and Electic Field Next Tuesday: Electic Potential

### Gauss Law. Physics 231 Lecture 2-1

Gauss Law Physics 31 Lectue -1 lectic Field Lines The numbe of field lines, also known as lines of foce, ae elated to stength of the electic field Moe appopiately it is the numbe of field lines cossing

### Revision Guide for Chapter 11

Revision Guide fo Chapte 11 Contents Student s Checklist Revision Notes Momentum... 4 Newton's laws of motion... 4 Gavitational field... 5 Gavitational potential... 6 Motion in a cicle... 7 Summay Diagams

### Voltage ( = Electric Potential )

V-1 of 9 Voltage ( = lectic Potential ) An electic chage altes the space aound it. Thoughout the space aound evey chage is a vecto thing called the electic field. Also filling the space aound evey chage

### Physics 235 Chapter 5. Chapter 5 Gravitation

Chapte 5 Gavitation In this Chapte we will eview the popeties of the gavitational foce. The gavitational foce has been discussed in geat detail in you intoductoy physics couses, and we will pimaily focus

### Chapter 22. Outside a uniformly charged sphere, the field looks like that of a point charge at the center of the sphere.

Chapte.3 What is the magnitude of a point chage whose electic field 5 cm away has the magnitude of.n/c. E E 5.56 1 11 C.5 An atom of plutonium-39 has a nuclea adius of 6.64 fm and atomic numbe Z94. Assuming

### The Electric Potential, Electric Potential Energy and Energy Conservation. V = U/q 0. V = U/q 0 = -W/q 0 1V [Volt] =1 Nm/C

Geneal Physics - PH Winte 6 Bjoen Seipel The Electic Potential, Electic Potential Enegy and Enegy Consevation Electic Potential Enegy U is the enegy of a chaged object in an extenal electic field (Unit

### mv2. Equating the two gives 4! 2. The angular velocity is the angle swept per GM (2! )2 4! 2 " 2 = GM . Combining the results we get !

Chapte. he net foce on the satellite is F = G Mm and this plays the ole of the centipetal foce on the satellite i.e. mv mv. Equating the two gives = G Mm i.e. v = G M. Fo cicula motion we have that v =!

### Voltage ( = Electric Potential )

V-1 Voltage ( = Electic Potential ) An electic chage altes the space aound it. Thoughout the space aound evey chage is a vecto thing called the electic field. Also filling the space aound evey chage is

### 2 - ELECTROSTATIC POTENTIAL AND CAPACITANCE Page 1

- ELECTROSTATIC POTENTIAL AND CAPACITANCE Page. Line Integal of Electic Field If a unit positive chage is displaced by `given by dw E. dl dl in an electic field of intensity E, wok done is Line integation

### Chapter 26 - Electric Field. A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University

Chapte 6 lectic Field A PowePoint Pesentation by Paul. Tippens, Pofesso of Physics Southen Polytechnic State Univesity 7 Objectives: Afte finishing this unit you should be able to: Define the electic field

### Problems on Force Exerted by a Magnetic Fields from Ch 26 T&M

Poblems on oce Exeted by a Magnetic ields fom Ch 6 TM Poblem 6.7 A cuent-caying wie is bent into a semicicula loop of adius that lies in the xy plane. Thee is a unifom magnetic field B Bk pependicula to

### Chapter 13 Gravitation. Problems: 1, 4, 5, 7, 18, 19, 25, 29, 31, 33, 43

Chapte 13 Gavitation Poblems: 1, 4, 5, 7, 18, 19, 5, 9, 31, 33, 43 Evey object in the univese attacts evey othe object. This is called gavitation. We e use to dealing with falling bodies nea the Eath.

### 2 r2 θ = r2 t. (3.59) The equal area law is the statement that the term in parentheses,

3.4. KEPLER S LAWS 145 3.4 Keple s laws You ae familia with the idea that one can solve some mechanics poblems using only consevation of enegy and (linea) momentum. Thus, some of what we see as objects

### Chapter 2. Electrostatics

Chapte. Electostatics.. The Electostatic Field To calculate the foce exeted by some electic chages,,, 3,... (the souce chages) on anothe chage Q (the test chage) we can use the pinciple of supeposition.

### Gauss Law in dielectrics

Gauss Law in dielectics We fist deive the diffeential fom of Gauss s law in the pesence of a dielectic. Recall, the diffeential fom of Gauss Law is This law is always tue. E In the pesence of dielectics,

### 12. Rolling, Torque, and Angular Momentum

12. olling, Toque, and Angula Momentum 1 olling Motion: A motion that is a combination of otational and tanslational motion, e.g. a wheel olling down the oad. Will only conside olling with out slipping.

### 2008 Quarter-Final Exam Solutions

2008 Quate-final Exam - Solutions 1 2008 Quate-Final Exam Solutions 1 A chaged paticle with chage q and mass m stats with an initial kinetic enegy K at the middle of a unifomly chaged spheical egion of

### Vector Calculus: Are you ready? Vectors in 2D and 3D Space: Review

Vecto Calculus: Ae you eady? Vectos in D and 3D Space: Review Pupose: Make cetain that you can define, and use in context, vecto tems, concepts and fomulas listed below: Section 7.-7. find the vecto defined

### Chapter 4. Gauss s Law

Chapte 4 Gauss s Law 4.1 lectic Flux...4-4. Gauss s Law...4-3 xample 4.1: Infinitely Long Rod of Unifom Chage Density...4-8 xample 4.: Infinite Plane of Chage...4-9 xample 4.3: Spheical Shell...4-1 xample

### F = kq 1q 2 r 2. F 13 = k( q)(2q) 2a 2 cosθˆx + sinθŷ F 14 = k( 2q)(2q) F 12 = k(q)(2q) a 2. tanθ = a a

.1 What ae the hoizontal and vetical components of the esultant electostatic foce on the chage in the lowe left cone of the squae if q =1. 1 7 and a =5.cm? +q -q a +q a -q F = kq 1q F 1 = k(q)(q) a F 13

### FXA 2008. Candidates should be able to : Describe how a mass creates a gravitational field in the space around it.

Candidates should be able to : Descibe how a mass ceates a gavitational field in the space aound it. Define gavitational field stength as foce pe unit mass. Define and use the peiod of an object descibing

### Gravitation and Kepler s Laws Newton s Law of Universal Gravitation in vectorial. Gm 1 m 2. r 2

F Gm Gavitation and Keple s Laws Newton s Law of Univesal Gavitation in vectoial fom: F 12 21 Gm 1 m 2 12 2 ˆ 12 whee the hat (ˆ) denotes a unit vecto as usual. Gavity obeys the supeposition pinciple,

### Chapter 22 The Electric Field II: Continuous Charge Distributions

Chapte The lectic Field II: Continuous Chage Distibutions 1 [M] A unifom line chage that has a linea chage density l equal to.5 nc/m is on the x axis between x and x 5. m. (a) What is its total chage?

### Gravitation. AP Physics C

Gavitation AP Physics C Newton s Law of Gavitation What causes YOU to be pulled down? THE EARTH.o moe specifically the EARTH S MASS. Anything that has MASS has a gavitational pull towads it. F α Mm g What

### Episode 401: Newton s law of universal gravitation

Episode 401: Newton s law of univesal gavitation This episode intoduces Newton s law of univesal gavitation fo point masses, and fo spheical masses, and gets students pactising calculations of the foce

### The force between electric charges. Comparing gravity and the interaction between charges. Coulomb s Law. Forces between two charges

The foce between electic chages Coulomb s Law Two chaged objects, of chage q and Q, sepaated by a distance, exet a foce on one anothe. The magnitude of this foce is given by: kqq Coulomb s Law: F whee

### Forces & Magnetic Dipoles. r r τ = μ B r

Foces & Magnetic Dipoles x θ F θ F. = AI τ = U = Fist electic moto invented by Faaday, 1821 Wie with cuent flow (in cup of Hg) otates aound a a magnet Faaday s moto Wie with cuent otates aound a Pemanent

### Lesson 7 Gauss s Law and Electric Fields

Lesson 7 Gauss s Law and Electic Fields Lawence B. Rees 7. You may make a single copy of this document fo pesonal use without witten pemission. 7. Intoduction While it is impotant to gain a solid conceptual

### Samples of conceptual and analytical/numerical questions from chap 21, C&J, 7E

CHAPTER 1 Magnetism CONCEPTUAL QUESTIONS Cutnell & Johnson 7E 3. ssm A chaged paticle, passing though a cetain egion of space, has a velocity whose magnitude and diection emain constant, (a) If it is known

### PHYSICS 111 HOMEWORK SOLUTION #13. May 1, 2013

PHYSICS 111 HOMEWORK SOLUTION #13 May 1, 2013 0.1 In intoductoy physics laboatoies, a typical Cavendish balance fo measuing the gavitational constant G uses lead sphees with masses of 2.10 kg and 21.0

### Fluids Lecture 15 Notes

Fluids Lectue 15 Notes 1. Unifom flow, Souces, Sinks, Doublets Reading: Andeson 3.9 3.12 Unifom Flow Definition A unifom flow consists of a velocit field whee V = uî + vĵ is a constant. In 2-D, this velocit

### Notes on Electric Fields of Continuous Charge Distributions

Notes on Electic Fields of Continuous Chage Distibutions Fo discete point-like electic chages, the net electic field is a vecto sum of the fields due to individual chages. Fo a continuous chage distibution

### CHAPTER 5 GRAVITATIONAL FIELD AND POTENTIAL

CHATER 5 GRAVITATIONAL FIELD AND OTENTIAL 5. Intoduction. This chapte deals with the calculation of gavitational fields and potentials in the vicinity of vaious shapes and sizes of massive bodies. The

### 14. Gravitation Universal Law of Gravitation (Newton):

14. Gavitation 1 Univesal Law of Gavitation (ewton): The attactive foce between two paticles: F = G m 1m 2 2 whee G = 6.67 10 11 m 2 / kg 2 is the univesal gavitational constant. F m 2 m 1 F Paticle #1

### PY1052 Problem Set 3 Autumn 2004 Solutions

PY1052 Poblem Set 3 Autumn 2004 Solutions C F = 8 N F = 25 N 1 2 A A (1) A foce F 1 = 8 N is exeted hoizontally on block A, which has a mass of 4.5 kg. The coefficient of static fiction between A and the

### So we ll start with Angular Measure. Consider a particle moving in a circular path. (p. 220, Figure 7.1)

Lectue 17 Cicula Motion (Chapte 7) Angula Measue Angula Speed and Velocity Angula Acceleation We ve aleady dealt with cicula motion somewhat. Recall we leaned about centipetal acceleation: when you swing

### In the lecture on double integrals over non-rectangular domains we used to demonstrate the basic idea

Double Integals in Pola Coodinates In the lectue on double integals ove non-ectangula domains we used to demonstate the basic idea with gaphics and animations the following: Howeve this paticula example

### Exam 3: Equation Summary

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Depatment of Physics Physics 8.1 TEAL Fall Tem 4 Momentum: p = mv, F t = p, Fext ave t= t f t= Exam 3: Equation Summay total = Impulse: I F( t ) = p Toque: τ = S S,P

### Introduction to Electric Potential

Univesiti Teknologi MARA Fakulti Sains Gunaan Intoduction to Electic Potential : A Physical Science Activity Name: HP: Lab # 3: The goal of today s activity is fo you to exploe and descibe the electic

### Learning Objectives. Decreasing size. ~10 3 m. ~10 6 m. ~10 10 m 1/22/2013. Describe ionic, covalent, and metallic, hydrogen, and van der Waals bonds.

Lectue #0 Chapte Atomic Bonding Leaning Objectives Descibe ionic, covalent, and metallic, hydogen, and van de Waals bonds. Which mateials exhibit each of these bonding types? What is coulombic foce of

### 8-1 Newton s Law of Universal Gravitation

8-1 Newton s Law of Univesal Gavitation One of the most famous stoies of all time is the stoy of Isaac Newton sitting unde an apple tee and being hit on the head by a falling apple. It was this event,

### Chapter F. Magnetism. Blinn College - Physics Terry Honan

Chapte F Magnetism Blinn College - Physics 46 - Tey Honan F. - Magnetic Dipoles and Magnetic Fields Electomagnetic Duality Thee ae two types of "magnetic chage" o poles, Noth poles N and South poles S.

### Mechanics 1: Motion in a Central Force Field

Mechanics : Motion in a Cental Foce Field We now stud the popeties of a paticle of (constant) ass oving in a paticula tpe of foce field, a cental foce field. Cental foces ae ve ipotant in phsics and engineeing.

### Deflection of Electrons by Electric and Magnetic Fields

Physics 233 Expeiment 42 Deflection of Electons by Electic and Magnetic Fields Refeences Loain, P. and D.R. Coson, Electomagnetism, Pinciples and Applications, 2nd ed., W.H. Feeman, 199. Intoduction An

### XIIth PHYSICS (C2, G2, C, G) Solution

XIIth PHYSICS (C, G, C, G) -6- Solution. A 5 W, 0 V bulb and a 00 W, 0 V bulb ae connected in paallel acoss a 0 V line nly 00 watt bulb will fuse nly 5 watt bulb will fuse Both bulbs will fuse None of

### PY1052 Problem Set 8 Autumn 2004 Solutions

PY052 Poblem Set 8 Autumn 2004 Solutions H h () A solid ball stats fom est at the uppe end of the tack shown and olls without slipping until it olls off the ight-hand end. If H 6.0 m and h 2.0 m, what

### Exam I. Spring 2004 Serway & Jewett, Chapters 1-5. Fill in the bubble for the correct answer on the answer sheet. next to the number.

Agin/Meye PART I: QUALITATIVE Exam I Sping 2004 Seway & Jewett, Chaptes 1-5 Assigned Seat Numbe Fill in the bubble fo the coect answe on the answe sheet. next to the numbe. NO PARTIAL CREDIT: SUBMIT ONE

### Poynting Vector and Energy Flow in a Capacitor Challenge Problem Solutions

Poynting Vecto an Enegy Flow in a Capacito Challenge Poblem Solutions Poblem 1: A paallel-plate capacito consists of two cicula plates, each with aius R, sepaate by a istance. A steay cuent I is flowing

### General Physics (PHY 2130)

Geneal Physics (PHY 130) Lectue 11 Rotational kinematics and unifom cicula motion Angula displacement Angula speed and acceleation http://www.physics.wayne.edu/~apetov/phy130/ Lightning Review Last lectue:

### Gauss s law relates to total electric flux through a closed surface to the total enclosed charge.

Chapte : Gauss s Law Gauss s Law is an altenative fomulation of the elation between an electic field and the souces of that field in tems of electic flu. lectic Flu Φ though an aea ~ Numbe of Field Lines

### Chapter 2 Coulomb s Law

Chapte Coulomb s Law.1 lectic Chage...-3. Coulomb's Law...-3 Animation.1: Van de Gaaff Geneato...-4.3 Pinciple of Supeposition...-5 xample.1: Thee Chages...-5.4 lectic Field...-7 Animation.: lectic Field

### 4.1 Cylindrical and Polar Coordinates

4.1 Cylindical and Pola Coodinates 4.1.1 Geometical Axisymmety A lage numbe of pactical engineeing poblems involve geometical featues which have a natual axis of symmety, such as the solid cylinde, shown

### Today in Physics 217: multipole expansion

Today in Physics 17: multipole expansion Multipole expansions Electic multipoles and thei moments Monopole and dipole, in detail Quadupole, octupole, Example use of multipole expansion as appoximate solution

### Chapter 17 The Kepler Problem: Planetary Mechanics and the Bohr Atom

Chapte 7 The Keple Poblem: Planetay Mechanics and the Boh Atom Keple s Laws: Each planet moves in an ellipse with the sun at one focus. The adius vecto fom the sun to a planet sweeps out equal aeas in

### Problem Set 6: Solutions

UNIVESITY OF ALABAMA Depatment of Physics and Astonomy PH 16-4 / LeClai Fall 28 Poblem Set 6: Solutions 1. Seway 29.55 Potons having a kinetic enegy of 5. MeV ae moving in the positive x diection and ente

### Magnetic Field and Magnetic Forces. Young and Freedman Chapter 27

Magnetic Field and Magnetic Foces Young and Feedman Chapte 27 Intoduction Reiew - electic fields 1) A chage (o collection of chages) poduces an electic field in the space aound it. 2) The electic field

### 1240 ev nm 2.5 ev. (4) r 2 or mv 2 = ke2

Chapte 5 Example The helium atom has 2 electonic enegy levels: E 3p = 23.1 ev and E 2s = 20.6 ev whee the gound state is E = 0. If an electon makes a tansition fom 3p to 2s, what is the wavelength of the

### A) 2 B) 2 C) 2 2 D) 4 E) 8

Page 1 of 8 CTGavity-1. m M Two spheical masses m and M ae a distance apat. The distance between thei centes is halved (deceased by a facto of 2). What happens to the magnitude of the foce of gavity between

### LINES AND TANGENTS IN POLAR COORDINATES

LINES AND TANGENTS IN POLAR COORDINATES ROGER ALEXANDER DEPARTMENT OF MATHEMATICS 1. Pola-coodinate equations fo lines A pola coodinate system in the plane is detemined by a point P, called the pole, and

### Charges, Coulomb s Law, and Electric Fields

Q&E -1 Chages, Coulomb s Law, and Electic ields Some expeimental facts: Expeimental fact 1: Electic chage comes in two types, which we call (+) and ( ). An atom consists of a heavy (+) chaged nucleus suounded

### Hour Exam No.1. p 1 v. p = e 0 + v^b. Note that the probe is moving in the direction of the unit vector ^b so the velocity vector is just ~v = v^b and

Hou Exam No. Please attempt all of the following poblems befoe the due date. All poblems count the same even though some ae moe complex than othes. Assume that c units ae used thoughout. Poblem A photon

### The Role of Gravity in Orbital Motion

! The Role of Gavity in Obital Motion Pat of: Inquiy Science with Datmouth Developed by: Chistophe Caoll, Depatment of Physics & Astonomy, Datmouth College Adapted fom: How Gavity Affects Obits (Ohio State

### Physics 111 Fall 2007 Electrostatic Forces and the Electric Field - Solutions

Physics 111 Fall 007 Electostatic Foces an the Electic Fiel - Solutions 1. Two point chages, 5 µc an -8 µc ae 1. m apat. Whee shoul a thi chage, equal to 5 µc, be place to make the electic fiel at the

### (a) The centripetal acceleration of a point on the equator of the Earth is given by v2. The velocity of the earth can be found by taking the ratio of

Homewok VI Ch. 7 - Poblems 15, 19, 22, 25, 35, 43, 51. Poblem 15 (a) The centipetal acceleation of a point on the equato of the Eath is given by v2. The velocity of the eath can be found by taking the

### Review Module: Dot Product

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Depatment of Physics 801 Fall 2009 Review Module: Dot Poduct We shall intoduce a vecto opeation, called the dot poduct o scala poduct that takes any two vectos and

### VISCOSITY OF BIO-DIESEL FUELS

VISCOSITY OF BIO-DIESEL FUELS One of the key assumptions fo ideal gases is that the motion of a given paticle is independent of any othe paticles in the system. With this assumption in place, one can use

### Solution Derivations for Capa #8

Solution Deivations fo Capa #8 1) A ass spectoete applies a voltage of 2.00 kv to acceleate a singly chaged ion (+e). A 0.400 T field then bends the ion into a cicula path of adius 0.305. What is the ass

### AP Physics Electromagnetic Wrap Up

AP Physics Electomagnetic Wap Up Hee ae the gloious equations fo this wondeful section. F qsin This is the equation fo the magnetic foce acting on a moing chaged paticle in a magnetic field. The angle

### Magnetism: a new force!

-1 Magnetism: a new foce! o fa, we'e leaned about two foces: gaity and the electic field foce. F E = E, FE = E Definition of E-field kq E-fields ae ceated by chages: E = 2 E-field exets a foce on othe

### A r. (Can you see that this just gives the formula we had above?)

24-1 (SJP, Phys 1120) lectic flux, and Gauss' law Finding the lectic field due to a bunch of chages is KY! Once you know, you know the foce on any chage you put down - you can pedict (o contol) motion

Chapte 5. Foce and Motion In this chapte we study causes of motion: Why does the windsufe blast acoss the wate in the way he does? The combined foces of the wind, wate, and gavity acceleate him accoding

### In this section we shall look at the motion of a projectile MOTION IN FIELDS 9.1 PROJECTILE MOTION PROJECTILE MOTION

MOTION IN FIELDS MOTION IN FIELDS 9 9. Pojectile motion 9. Gavitational field, potential and enegy 9.3 Electic field, potential and enegy 9. PROJECTILE MOTION 9.. State the independence of the vetical

### PHYSICS 111 HOMEWORK SOLUTION #5. March 3, 2013

PHYSICS 111 HOMEWORK SOLUTION #5 Mach 3, 2013 0.1 You 3.80-kg physics book is placed next to you on the hoizontal seat of you ca. The coefficient of static fiction between the book and the seat is 0.650,

### UNIT 21: ELECTRICAL AND GRAVITATIONAL POTENTIAL Approximate time two 100-minute sessions

Name St.No. - Date(YY/MM/DD) / / Section Goup# UNIT 21: ELECTRICAL AND GRAVITATIONAL POTENTIAL Appoximate time two 100-minute sessions OBJECTIVES I began to think of gavity extending to the ob of the moon,

### Lab #7: Energy Conservation

Lab #7: Enegy Consevation Photo by Kallin http://www.bungeezone.com/pics/kallin.shtml Reading Assignment: Chapte 7 Sections 1,, 3, 5, 6 Chapte 8 Sections 1-4 Intoduction: Pehaps one of the most unusual

### Solutions to Homework Set #5 Phys2414 Fall 2005

Solution Set #5 1 Solutions to Homewok Set #5 Phys414 Fall 005 Note: The numbes in the boxes coespond to those that ae geneated by WebAssign. The numbes on you individual assignment will vay. Any calculated

### Carter-Penrose diagrams and black holes

Cate-Penose diagams and black holes Ewa Felinska The basic intoduction to the method of building Penose diagams has been pesented, stating with obtaining a Penose diagam fom Minkowski space. An example

### rotation -- Conservation of mechanical energy for rotation -- Angular momentum -- Conservation of angular momentum

Final Exam Duing class (1-3:55 pm) on 6/7, Mon Room: 41 FMH (classoom) Bing scientific calculatos No smat phone calculatos l ae allowed. Exam coves eveything leaned in this couse. Review session: Thusday

### Multiple choice questions [60 points]

1 Multiple choice questions [60 points] Answe all o the ollowing questions. Read each question caeully. Fill the coect bubble on you scanton sheet. Each question has exactly one coect answe. All questions

### L19 Geomagnetic Field Part I

Intoduction to Geophysics L19-1 L19 Geomagnetic Field Pat I 1. Intoduction We now stat the last majo topic o this class which is magnetic ields and measuing the magnetic popeties o mateials. As a way o

### Magnetic Field in a Time-Dependent Capacitor

Magnetic Field in a Time-Dependent Capacito 1 Poblem Kik T. McDonald Joseph Heny Laboatoies, Pinceton Univesity, Pinceton, NJ 8544 (Octobe 3, 23) Reconside the classic example of the use of Maxwell s displacement

### Gravity and the figure of the Earth

Gavity and the figue of the Eath Eic Calais Pudue Univesity Depatment of Eath and Atmospheic Sciences West Lafayette, IN 47907-1397 ecalais@pudue.edu http://www.eas.pudue.edu/~calais/ Objectives What is

### Introduction to Fluid Mechanics

Chapte 1 1 1.6. Solved Examples Example 1.1 Dimensions and Units A body weighs 1 Ibf when exposed to a standad eath gavity g = 3.174 ft/s. (a) What is its mass in kg? (b) What will the weight of this body

### Physics 107 HOMEWORK ASSIGNMENT #14

Physics 107 HOMEWORK ASSIGNMENT #14 Cutnell & Johnson, 7 th edition Chapte 17: Poblem 44, 60 Chapte 18: Poblems 14, 18, 8 **44 A tube, open at only one end, is cut into two shote (nonequal) lengths. The

### Chapter 30: Magnetic Fields Due to Currents

d Chapte 3: Magnetic Field Due to Cuent A moving electic chage ceate a magnetic field. One of the moe pactical way of geneating a lage magnetic field (.1-1 T) i to ue a lage cuent flowing though a wie.

### Mechanics 1: Work, Power and Kinetic Energy

Mechanics 1: Wok, Powe and Kinetic Eneg We fist intoduce the ideas of wok and powe. The notion of wok can be viewed as the bidge between Newton s second law, and eneg (which we have et to define and discuss).

### Coordinate Systems L. M. Kalnins, March 2009

Coodinate Sstems L. M. Kalnins, Mach 2009 Pupose of a Coodinate Sstem The pupose of a coodinate sstem is to uniquel detemine the position of an object o data point in space. B space we ma liteall mean

### Physics 505 Homework No. 5 Solutions S5-1. 1. Angular momentum uncertainty relations. A system is in the lm eigenstate of L 2, L z.

Physics 55 Homewok No. 5 s S5-. Angula momentum uncetainty elations. A system is in the lm eigenstate of L 2, L z. a Show that the expectation values of L ± = L x ± il y, L x, and L y all vanish. ψ lm

### Theory and measurement

Gavity: Theoy and measuement Reading: Today: p11 - Theoy of gavity Use two of Newton s laws: 1) Univesal law of gavitation: ) Second law of motion: Gm1m F = F = mg We can combine them to obtain the gavitational

### Experiment 6: Centripetal Force

Name Section Date Intoduction Expeiment 6: Centipetal oce This expeiment is concened with the foce necessay to keep an object moving in a constant cicula path. Accoding to Newton s fist law of motion thee

### ELECTRIC CHARGES AND FIELDS

Chapte One ELECTRIC CHARGES AND FIELDS 1.1 INTRODUCTION All of us have the expeience of seeing a spak o heaing a cackle when we take off ou synthetic clothes o sweate, paticulaly in dy weathe. This is

### Lab M4: The Torsional Pendulum and Moment of Inertia

M4.1 Lab M4: The Tosional Pendulum and Moment of netia ntoduction A tosional pendulum, o tosional oscillato, consists of a disk-like mass suspended fom a thin od o wie. When the mass is twisted about the

### GAUSS S LAW APPLIED TO CYLINDRICAL AND PLANAR CHARGE DISTRIBUTIONS ` E MISN-0-133. CHARGE DISTRIBUTIONS by Peter Signell, Michigan State University

MISN-0-133 GAUSS S LAW APPLIED TO CYLINDRICAL AND PLANAR CHARGE DISTRIBUTIONS GAUSS S LAW APPLIED TO CYLINDRICAL AND PLANAR CHARGE DISTRIBUTIONS by Pete Signell, Michigan State Univesity 1. Intoduction..............................................

### Chapter 3 Savings, Present Value and Ricardian Equivalence

Chapte 3 Savings, Pesent Value and Ricadian Equivalence Chapte Oveview In the pevious chapte we studied the decision of households to supply hous to the labo maket. This decision was a static decision,

### Chapter 19: Electric Charges, Forces, and Fields ( ) ( 6 )( 6

Chapte 9 lectic Chages, Foces, an Fiels 6 9. One in a million (0 ) ogen molecules in a containe has lost an electon. We assume that the lost electons have been emove fom the gas altogethe. Fin the numbe

### Lesson 8 Ampère s Law and Differential Operators

Lesson 8 Ampèe s Law and Diffeential Opeatos Lawence Rees 7 You ma make a single cop of this document fo pesonal use without witten pemission 8 Intoduction Thee ae significant diffeences between the electic