Poynting Vector and Energy Flow in a Capacitor Challenge Problem Solutions
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1 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 towas the lowe plate an away fom the uppe plate, chaging the plates. a) What is the iection an magnitue of the electic fiel E between the plates? You may neglect any finging fiels ue to ege effects. b) What is the total enegy stoe in the electic fiel of the capacito? c) What is the ate of change of the enegy stoe in the electic fiel? ) What is the magnitue of the magnetic fiel B at point P locate between the plates at aius < R (see figue above). As seen fom above, is the iection of the magnetic fiel clockwise o counteclockwise. Explain you answe. e) Make a sketch of the electic an magnetic fiel insie the capacito. f) What is the iection an magnitue of the Pointing vecto S at a istance = R fom the cente of the capacito. g) By integating S ove an appopiate suface, fin the powe that flows into the capacito. h) How oes you answe in pat g) compae to you answe in pat c)?
2 Poblem 1 Solutions: (a) If we ignoe finging fiels then we can calculate the electic fiel using Gauss s Law, Q enc E a =. ε close suface By supeposition, the electic fiel is non-zeo between the plates an zeo eveywhee else. Choose a Gaussian cyline passing though the lowe plate with its en faces paallel to the plates. Let A enote the aea of the enface. The suface chage ensity cap is given by σ = Q/ π R. Let ˆk enote the unit vecto pointing fom the lowe plate to the uppe plate. Then Gauss Law becomes E A cap σ A = ε cap which we can solve fo the electic fiel σ ˆ Q E= k = kˆ. ε πr ε (b) The total enegy stoe in the electic fiel is given by 1 1 Uelec = ε E V = ε E π R. volume Substitute the esult fo the electic fiel intot he enegy equation yiels U elec Q Q R π π R ε πr 1 1 = ε = ε. (c) The ate of change of the stoe electic enegy is foun by taking the time eivative of the enegy equation U = Q Q. elec πr ε The cuent flowing to the plate is equal to
3 I Q =. Substitute the expession fo the cuent into the expession fo the ate of change of the stoe electic enegy yiels U elec = QI π R ε. () We shall calculate the magnetic fiel by using the genealize Ampee s Law, close path B s = μ J a open suface + μ ε open suface E a We choose a cicle of aius < R passing though the point P as the Ampeian loop an the isk efine by the cicle as the open suface with the cicle as its bounay. We choose to ciculate aoun the loop in the counteclockwise iection as seen fom above. This means that flux in the positive ˆk -iection is positive. The left han sie (LHS) of the genealize Ampee s Law becomes LHS = cicle B s = B π. The conuction cuent is zeo passing though the isk, since no chages ae moving between the plates. Thee is an electic flux passing though the isk. So the ight han sie (RHS) of the genealize Ampee s Law becomes E RHS = μ ε E a= μ ε π isk Take the time eivative of the expession fo the electic fiel an the expession fo the cuent, an substitute it into the RHS of the genealize Ampee s Law: E μ Iπ RHS = με = π R π. Equating the two sies of the genealize Ampee s Law yiels μiπ B π = π R
4 Finally the magnetic fiel between the plates is then μ I ; π R B = < <. The sign of the magnetic fiel is positive theefoe the magnetic fiel points in the counteclockwise iection (consistent with ou sign convention fo the integation iection fo the cicle) as seen fom above. Define the unit vecto ˆθ such that is it tangent to the cicle pointing in the counteclockwise iection, then R (e) B μ I π R ˆ ; = θ < < R. (f) The Poynting vecto at a istance = R is given by 1 S( = R) = E B μ = R Substituting the electic fiel an the magnetic fiel (setting = R) into the above equation, an noting that kˆ θˆ = ˆ, yiels 1 Q ( ) ˆ μ I ˆ Q I S = R = k θ = ( ˆ ). μ πr ε πr πr ε πr. So the Poynting vecto points inwa with magnitue Q I S ( = R) =. R π R ε π
5 (g) The powe flowing into the capacito is the close suface integal P = close suface S( = R) a. The Poynting vecto points aially inwa so the only contibution to this integal is fom the cylinical boy of the capacito. The unit nomal associate with the aea vecto fo a close suface integal always points outwa, so on the cylinical boy a = a. ˆ Use this efinition fo the aea element an the powe is then Q I P= S( = R) a= ( ˆ) aˆ R R cylinical boy π ε cylinical π boy The Poynting vecto is constant an the aea of the cylinical boy is π R Q I Q I QI P= ( ˆ) aˆ = π R =. π R ε πr πr ε πr πr ε cylinical boy, so The minus sign coespon to powe flowing into the egion. (h) The two expessions fo powe ae equal so the powe flowing in is equal to the change of enegy stoe in the electic fiels.
6 Poblem : A coaxial cable consists of two concentic long hollow cylines of zeo esistance; the inne has aius a, the oute has aius b, an the length of both is l, with l >> b, as shown in the figue. The cable tansmits DC powe fom a battey to a loa. The battey povies an electomotive foce ε between the two conuctos at one en of the cable, an the loa is a esistance R connecte between the two conuctos at the othe en of the cable. A cuent I flows own the inne conucto an back up the oute one. The battey chages the inne conucto to a chage Q an the oute conucto to a chage + Q. (a) Fin the iection an magnitue of the electic fiel E eveywhee. (b) Fin the iection an magnitue of the magnetic fiel B eveywhee. (c) Calculate the Poynting vecto S in the cable. () By integating S ove appopiate suface, fin the powe that flows into the coaxial cable. (e) How oes you esult in () compae to the powe issipate in the esisto? Poblem Solutions: (a) Consie a Gaussian suface in the fom of a cyline with aius an length l, coaxial with the cylines. Insie the inne cyline (<a) an outsie the oute cyline (>b) no chage is enclose an hence the fiel is. In between the two cylines (a<<b) the chage enclose by the Gaussian suface is Q, the total flux though the Gaussian cyline is Φ E = E A = E(π l) qenc Thus, Gauss s law leas to E( π l ) =, o ε qenc Q E = ˆ ˆ (inwa) fo a b, elsewhee πl = πε l < < (b) Just as with the E fiel, the enclose cuent I enc in the Ampee s loop with aius is zeo insie the inne cyline (<a) an outsie the oute cyline (>b) an hence the fiel thee is. In between the two cylines (a<<b) the cuent enclose is I.
7 Applying Ampee s law, B s = B(π ) = μ I enc, we obtain μi B = ϕ ˆ (clockwise viewing fom the left sie) fo a< < b, elsewhee π (c) Fo a < < b, the Poynting vecto is Q μ I QI S= E B= = μ μ πε π π ε 1 1 ˆ ˆ ϕˆ l 4 l k (fom ight to left) On the othe han, fo < a an > b, we have S =. () With A = ( π )kˆ, the powe is P = S A = QI b 1 4π ε l (π) = QI a πε l ln b a S (e) Since b Q Q b ε = E s = = ln = a IR πlε πlε a lir the chage Q is elate to the esistance R by Q = π ε. The above expession fo P ln( b/ a) becomes π ε lir I b P = ln = I ln( b / a) π ε l a R which is equal to the ate of enegy issipation in a esisto with esistance R.
8 Poblem 3: A capacito consists of two cicula plates of aius a sepaate by a istance (assume << a). The cente of each plate is connecte to the teminals of a voltage souce by a thin wie. A switch in the cicuit is close at time t = an a cuent I(t) flows in the cicuit. The chage on the plate is elate Q() t to the cuent accoing to It () =. We begin by calculating the electic fiel between the plates. Thoughout this poblem you may ignoe ege effects. We assume that the electic fiel is zeo fo > a. (a) Use Gauss Law to fin the electic fiel between the plates as a function of time t, in tems of Q(t), a, ε, an π. The vetical iection is the ˆk iection. (b) Now take an imaginay flat isk of aius < a insie the capacito, as shown below. Using you expession fo E above, calculate the electic flux though this flat isk of aius < a in the plane miway between the plates, in tems of, Q(t), a, anε. Take the suface nomal to the imaginay isk to be in the ˆk iection. (c) Calculate the Maxwell isplacement cuent, I = Φ E ε = ε E A isk though the flat isc of aius < a in the plane miway between the plates, in tems of, I(t), an a. Remembe, thee is eally not a cuent thee, we just call it that to confuse you. () What is the conuction cuent J S A though the flat isk of aius < a? Conuction cuent just means the cuent ue to the flow of eal chage acoss the suface (e.g. electons o ions). (e) Since the capacito plates have an axial symmety an we know that the magnetic fiel ue to a wie uns in azimuthal cicles about the wie, we assume that the magnetic fiel between the plates is non-zeo, an also uns in azimuthal cicles.
9 Choose fo an Ampeian loop, a cicle of aius < a in the plane miway between the plates. Calculate the line integal of the magnetic fiel aoun the cicle, B s Expess you answe in tems of B, π, an. The line element espect to A, that is counteclockwise as seen fom the top.. cicle s is ight-hane with (f) Now use the esults of you answes above, an apply the genealize Ampee Law Equation to fin the magnitue of the magnetic fiel at a istance < a fom the axis. You answe shoul be in tems of, I(t), μ, π, an a. Poblem 3 Solutions: (a) The electic fiel between the plates is (b) The electic flux though the isk of aius is isk o Q Q Φ E = E A π = E( π ) = = επa ε a Qt () Qt () EA ˆ ˆ E A = = E = = a k σ ε π ε ε k (c) Using the above equation, the isplacement cuent is Q() t Q() t I = ε = = = I () E A ε t ε isk a a a () The conuction cuent though the flat isk is zeo. (e) The line integal of the magnetic fiel aoun the cicle is (f) The magnetic fiel at a istance < a is B s = B(π). cicle
10 μi B( π) = μi = μ I B = a π a
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