# Chapter 4. Gauss s Law

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 Chapte 4 Gauss s Law 4.1 lectic Flux Gauss s Law xample 4.1: Infinitely Long Rod of Unifom Chage Density xample 4.: Infinite Plane of Chage xample 4.3: Spheical Shell xample 4.4: Non-Conducting Solid Sphee Conductos xample 4.5: Conducto with Chage Inside a Cavity xample 4.6: lectic Potential Due to a Spheical Shell Foce on a Conducto Summay Appendix: Tensions and Pessues Animation 4.1: Chaged Paticle Moving in a Constant lectic Field Animation 4.: Chaged Paticle at Rest in a Time-Vaying Field Animation 4.3: Like and Unlike Chages Hanging fom Pendulums Poblem-Solving Stategies Solved Poblems Two Paallel Infinite Non-Conducting Planes lectic Flux Though a Squae Suface Gauss s Law fo Gavity lectic Potential of a Unifomly Chaged Sphee Conceptual Questions Additional Poblems Non-Conducting Solid Sphee with a Cavity P-N Junction Sphee with Non-Unifom Chage Distibution Thin Slab lectic Potential negy of a Solid Sphee Calculating lectic Field fom lectical Potential

2 Gauss s Law 4.1 lectic Flux In Chapte we showed that the stength of an electic field is popotional to the numbe of field lines pe aea. The numbe of electic field lines that penetates a given suface is called an electic flux, which we denote as Φ. The electic field can theefoe be thought of as the numbe of lines pe unit aea. Figue lectic field lines passing though a suface of aea A. Conside the suface shown in Figue Let A= Anˆ be defined as the aea vecto having a magnitude of the aea of the suface, A, and u pointing in the nomal diection, ˆn. If the suface is placed in a unifom electic field that points in the same diection as ˆn, i.e., pependicula to the suface A, the flux though the suface is Φ ˆ = A= na= A (4.1.1) On the othe hand, if the electic field u makes an angle θ with ˆn (Figue 4.1.), the electic flux becomes whee Φ = A = = (4.1.) Acosθ n A = n ˆ is the component of pependicula to the suface. n Figue 4.1. lectic field lines passing though a suface of aea A whose nomal makes an angle θ with the field. 4-

3 Note that with the definition fo the nomal vecto ˆn, the electic flux Φ is positive if the electic field lines ae leaving the suface, and negative if enteing the suface. In geneal, a suface S can be cuved and the electic field u may vay ove the suface. We shall be inteested in the case whee the suface is closed. A closed suface is a suface which completely encloses a volume. In ode to compute the electic flux, we divide the suface into a lage numbe of infinitesimal aea elements A ˆ i = Aini, as shown in Figue Note that fo a closed suface the unit vecto nˆ i is chosen to point in the outwad nomal diection. Figue lectic field passing though an aea element the nomal of the suface. A i, making an angle θ with The electic flux though A i is Φ = A = A cosθ (4.1.3) i i i i The total flux though the entie suface can be obtained by summing ove all the aea elements. Taking the limit A and the numbe of elements to infinity, we have whee the symbol S i Φ = lim da = d Ò A (4.1.4) i A i i S denotes a double integal ove a closed suface S. In ode to evaluate the above integal, we must fist specify the suface and then sum ove the dot u poduct da. 4. Gauss s Law Conside a positive point chage Q located at the cente of a sphee of adius, as shown u in Figue The electic field due to the chage Q is = ( Q/4 πε ˆ ), which points in the adial diection. We enclose the chage by an imaginay sphee of adius called the Gaussian suface. 4-3

4 Figue 4..1 A spheical Gaussian suface enclosing a chage Q. In spheical coodinates, a small suface aea element on the sphee is given by (Figue 4..) da= sin θ dθ dφ ˆ (4..1) Figue 4.. A small aea element on the suface of a sphee of adius. Thus, the net electic flux though the aea element is 1 Q Q dφ = da = da= ( sin θ dθ dφ) = sin θ dθ dφ 4πε 4πε (4..) The total flux though the entie suface is Q π π Q d sin θ dθ dφ = (4..3) 4πε S ε Φ = A = The same esult can also be obtained by noting that a sphee of adius has a suface aea A= 4π, and since the magnitude of the electic field at any point on the spheical suface is = Q/4πε, the electic flux though the suface is 1 Q Q A (4..4) Φ = d = da= A= 4π = 4πε S S ε 4-4

5 In the above, we have chosen a sphee to be the Gaussian suface. Howeve, it tuns out that the shape of the closed suface can be abitaily chosen. Fo the sufaces shown in Figue 4..3, the same esult ( Φ = Q / ε ) is obtained. whethe the choice is S1, S o. S 3 Figue 4..3 Diffeent Gaussian sufaces with the same outwad electic flux. The statement that the net flux though any closed suface is popotional to the net chage enclosed is known as Gauss s law. Mathematically, Gauss s law is expessed as u q Φ = = enc Ò d A (Gauss s law) (4..5) ε S whee q enc is the net chage inside the suface. One way to explain why Gauss s law holds is due to note that the numbe of field lines that leave the chage is independent of the shape of the imaginay Gaussian suface we choose to enclose the chage. To pove Gauss s law, we intoduce the concept of the solid angle. Let A = A 1 1ˆ an aea element on the suface of a sphee of adius, as shown in Figue S1 1 be Figue 4..4 The aea element A subtends a solid angle Ω. The solid angle Ω subtended by A = A 1 1ˆ at the cente of the sphee is defined as A Ω (4..6)

6 Solid angles ae dimensionless quantities measued in steadians (s). Since the suface aea of the sphee is 4π 1, the total solid angle subtended by the sphee is S 1 4π Ω= = 4π (4..7) 1 1 The concept of solid angle in thee dimensions is analogous to the odinay angle in two dimensions. As illustated in Figue 4..5, an angle ϕ is the atio of the length of the ac to the adius of a cicle: s ϕ = (4..8) Figue 4..5 The ac s subtends an angle ϕ. Since the total length of the ac is s = π, the total angle subtended by the cicle is π ϕ = = π (4..9) In Figue 4..4, the aea element A makes an angle θ with the adial unit vecto ˆ, then the solid angle subtended by A is A ˆ A cosθ An Ω = = = (4..1) whee A = A cosθ is the aea of the adial pojection of n A onto a second sphee of adius, concentic with S. S 1 As shown in Figue 4..4, the solid angle subtended is the same fo both and A : A Acosθ Ω = = (4..11) 1 1 A 1 n 4-6

7 Now suppose a point chage Q is placed at the cente of the concentic sphees. The electic field stengths and 1 at the cente of the aea elements A1 and A ae elated by Coulomb s law: i 1 Q = 4πε 1 = i 1 (4..1) The electic flux though A 1 on S1 is Φ = A = A (4..13) On the othe hand, the electic flux though A on S is 1 Φ = A = Acosθ = 1 A 1 = 1 A1 = Φ (4..14) 1 1 Thus, we see that the electic flux though any aea element subtending the same solid angle is constant, independent of the shape o oientation of the suface. In summay, Gauss s law povides a convenient tool fo evaluating electic field. Howeve, its application is limited only to systems that possess cetain symmety, namely, systems with cylindical, plana and spheical symmety. In the table below, we give some examples of systems in which Gauss s law is applicable fo detemining electic field, with the coesponding Gaussian sufaces: Symmety System Gaussian Suface xamples Cylindical Infinite od Coaxial Cylinde xample 4.1 Plana Infinite plane Gaussian Pillbox xample 4. Spheical Sphee, Spheical shell Concentic Sphee xamples 4.3 & 4.4 The following steps may be useful when applying Gauss s law: (1) Identify the symmety associated with the chage distibution. () Detemine the diection of the electic field, and a Gaussian suface on which the magnitude of the electic field is constant ove potions of the suface. (3) Divide the space into diffeent egions associated with the chage distibution. Fo each egion, calculate, the chage enclosed by the Gaussian suface. q enc 4-7

8 (4) Calculate the electic flux Φ though the Gaussian suface fo each egion. Φ enc / (5) quate with q ε, and deduce the magnitude of the electic field. xample 4.1: Infinitely Long Rod of Unifom Chage Density An infinitely long od of negligible adius has a unifom chage density λ. Calculate the electic field at a distance fom the wie. Solution: We shall solve the poblem by following the steps outlined above. (1) An infinitely long od possesses cylindical symmety. () The chage density is unifomly distibuted thoughout the length, and the electic field must be point adially away fom the symmety axis of the od (Figue 4..6). The magnitude of the electic field is constant on cylindical sufaces of adius. Theefoe, we choose a coaxial cylinde as ou Gaussian suface. Figue 4..6 Field lines fo an infinite unifomly chaged od (the symmety axis of the od and the Gaussian cylinde ae pependicula to plane of the page.) (3) The amount of chage enclosed by the Gaussian suface, a cylinde of adius and length l (Figue 4..7), is qenc = λl. Figue 4..7 Gaussian suface fo a unifomly chaged od. 4-8

9 (4) As indicated in Figue 4..7, the Gaussian suface consists of thee pats: a two ends and plus the cuved side wall S. The flux though the Gaussian suface is S1 S 3 whee we have set Φ = da= da + da + da 3 Ò S S1 S S3 3 3 ( π l) = + + A= 3 (4..15) =. As can be seen fom the figue, no flux passes though the ends since the aea vectos da 1 and da ae pependicula to the electic field which points in the adial diection. (5) Applying Gauss s law gives ( π ) = λ / ε l l, o λ = (4..16) πε The esult is in complete ageement with that obtained in q. (.1.11) using Coulomb s law. Notice that the esult is independent of the length l of the cylinde, and only depends on the invese of the distance fom the symmety axis. The qualitative behavio of as a function of is plotted in Figue Figue 4..8 lectic field due to a unifomly chaged od as a function of xample 4.: Infinite Plane of Chage Conside an infinitely lage non-conducting plane in the xy-plane with unifom suface chage density σ. Detemine the electic field eveywhee in space. Solution: (1) An infinitely lage plane possesses a plana symmety. 4-9

10 () Since the chage is unifomly distibuted on the suface, the electic field must point pependiculaly away fom the plane, = kˆ. The magnitude of the electic field is constant on planes paallel to the non-conducting plane. Figue 4..9 lectic field fo unifom plane of chage We choose ou Gaussian suface to be a cylinde, which is often efeed to as a pillbox (Figue 4..1). The pillbox also consists of thee pats: two end-caps and S, and a cuved side. S 3 S1 Figue 4..1 A Gaussian pillbox fo calculating the electic field due to a lage plane. (3) Since the suface chage distibution on is unifom, the chage enclosed by the Gaussian pillbox is qenc = σ A, whee A= A1 = A is the aea of the end-caps. (4) The total flux though the Gaussian pillbox flux is Φ = da= da + da + da 3 Ò S S1 S S3 = A+ A = ( + ) A 1 (4..17) 4-1

11 Since the two ends ae at the same distance fom the plane, by symmety, the magnitude of the electic field must be the same: 1 = =. Hence, the total flux can be ewitten as (5) By applying Gauss s law, we obtain A Φ = A (4..18) q ε enc = = σ A ε which gives σ = (4..19) ε In unit-vecto notation, we have σ ˆ k, z > ε = σ kˆ, z < ε (4..) Thus, we see that the electic field due to an infinite lage non-conducting plane is unifom in space. The esult, plotted in Figue 4..11, is the same as that obtained in q. (.1.1) using Coulomb s law. Figue lectic field of an infinitely lage non-conducting plane. Note again the discontinuity in electic field as we coss the plane: σ σ σ z = z+ z = = ε ε ε (4..1) 4-11

12 xample 4.3: Spheical Shell A thin spheical shell of adius a has a chage + Q evenly distibuted ove its suface. Find the electic field both inside and outside the shell. Solutions: The chage distibution is spheically symmetic, with a suface chage density σ = Q/ A = Q/4π a, whee A = 4π a is the suface aea of the sphee. The electic field s s must be adially symmetic and diected outwad (Figue 4..1). We teat the egions a and asepaately. Figue 4..1 lectic field fo unifom spheical shell of chage Case 1: a We choose ou Gaussian suface to be a sphee of adius a, as shown in Figue 4..13(a). (a) (b) Figue Gaussian suface fo unifomly chaged spheical shell fo (a) < a, and (b) a The chage enclosed by the Gaussian suface is q enc = since all the chage is located on the suface of the shell. Thus, fom Gauss s law, Φ = qenc / ε, we conclude Case : a =, < a (4..) 4-1

13 In this case, the Gaussian suface is a sphee of adius a, as shown in Figue 4..13(b). Since the adius of the Gaussian sphee is geate than the adius of the spheical shell, all the chage is enclosed: qenc = Q Since the flux though the Gaussian suface is by applying Gauss s law, we obtain Φ = d A = A= Ò S e (4 π ) Q Q = = k, a (4..3) 4πε Note that the field outside the sphee is the same as if all the chages wee concentated at the cente of the sphee. The qualitative behavio of as a function of is plotted in Figue Figue lectic field as a function of due to a unifomly chaged spheical shell. As in the case of a non-conducting chaged plane, we again see a discontinuity in as we coss the bounday at = a. The change, fom oute to the inne suface, is given by Q σ = + = = 4πε a ε xample 4.4: Non-Conducting Solid Sphee An electic chage + Q is unifomly distibuted thoughout a non-conducting solid sphee of adius a. Detemine the electic field eveywhee inside and outside the sphee. Solution: 4-13

14 The chage distibution is spheically symmetic with the chage density given by Q Q ρ = = (4..4) 3 V (4/3) π a whee V is the volume of the sphee. In this case, the electic field is adially symmetic and diected outwad. The magnitude of the electic field is constant on spheical sufaces of adius. The egions a and ashall be studied sepaately. Case 1: a. We choose ou Gaussian suface to be a sphee of adius a, as shown in Figue 4..15(a). (a) (b) Figue Gaussian suface fo unifomly chaged solid sphee, fo (a) a, and (b) > a. The flux though the Gaussian suface is Φ = d A = A= Ò With unifom chage distibution, the chage enclosed is S (4 π ) qenc = ρdv = ρv = ρ π = Q 3 (4..5) 3 V a which is popotional to the volume enclosed by the Gaussian suface. Applying Gauss s Φ = q / ε, we obtain law enc ρ 4 ( 4π ) = ε 3 π

15 o Q = ρ, a 3 3ε = 4πε a (4..6) Case : a. In this case, ou Gaussian suface is a sphee of adius a, as shown in Figue 4..15(b). Since the adius of the Gaussian suface is geate than the adius of the sphee all the chage is enclosed in ou Gaussian suface: q = Q. With the electic flux though the Gaussian suface given by Φ = (4 π ), upon applying Gauss s law, we obtain (4 π ) = Q/ ε, o enc Q Q = = k, > a (4..7) 4πε e The field outside the sphee is the same as if all the chages wee concentated at the cente of the sphee. The qualitative behavio of as a function of is plotted in Figue Figue lectic field due to a unifomly chaged sphee as a function of. 4.3 Conductos An insulato such as glass o pape is a mateial in which electons ae attached to some paticula atoms and cannot move feely. On the othe hand, inside a conducto, electons ae fee to move aound. The basic popeties of a conducto ae the following: (1) The electic field is zeo inside a conducto. If we place a solid spheical conducto in a constant extenal field u, the positive and negative chages will move towad the pola egions of the sphee (the egions on the left and ight of the sphee in Figue below), theeby inducing an electic field. Inside the conducto, points in the opposite diection of u. Since chages ae mobile, they will continue to move until completely cancels u inside the conducto. At electostatic equilibium, must vanish inside a conducto. Outside the conducto, the 4-15

16 electic field due to the induced chage distibution coesponds to a dipole field, and u u the total electic field is simply = +. The field lines ae depicted in Figue Figue Placing a conducto in a unifom electic field u. () Any net chage must eside on the suface. u If thee wee a net chage inside the conducto, then by Gauss s law (q. 4.3.), would no longe be zeo thee. Theefoe, all the net excess chage must flow to the suface of the conducto. Figue 4.3. Gaussian suface inside a conducto. The enclosed chage is zeo. (3) The tangential component of is zeo on the suface of a conducto. We have aleady seen that fo an isolated conducto, the electic field is zeo in its inteio. Any excess chage placed on the conducto must then distibute itself on the suface, as implied by Gauss s law. Conside the line integal Ñ d s aound a closed path shown in Figue 4.3.3: Figue Nomal and tangential components of electic field outside the conducto 4-16

17 Since the electic field u vanishes: is consevative, the line integal aound the closed path abcda u Ñ ds = t( l) n( x') + ( l') + n( x) = abcda whee t and n ae the tangential and the nomal components of the electic field, espectively, and we have oiented the segment ab so that it is paallel to t. In the limit whee both x and x ', we have l =. Howeve, since the length element l is t finite, we conclude that the tangential component of the electic field on the suface of a conducto vanishes: = (on the suface of a conducto) (4.3.1) t 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 t =, the potential diffeence is because is pependicula to d V = V A B. B VB VA = u d s = u (4) is nomal to the suface just outside the conducto. A s. Thus, points A and B ae at the same potential with If the tangential component of u is initially non-zeo, chages will then move aound until it vanishes. Hence, only the nomal component suvives. Figue Gaussian pillbox fo computing the electic field outside the conducto. To compute the field stength just outside the conducto, conside the Gaussian pillbox dawn in Figue Using Gauss s law, we obtain 4-17

18 A Φ = Ò d A = na+ () A= σ (4.3.) ε S o n σ = (4.3.3) ε The above esult holds fo a conducto of abitay shape. The patten of the electic field line diections fo the egion nea a conducto is shown in Figue Figue Just outside the conducto, u is always pependicula to the suface. As in the examples of an infinitely lage non-conducting plane and a spheical shell, the nomal component of the electic field exhibits a discontinuity at the bounday: ( + ) ( ) σ σ n = n n = = ε ε xample 4.5: Conducto with Chage Inside a Cavity Conside a hollow conducto shown in Figue below. Suppose the net chage caied by the conducto is +Q. In addition, thee is a chage q inside the cavity. What is the chage on the oute suface of the conducto? Figue Conducto with a cavity 4-18

19 Since the electic field inside a conducto must be zeo, the net chage enclosed by the Gaussian suface shown in Figue must be zeo. This implies that a chage q must have been induced on the cavity suface. Since the conducto itself has a chage +Q, the amount of chage on the oute suface of the conducto must be Q+ q. xample 4.6: lectic 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 xample 4.3, we showed that the electic field fo a spheical shell of is given by Q 4πε ˆ, > a =, < a The electic potential may be calculated by using q. (3.1.9): Fo > a, we have B VB VA = u d s A Q 1 Q V() V( ) = d = = k 4πε 4πε e Q (4.3.4) 4-19

20 whee we have chosen V ( ) = potential becomes as ou efeence point. On the othe hand, fo < a, the a ( ) ( ) V() V( ) = d > a < a a Q 1 Q = d = = k 4πε 4πε a a e Q a (4.3.5) A plot of the electic potential is shown in Figue Note that the potential V is constant inside a conducto. Figue lectic 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/4πεa. 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 dwext = Vdq = dq 4πεa (4.3.6) Theefoe, the total amount of wok needed to chage the sphee to Q is W ext Q q Q = dq = (4.3.7) 4πε a 8πε a Since V = Q/4πεa and W ext = U, the above expession is simplified to U 1 = QV (4.3.8) 4-

21 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 would be W = QV. Theefoe, fo a point chage Q, the potential enegy is U=QV. ext Now, suppose two metal sphees with adii 1 and 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 V1 = V = V. Suppose the chages on the sphees at equilibium ae q1 and q. Neglecting the effect of the wie that connects the two sphees, the equipotential condition implies 1 q1 1 q V = = 4πε 4πε 1 o q 1 q = (4.3.9) 1 assuming 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 1 q σ 1 q σ = =, = = (4.3.1) ε πε 1 ε 4πε whee σ 1 and σ ae the suface chage densities on sphees 1 and, espectively. The two equations can be combined to yield σ = = (4.3.11) σ 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. 4-1

22 4.4 Foce on a Conducto We have seen that at the bounday suface of a conducto with a unifom chage density σ, the tangential component of the electic field is zeo, and hence, continuous, while the nomal component of the electic field exhibits discontinuity, with n = σ / ε. Conside a small patch of chage on a conducting suface, as shown in Figue Figue Foce on a conducto What is the foce expeienced by this patch? To answe this question, let s wite the total electic field anywhee outside the suface as = patch + (4.4.1) whee patch is the electic field due to chage on the patch, and is the electic field due to all othe chages. Since by Newton s thid law, the patch cannot exet a foce on itself, the foce on the patch must come solely fom. Assuming the patch to be a flat suface, fom Gauss s law, the electic field due to the patch is patch σ ˆ + k, z > = ε σ kˆ, z < ε (4.4.) By supeposition pinciple, the electic field above the conducting suface is σ ˆ above = k+ ε (4.4.3) Similaly, below the conducting suface, the electic field is σ ˆ below = k+ ε (4.4.4) 4-

23 Notice that is continuous acoss the bounday. This is due to the fact that if the patch wee emoved, the field in the emaining hole exhibits no discontinuity. Using the two equations above, we find 1 = ( + ) = In the case of a conducto, with above = ( σ / ε)ˆ k above below avg and below =, we have (4.4.5) 1 σ ˆ σ avg = k + = k ˆ (4.4.6) ε ε Thus, the foce acting on the patch is F= qavg = A σ k = σ ε ε ( ) ˆ A σ ˆk (4.4.7) whee A is the aea of the patch. This is pecisely the foce needed to dive the chages on the suface of a conducto to an equilibium state whee the electic field just outside the conducto takes on the value σ / ε and vanishes inside. Note that iespective of the sign of σ, the foce tends to pull the patch into the field. Using the esult obtained above, we may define the electostatic pessue on the patch as F σ 1 σ 1 P= = = ε = A ε ε ε (4.4.8) whee is the magnitude of the field just above the patch. The pessue is being tansmitted via the electic field. 4.5 Summay The electic flux that passes though a suface chaacteized by the aea vecto A = Anˆ is Φ = A = Acosθ whee θ is the angle between the electic field and the unit vecto ˆn. 4-3

24 In geneal, the electic flux though a suface is Φ = d A S Gauss s law states that the electic flux though any closed Gaussian suface is popotional to the total chage enclosed by the suface: q Φ = u d A Ò = ε S enc Gauss s law can be used to calculate the electic field fo a system that possesses plana, cylindical o spheical symmety. The nomal component of the electic field exhibits discontinuity, with = σ / ε, when cossing a bounday with suface chage density σ. n The basic popeties of a conducto ae (1) The electic field inside a conducto is zeo; () any net chage must eside on the suface of the conducto; (3) the suface of a conducto is an equipotential suface, and the tangential component of the electic field on the suface is zeo; and (4) just outside the conducto, the electic field is nomal to the suface. lectostatic pessue on a conducting suface is F σ 1 σ 1 P= = = ε = ε A ε ε 4.6 Appendix: Tensions and Pessues In Section 4.4, the pessue tansmitted by the electic field on a conducting suface was deived. We now conside a moe geneal case whee a closed suface (an imaginay box) is placed in an electic field, as shown in Figue If we look at the top face of the imaginay box, thee is an electic field pointing in the outwad nomal diection of that face. Fom Faaday s field theoy pespective, we would say that the field on that face tansmits a tension along itself acoss the face, theeby esulting in an upwad pull, just as if we had attached a sting unde tension to that face to pull it upwad. Similaly, if we look at the bottom face of the imaginay box, the field on that face is anti-paallel to the outwad nomal of the face, and accoding to Faaday s intepetation, we would again say that the field on the bottom face tansmits a tension 4-4

27 The kinetic enegy of the upwadly moving chage is deceasing as moe and moe enegy is stoed in the compessed electostatic field, and convesely when the chage is moving downwad. Moeove, because the field line motion in the animation is in the diection of the enegy flow, we can explicitly see the electomagnetic enegy flow away fom the chage into the suounding field when the chage is slowing. Convesely, we see the electomagnetic enegy flow back to the chage fom the suounding field when the chage is being acceleated back down the z-axis by the enegy eleased fom the field. Finally, conside momentum consevation. The moving chage in the animation of Figue 4.6. completely eveses its diection of motion ove the couse of the animation. How do we conseve momentum in this pocess? Momentum is conseved because momentum in the positive z-diection is tansmitted fom the moving chage to the chages that ae geneating the constant downwad electic field (not shown). This is obvious fom the field configuation shown in Figue The field stess, which pushes downwad on the chage, is accompanied by a stess pushing upwad on the chages geneating the constant field. Animation 4.: Chaged Paticle at Rest in a Time-Vaying Field As a second example of the stesses tansmitted by electic fields, conside a positive point chage sitting at est at the oigin in an extenal field which is constant in space but vaies in time. This extenal field is unifom vaies accoding to the equation 4 π t = ˆ sin k (4.6.1) T (a) (b) Figue Two fames of an animation of the electic field aound a positive chage sitting at est in a time-changing electic field that points downwad. The oange vecto is the electic field and the white vecto is the foce on the point chage. Figue shows two fames of an animation of the total electic field configuation fo this situation. Figue 4.6.4(a) is at t =, when the vetical electic field is zeo. Fame 4.6.4(b) is at a quate peiod late, when the downwad electic field is at a maximum. As in Figue above, we intepet the field configuation in Figue 4.6.4(b) as 4-7

28 indicating a net downwad foce on the stationay chage. The animation of Figue shows damatically the inflow of enegy into the neighbohood of the chage as the extenal electic field gows in time, with a esulting build-up of stess that tansmits a downwad foce to the positive chage. We can estimate the magnitude of the foce on the chage in Figue 4.6.4(b) as follows. At the time shown in Figue 4.6.4(b), the distance above the chage at which the electic field of the chage is equal and opposite to the constant electic field is detemined by the equation = q (4.6.) 4πε The suface aea of a sphee of this adius is A= 4 π = q/ ε. Now accoding to q. (4.4.8) the pessue (foce pe unit aea) and/o tension tansmitted acoss the suface of this sphee suounding the chage is of the ode of ε /. Since the electic field on the suface of the sphee is of ode, the total foce tansmitted by the field is of ode ε / times the aea of the sphee, o ( ε /)(4 π ) = ( ε /)( q/ ε ) q we expect. Of couse this net foce is a combination of a pessue pushing down on the top of the sphee and a tension pulling down acoss the bottom of the sphee. Howeve, the ough estimate that we have just made demonstates that the pessues and tensions tansmitted acoss the suface of this sphee suounding the chage ae plausibly of ode ε /, as we claimed in q. (4.4.8)., as Animation 4.3: Like and Unlike Chages Hanging fom Pendulums Conside two chages hanging fom pendulums whose suppots can be moved close o futhe apat by an extenal agent. Fist, suppose the chages both have the same sign, and theefoe epel. Figue Two pendulums fom which ae suspended chages of the same sign. 4-8

29 Figue shows the situation when an extenal agent ties to move the suppots (fom which the two positive chages ae suspended) togethe. The foce of gavity is pulling the chages down, and the foce of electostatic epulsion is pushing them apat on the adial line joining them. The behavio of the electic fields in this situation is an example of an electostatic pessue tansmitted pependicula to the field. That pessue ties to keep the two chages apat in this situation, as the extenal agent contolling the pendulum suppots ties to move them togethe. When we move the suppots togethe the chages ae pushed apat by the pessue tansmitted pependicula to the electic field. We atificially teminate the field lines at a fixed distance fom the chages to avoid visual confusion. In contast, suppose the chages ae of opposite signs, and theefoe attact. Figue shows the situation when an extenal agent moves the suppots (fom which the two positive chages ae suspended) togethe. The foce of gavity is pulling the chages down, and the foce of electostatic attaction is pulling them togethe on the adial line joining them. The behavio of the electic fields in this situation is an example of the tension tansmitted paallel to the field. That tension ties to pull the two unlike chages togethe in this situation. Figue Two pendulums with suspended chages of opposite sign. When we move the suppots togethe the chages ae pulled togethe by the tension tansmitted paallel to the electic field. We atificially teminate the field lines at a fixed distance fom the chages to avoid visual confusion. 4.7 Poblem-Solving Stategies In this chapte, we have shown how electic field can be computed using Gauss s law: u Φ = d A = Ò S q ε enc The pocedues ae outlined in Section 4.. Below we summaize how the above pocedues can be employed to compute the electic field fo a line of chage, an infinite plane of chage and a unifomly chaged solid sphee. 4-9

30 System Infinite line of chage Infinite plane of chage Unifomly chaged solid sphee Figue Identify symmety the Cylindical Plana Spheical Detemine diection of the Divide the space into diffeent egions > z > and z < a and a Choose suface Gaussian Coaxial cylinde Gaussian pillbox Concentic sphee Calculate flux electic Φ = ( π l) Φ = A+ A= A Φ = (4 π ) Calculate enclosed chage qin qenc Apply Gauss s law λ Φ = qin / ε to = πε find = λl qenc = σ A σ = ε q enc = Q 3 Q ( / a) a a Q, a 3 4πεa = Q, a 4πε 4-3

31 4.8 Solved Poblems Two Paallel Infinite Non-Conducting Planes Two paallel infinite non-conducting planes lying in the xy-plane ae sepaated by a distance d. ach plane is unifomly chaged with equal but opposite suface chage densities, as shown in Figue Find the electic field eveywhee in space. Solution: Figue Positive and negative unifomly chaged infinite planes The electic field due to the two planes can be found by applying the supeposition pinciple to the esult obtained in xample 4. fo one plane. Since the planes cay equal but opposite suface chage densities, both fields have equal magnitude: σ = = ε + The field of the positive plane points away fom the positive plane and the field of the negative plane points towads the negative plane (Figue 4.8.) Figue 4.8. lectic field of positive and negative planes Theefoe, when we add these fields togethe, we see that the field outside the paallel planes is zeo, and the field between the planes has twice the magnitude of the field of eithe plane. 4-31

32 Figue lectic field of two paallel planes The electic field of the positive and the negative planes ae given by σ ˆ σ, / ˆ + k z > d, z d ε k > / ε + =, = σ ˆ σ k, z< d/ + kˆ, z< d/ ε ε Adding these two fields togethe then yields kˆ, z > d/ σ = kˆ, d / > z > d / (4.8.1) ε kˆ, z < d/ Note that the magnitude of the electic field between the plates is = σ / ε, which is twice that of a single plate, and vanishes in the egions z > d /and z < d / lectic Flux Though a Squae Suface (a) Compute the electic flux though a squae suface of edges l due to a chage +Q located at a pependicula distance l fom the cente of the squae, as shown in Figue Figue lectic flux though a squae suface 4-3

33 (b) Using the esult obtained in (a), if the chage +Q is now at the cente of a cube of side l (Figue 4.8.5), what is the total flux emeging fom all the six faces of the closed suface? Figue lectic flux though the suface of a cube Solutions: (a) The electic field due to the chage +Q is 1 Q 1 Q xˆi + yˆj + zkˆ ˆ = = 4πε 4πε 1/ whee = ( x + y + z ) in Catesian coodinates. On the suface S, y = l and the aea element is da = daˆj= ( dxdz) ˆj. Since ˆˆ i j= ˆj kˆ = and ˆˆ jj= 1, we have Q xˆi+ yˆj+ zkˆ Ql da= ( dxdz) ˆ j= 3 4πε 4πε dxdz Thus, the electic flux though S is Ql l l dz Ql l z Φ = Ò da= dx = dx ( x + l )( x + l + z ) 3/ 4 πε l l 1 ( x + l + z ) 4πε l / S Ql l l dx Q 1 x = 1/ tan πε = l ( x + l )( x + l ) πε x + l Q 1 1 Q = tan (1 / 3) tan ( 1/ 3) πε = 6 ε l l l l whee the following integals have been used: 4-33

34 dx x = ( x + a ) a ( x + a ) 3/ 1/ dx 1 1 b a = tan, b > a 1/ 1/ ( x + a )( x + b ) a( b a ) a ( x + b ) (b) Fom symmety aguments, the flux though each face must be the same. Thus, the total flux though the cube is just six times that though one face: Q Q Φ = 6 = 6 ε ε The esult shows that the electic flux Φ passing though a closed suface is popotional to the chage enclosed. In addition, the esult futhe einfoces the notion that Φ is independent of the shape of the closed suface Gauss s Law fo Gavity What is the gavitational field inside a spheical shell of adius a and mass m? Solution: Since the gavitational foce is also an invese squae law, thee is an equivalent Gauss s law fo gavitation: Φ = 4π Gm (4.8.) g enc The only changes ae that we calculate gavitational flux, the constant 1/ ε 4π G, and qenc menc. Fo a, the mass enclosed in a Gaussian suface is zeo because the mass is all on the shell. Theefoe the gavitational flux on the Gaussian suface is zeo. This means that the gavitational field inside the shell is zeo! lectic 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: 4-34

35 Using Gauss s law, we showed in xample 4.4 that the electic field due to the chage distibution is Q ˆ, > a 4πε = Q ˆ, < a 3 4πεa (4.8.3) The electic potential at P 1 Figue (indicated in Figue 4.8.6) outside the sphee is Q 1 Q V () V( ) = d = = k 1 4πε 4πε e Q (4.8.4) On the othe hand, the electic potential at P inside the sphee is given by a a Q Q V () V( ) = d > a < a = d d ( ) ( ) a 3 4πε a 4πε a 1 Q 1 Q 1 1 Q = ( a ) = 3 4πε a 4πε a 8πε a a 3 Q = ke 3 a a (4.8.5) A plot of electic potential as a function of is given in Figue 4.8.7: Figue lectic potential due to a unifomly chaged sphee as a function of. 4-35

36 4.9 Conceptual Questions 1. If the electic field in some egion of space is zeo, does it imply that thee is no electic chage in that egion?. Conside the electic field due to a non-conducting infinite plane having a unifom chage density. Why is the electic field independent of the distance fom the plane? xplain in tems of the spacing of the electic field lines. 3. If we place a point chage inside a hollow sealed conducting pipe, descibe the electic field outside the pipe. 4. Conside two isolated spheical conductos each having net chage Q >. The sphees have adii a and b, whee b>a. Which sphee has the highe potential? 4.1 Additional Poblems Non-Conducting Solid Sphee with a Cavity A sphee of adius R is made of a non-conducting mateial that has a unifom volume chage density ρ. (Assume that the mateial does not affect the electic field.) A spheical cavity of adius R is then caved out fom the sphee, as shown in the figue below. Compute the electic field within the cavity. Figue Non-conducting solid sphee with a cavity 4.1. 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 =. The N-type mateial lies in the ange < z< a and has unifom 4-36

37 chage density +ρ. The adjacent P-type mateial lies in the ange a < z< unifom chage density ρ. Thus: and has + ρ < z< a ρ( x, yz, ) = ρ( z) = ρ a< z< z > a (a) Find the electic field eveywhee. (b) Find the potential diffeence between the points P and P 1.. 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. is located on plane paallel to the slab a distance z < 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? (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 Thin Slab Let some chage be unifomly distibuted thoughout the volume of a lage plana slab of plastic of thickness d. The chage density is ρ. The mid-plane of the slab is the y-z plane. (a) What is the electic field at a distance x fom the mid-plane when x < d? (b) What is the electic field at a distance x fom the mid-plane when x > d? [Hint: put pat of you Gaussian suface whee the electic field is zeo.] 4-37

38 4.1.5 lectic Potential negy of a Solid Sphee Calculate the electic potential enegy of a solid sphee of adius R filled with chage of unifom density ρ. xpess you answe in tems of Q, the total chage on the sphee Calculating lectic Field fom lectical Potential Figue 4.1. 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. Outside of this egion, the electic potential vaies linealy with z, as indicated in the gaph. Figue 4.1. (a) Find an equation fo the z-component of the electic field, 1m < z < 1m. z, in the egion (b) What is in the egion z > 1 m? Be caeful to indicate the sign of. z (c) What is in the egion z < 1 m? Be caeful to indicate the sign of. z (d) This potential is due a slab of chage with constant chage pe unit volume ρ. Whee is this slab of chage located (give the z-coodinates that bound the slab)? What is the chage density ρ of the slab in C/m 3? Be sue to give clealy both the sign and magnitude of ρ. z z 4-38

### 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

### 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

### 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

### 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

### Chapter 4. Electric Potential

Chapte 4 Electic Potential 4.1 Potential and Potential Enegy... 4-3 4.2 Electic Potential in a Unifom Field... 4-7 4.3 Electic Potential due to Point Chages... 4-8 4.3.1 Potential Enegy in a System of

### 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

### 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

### 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 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

### 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

### 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.

### 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

### 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.

### 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

### 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

### 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

### 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

### 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

### 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

### 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,

### 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.

### 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

### 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,

### 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

### 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,

### 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

### 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

### 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

### 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

### 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

### 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

### Chapter 3. Gauss s Law

3 3 3-0 Chapter 3 Gauss s Law 3.1 Electric Flux... 3-2 3.2 Gauss s Law (see also Gauss s Law Simulation in Section 3.10)... 3-4 Example 3.1: Infinitely Long Rod of Uniform Charge Density... 3-9 Example

### 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

### 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

### 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

### 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.

### 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

### 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,

### 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

### 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

### 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

### 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

### 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

### Gravity. A. Law of Gravity. Gravity. Physics: Mechanics. A. The Law of Gravity. Dr. Bill Pezzaglia. B. Gravitational Field. C.

Physics: Mechanics 1 Gavity D. Bill Pezzaglia A. The Law of Gavity Gavity B. Gavitational Field C. Tides Updated: 01Jul09 A. Law of Gavity 3 1a. Invese Squae Law 4 1. Invese Squae Law. Newton s 4 th law

### 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

### 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

### 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

### 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

### (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

### SELF-INDUCTANCE AND INDUCTORS

MISN-0-144 SELF-INDUCTANCE AND INDUCTORS SELF-INDUCTANCE AND INDUCTORS by Pete Signell Michigan State Univesity 1. Intoduction.............................................. 1 A 2. Self-Inductance L.........................................

### 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

### Ch. 8 Universal Gravitation. Part 1: Kepler s Laws. Johannes Kepler. Tycho Brahe. Brahe. Objectives: Section 8.1 Motion in the Heavens and on Earth

Ch. 8 Univesal Gavitation Pat 1: Keple s Laws Objectives: Section 8.1 Motion in the Heavens and on Eath Objectives Relate Keple s laws of planetay motion to Newton s law of univesal gavitation. Calculate

### 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

### 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

### 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

### 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

### 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

### 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.

### Nuclear models: Fermi-Gas Model Shell Model

Lectue Nuclea mode: Femi-Gas Model Shell Model WS/: Intoduction to Nuclea and Paticle Physics The basic concept of the Femi-gas model The theoetical concept of a Femi-gas may be applied fo systems of weakly

### 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,

### 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..............................................

### F G r. Don't confuse G with g: "Big G" and "little g" are totally different things.

G-1 Gavity Newton's Univesal Law of Gavitation (fist stated by Newton): any two masses m 1 and m exet an attactive gavitational foce on each othe accoding to m m G 1 This applies to all masses, not just

### 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

### 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).

### 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

### 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

### Algebra and Trig. I. A point is a location or position that has no size or dimension.

Algeba and Tig. I 4.1 Angles and Radian Measues A Point A A B Line AB AB A point is a location o position that has no size o dimension. A line extends indefinitely in both diections and contains an infinite

### Lecture 16: Color and Intensity. and he made him a coat of many colours. Genesis 37:3

Lectue 16: Colo and Intensity and he made him a coat of many colous. Genesis 37:3 1. Intoduction To display a pictue using Compute Gaphics, we need to compute the colo and intensity of the light at each

### Structure and evolution of circumstellar disks during the early phase of accretion from a parent cloud

Cente fo Tubulence Reseach Annual Reseach Biefs 2001 209 Stuctue and evolution of cicumstella disks duing the ealy phase of accetion fom a paent cloud By Olusola C. Idowu 1. Motivation and Backgound The

### TORQUE AND ANGULAR MOMENTUM IN CIRCULAR MOTION

MISN-0-34 TORQUE AND ANGULAR MOMENTUM IN CIRCULAR MOTION shaft TORQUE AND ANGULAR MOMENTUM IN CIRCULAR MOTION by Kiby Mogan, Chalotte, Michigan 1. Intoduction..............................................

### 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

### 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

### TECHNICAL DATA. JIS (Japanese Industrial Standard) Screw Thread. Specifications

JIS (Japanese Industial Standad) Scew Thead Specifications TECNICAL DATA Note: Although these specifications ae based on JIS they also apply to and DIN s. Some comments added by Mayland Metics Coutesy

### 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

### 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

### 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

### CHAPTER 10 Aggregate Demand I

CHAPTR 10 Aggegate Demand I Questions fo Review 1. The Keynesian coss tells us that fiscal policy has a multiplied effect on income. The eason is that accoding to the consumption function, highe income

### 1.4 Phase Line and Bifurcation Diag

Dynamical Systems: Pat 2 2 Bifucation Theoy In pactical applications that involve diffeential equations it vey often happens that the diffeential equation contains paametes and the value of these paametes

### dz + η 1 r r 2 + c 1 ln r + c 2 subject to the boundary conditions of no-slip side walls and finite force over the fluid length u z at r = 0

Poiseuille Flow Jean Louis Maie Poiseuille, a Fench physicist and physiologist, was inteested in human blood flow and aound 1840 he expeimentally deived a law fo flow though cylindical pipes. It s extemely

### Experiment MF Magnetic Force

Expeiment MF Magnetic Foce Intoduction The magnetic foce on a cuent-caying conducto is basic to evey electic moto -- tuning the hands of electic watches and clocks, tanspoting tape in Walkmans, stating

### UNIT CIRCLE TRIGONOMETRY

UNIT CIRCLE TRIGONOMETRY The Unit Cicle is the cicle centeed at the oigin with adius unit (hence, the unit cicle. The equation of this cicle is + =. A diagam of the unit cicle is shown below: + = - - -

### Physics HSC Course Stage 6. Space. Part 1: Earth s gravitational field

Physics HSC Couse Stage 6 Space Pat 1: Eath s gavitational field Contents Intoduction... Weight... 4 The value of g... 7 Measuing g...8 Vaiations in g...11 Calculating g and W...13 You weight on othe

### Displacement, Velocity And Acceleration

Displacement, Velocity And Acceleation Vectos and Scalas Position Vectos Displacement Speed and Velocity Acceleation Complete Motion Diagams Outline Scala vs. Vecto Scalas vs. vectos Scala : a eal numbe,

### EXPERIMENT 16 THE MAGNETIC MOMENT OF A BAR MAGNET AND THE HORIZONTAL COMPONENT OF THE EARTH S MAGNETIC FIELD

260 16-1. THEORY EXPERMENT 16 THE MAGNETC MOMENT OF A BAR MAGNET AND THE HORZONTAL COMPONENT OF THE EARTH S MAGNETC FELD The uose of this exeiment is to measue the magnetic moment μ of a ba magnet and

### 2. Orbital dynamics and tides

2. Obital dynamics and tides 2.1 The two-body poblem This efes to the mutual gavitational inteaction of two bodies. An exact mathematical solution is possible and staightfowad. In the case that one body

### 4a 4ab b 4 2 4 2 5 5 16 40 25. 5.6 10 6 (count number of places from first non-zero digit to

. Simplify: 0 4 ( 8) 0 64 ( 8) 0 ( 8) = (Ode of opeations fom left to ight: Paenthesis, Exponents, Multiplication, Division, Addition Subtaction). Simplify: (a 4) + (a ) (a+) = a 4 + a 0 a = a 7. Evaluate

### 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

### Solutions for Physics 1301 Course Review (Problems 10 through 18)

Solutions fo Physics 1301 Couse Review (Poblems 10 though 18) 10) a) When the bicycle wheel comes into contact with the step, thee ae fou foces acting on it at that moment: its own weight, Mg ; the nomal

### Determining solar characteristics using planetary data

Detemining sola chaacteistics using planetay data Intoduction The Sun is a G type main sequence sta at the cente of the Sola System aound which the planets, including ou Eath, obit. In this inestigation

### (References to) "Introduction to Electrodynamics" by David J. Griffiths 3rd ed., Prentice-Hall, 1999. ISBN 0-13-805326-X

lectic Fields in Matte page. Capstones in Physics: lectomagnetism. LCTRIC FILDS IN MATTR.. Multipole xpansion A. Multipole expansion of potential B. Dipole moment C. lectic field of dipole D. Dipole in

### Newton s Shell Theorem

Newton Shell Theoem Abtact One of the pincipal eaon Iaac Newton wa motivated to invent the Calculu wa to how that in applying hi Law of Univeal Gavitation to pheically-ymmetic maive bodie (like planet,

### Figure 2. So it is very likely that the Babylonians attributed 60 units to each side of the hexagon. Its resulting perimeter would then be 360!

1. What ae angles? Last time, we looked at how the Geeks intepeted measument of lengths. Howeve, as fascinated as they wee with geomety, thee was a shape that was much moe enticing than any othe : the

### 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

### Problems of the 2 nd and 9 th International Physics Olympiads (Budapest, Hungary, 1968 and 1976)

Poblems of the nd and 9 th Intenational Physics Olympiads (Budapest Hungay 968 and 976) Péte Vankó Institute of Physics Budapest Univesity of Technology and Economics Budapest Hungay Abstact Afte a shot

### Semipartial (Part) and Partial Correlation

Semipatial (Pat) and Patial Coelation his discussion boows heavily fom Applied Multiple egession/coelation Analysis fo the Behavioal Sciences, by Jacob and Paticia Cohen (975 edition; thee is also an updated

### Electrostatic properties of conductors and dielectrics

Unit Electostatic popeties of conductos and dielectics. Intoduction. Dielectic beaking. onducto in electostatic equilibium..3 Gound connection.4 Phenomena of electostatic influence. Electostatic shields.5

### Gravitational Mechanics of the Mars-Phobos System: Comparing Methods of Orbital Dynamics Modeling for Exploratory Mission Planning

Gavitational Mechanics of the Mas-Phobos System: Compaing Methods of Obital Dynamics Modeling fo Exploatoy Mission Planning Alfedo C. Itualde The Pennsylvania State Univesity, Univesity Pak, PA, 6802 This