Chapte 4 Gauss s Law 4.1 lectic Flux... 1 4. Gauss s Law... xample 4.1: Infinitely Long Rod of Unifom Chage Density... 7 xample 4.: Infinite Plane of Chage... 9 xample 4.3: Spheical Shell... 11 xample 4.4: Non-Conducting Solid Sphee... 13 4.3 Conductos... 14 xample 4.5: Conducto with Chage Inside a Cavity... 17 xample 4.6: lectic Potential Due to a Spheical Shell... 18 4.4 Foce on a Conducto... 1 4.5 Summay... 3 4.6 Appendix: Tensions and Pessues... 4 Animation 4.1: Chaged Paticle Moving in a Constant lectic Field... 5 Animation 4.: Chaged Paticle at Rest in a Time-Vaying Field... 6 Animation 4.3: Like and Unlike Chages Hanging fom Pendulums... 8 4.7 Poblem-Solving Stategies... 9 4.8 Solved Poblems... 31 4.8.1 Two Paallel Infinite Non-Conducting Planes... 31 4.8. lectic Flux Though a Squae Suface... 3 4.8.3 Gauss s Law fo Gavity... 34 4.8.4 lectic Potential of a Unifomly Chaged Sphee... 34 4.9 Conceptual Questions... 36 4.1 Additional Poblems... 36 4.1.1 Non-Conducting Solid Sphee with a Cavity... 36 4.1. P-N Junction... 36 4.1.3 Sphee with Non-Unifom Chage Distibution... 37 4.1.4 Thin Slab... 37 4.1.5 lectic Potential negy of a Solid Sphee... 38 4.1.6 Calculating lectic Field fom lectical Potential... 38
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 4.1.1 lectic field lines passing though a suface of aea A. Conside the suface shown in Figue 4.1.1. Let A = Anˆ be defined as the aea vecto having a magnitude of the aea of the suface, A, and 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 = n A = A (4.1.1) On the othe hand, if the electic field 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. 1
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 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 4.1.3. Note that fo a closed suface the unit vecto nˆ i is chosen to point in the outwad nomal diection. Figue 4.1.3 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 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 poduct da. 4. Gauss s Law Conside a positive point chage Q located at the cente o f a sphee of adius, as shown in Figue 4..1. The electic field due to the chage Q is = ( Q/4 πε ), which points ˆ
in the adial diection. We enclose the chage by an imaginay sphee of adius called the Gaussian suface. 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 3
Φ = d = da= A= 4π = 4πε S S ε 1 Q (4..4) A 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 S 1, S o. S 3 Q 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 q Φ = = enc 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 be an aea element on the suface of a sphee of adius, as shown in Figue 4..4. S1 1 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 4
A Ω (4..6) 1 1 Solid angles ae dimensionless quantities measued in steadians (s). Since the suface a ea 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 th en the solid angle subtended by A is ˆ, A ˆ A cosθ A Ω = = = n (4..1) whee A = A cosθ is the aea of the adial pojection of n A onto a s econd sphee S of adius, concentic with S 1. As shown in Figue 4..4, the solid angle subtended is the same fo both A and A : 1 n 5
A A cosθ Ω = = (4..11) 1 1 Now suppose a point chage Q is placed at the cente of the concentic sphees. The electic field stengths 1 and at the cente of the aea elements A1 and A ae elated by Coulomb s law: 1 Q i = 4πε = (4..1) 1 i 1 The electic flux though A 1 on S 1 is Φ = A = A (4..13) 1 1 1 1 O n the othe hand, the electic flux though A on S is 1 Φ = A = Acosθ = 1 A 1 = 1 A1 = Φ1 (4..14) 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 app licable fo detemining electic field, with the c oesponding 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. 6
(3) Divide the space into diffeent egions associated with the chage distibution. Fo each e gion, calculate, the chage enclosed by the Gaussian suface. q enc (4) Calculate the electic flux Φ though the Gaussian suface fo each egion. (5) quate Φ with qenc / ε, and deduce the magnitude of the electic field. xample 4.1: Infinitely Long Rod of Unifom Chage Density An infinite ly 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 (Figue 4..7), is q = λ. enc 7
Figue 4..7 Gaussian suface fo a unifomly chaged od. (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 3 Φ = da= da + da + da 1 1 3 3 S S1 S S3 = + + A= 3 3 ( π ) (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. ( π ) = λ/ ε (5) Applying Gauss s law gives, 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 of the cylinde, an d only d epends on the invese of the distance fom the symmety axis. The qualitative b ehavio of as a function of is plotted in Figue 4..8. Figue 4..8 lectic field due to a unifomly chaged od as a function of 8
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. () 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. 9
(4) The total flux though the Gaussian pillbox flux is Φ = da= da + da + da 1 1 3 S S1 S S3 = A+ A + 1 1 = ( + ) A 1 3 (4..17) Since the two ends ae at the same distance fom the plane, by symmety, the magnitude o f 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 4..11 lectic field of an infinitely lage non-conducting plane. 1
Note again the discontinuity in electic field as we coss the plane: σ σ σ z = z+ z = = ε ε ε (4..1) 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 4..13(a). a, as shown in Figue (a) (b) Figue 4..13 Gaussian suface fo unifomly chaged spheical shell fo (a) < a, and (b) a 11
T he 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 =, < a (4..) Case : a 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= (4 π ) S Q Q = = k, a (4..3) 4πε e 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 4..14. Figue 4..14 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 1
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: 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 4..15(a). a, as shown in Figue (a) (b) Figue 4..15 Gaussian suface fo unifomly chaged solid sphee, fo (a) a, and (b) > a. The flux though the Gaussian suface is Φ = d A = A= (4 π ) With unifom chage distibution, the chage enclosed is S 3 4 = Q 3 qenc = ρdv = ρv = ρ π 3 (4..5) 3 a V 13
which is popotional to the volume enclosed by the Gaussian suface. Applying Gauss s la w Φ = qenc / ε, we obtain ρ 4 3 ( 4π ) = π ε 3 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 4..16. Figue 4..16 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. 14
If we place a solid spheical conducto in a constant extenal field, 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 4.3.1 below), theeby inducing an electic field. Inside the conducto, points in the opposite diection of. Since chages ae mobile, they will continue to move until completely cancels inside the conducto. At electostatic equilibium, must vanish inside a conducto. Outside the conducto, the electic field due to the induced chage distibution coesponds to a dipole field, and the total electic field is simply = +. The field lines ae depicted in Figue 4.3.1. Figue 4.3.1 Placing a conducto in a unifom electic field () Any net chage must eside on the suface. 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 d s aound a closed path shown in Figue 4.3.3: 15
Figue 4.3.3 Nomal and tangential components of electic field outside the conducto Since the electic field is consevative, the line integal aound the closed path abcda vanishes: 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 t l =. Howeve, since the length element l is 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 V = V A B. B VB VA = d s = A is pependicula to d s. Thus, points A and B ae at the same potential with (4) is nomal to the suface just outside the conducto. If the tangential component of is initially non-zeo, chages will then move aound until it vanishes. Hence, only the nomal component suvives. 16
Figue 4.3.3 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 4.3.3. Using Gauss s law, we obtain A Φ = 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 4.3.4. Figue 4.3.4 Just outside the conducto, 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 17
Conside a hollow conducto shown in Figue 4.3.5 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 4.3.5 Conducto with a cavity Since the electic field inside a conducto must be zeo, the net chage enclosed by the Gaussian suface shown in Figue 4.3.5 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 4.3.6. Figue 4.3.6 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 18
Q 4πε ˆ, > a =, < a The electic potential may be calculated by using q. (3.1.9): Fo > a, we have B VB VA = d s A Q 1 Q V() V( ) = d = = k 4πε 4πε e Q (4.3.4) whee we have chosen V ( ) = potential becomes as ou efeence point. On the othe hand, fo < a, the V() a V( ) = d > a < a) ( ) ( a 1 = d Q = Q = k Q e 4πε 4πε a a a (4.3.5) A plot of the electic potential is shown in Figue 4.3.7. Note that the potential constant inside a conducto. V is Figue 4.3.7 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 su face 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 19
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 = 4πε a 8πε a (4.3.7) Since V = Q/4πεa and Wext = U, the above expession is simplified to U 1 = QV (4.3.8) 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 4.3.8. Figue 4.3.8 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 q 1 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 q = (4.3.9) 1 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 1 q = = σ σ, = = (4.3.1) 1 1 1 4πε 1 ε 4πε ε w hee 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 1 1 1 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. 4.4 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 4.4.1. Figue 4.4.1 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 m ust come solely fom. Assuming the patch to be a flat suface, fom Gauss s law, the electic field due to the patch is 1
σ + ˆ, z > k ε patch = (4.4.) σ kˆ, z < ε 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+ (4.4.4) ε 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 = ( + ) = above below avg In the case of a conducto, with above = ( σ / ε)ˆ k and below =, we have (4.4.5) 1 σ ˆ ˆ avg = k + = σ k (4.4.6) ε ε Thus, the foce acting on the patch is ˆ A F= q ˆ avg = ( σ A) σ k = σ k ε ε (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 ε ε A ε ε P = = = = (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. 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 = d A = 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 3
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 4.6.1. 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 along itself, giving ise to a downwad pull, just as if a sting has been attached to that face to pull it downwad. (The actual detemination of the diection of the foce equies an advanced teatment using the Maxwell s stess tenso.) Note that this is a pull paallel to the outwad nomal of the bottom face, egadless of whethe the field is into the suface o out of the suface. Figue 4.6.1 An imaginay box in an electic field (long oange vectos). The shot vectos indicate the diections of stesses tansmitted by the field, eithe pessues (on the left o ight faces of the box) o tensions (on the top and bottom faces of the box). Fo the left side of the imaginay box, the field on that face is pependicula to the outwad nomal of that face, and Faaday would have said that the field on that face tansmits a pessue pependicula to itself, causing a push to the ight. Similaly, fo the ight side of the imaginay box, the field on that face is pependicula to the outwad nomal of the face, and the field would tansmit a pessue pependicula to itself. In this case, thee is a push to the left. 4
Note that the tem tension is used when the stess tansmitted by the field is paallel (o anti-paallel) to the outwad nomal of the suface, and pessue when it is pependicula to the outwad nomal. The magnitude of these pessues and tensions on the vaious faces of the imaginay suface in Figue 4.6.1 is given by ε / fo the electic field. This quantity has units of foce pe unit aea, o pessue. It is also the enegy density stoed in the electic field since enegy pe unit volume has the same units as pessue. Animation 4.1: Chaged Paticle Moving in a Constant lectic Field As an example of the stesses tansmitted by electic fields, and of the intechange of e negy between fields and paticles, conside a positive electic chage q > moving in a constant electic field. Suppose the chage is initially moving upwad along the positive z-axis in a constant backgound field = k ˆ. Since the chage expeiences a constant downwad foce Fe = q= q k, ˆ it eventually comes to est (say, at the oigin z = ), and then moves back down the negative z-axis. This motion and the fields that accompany it ae shown in Figue 4.6., at two diffeent t imes. (a) (b) Figue 4.6. A positive chage moving in a constant electic field which points downwad. (a) The total field configuation when the chage is still out of sight on the negative z-axis. (b) The total field configuation when the chage comes to est at the oigin, befoe it moves back down the negative z-axis. How do we intepet the motion of the chage in tems of the stesses tansmitted by the fields? Faaday would have descibed the downwad foce on the chage in Figue 4.6.(b) as follows: Let the chage be suounded by an imaginay sphee centeed on it, as shown in Figue 4.6.3. The field lines piecing the lowe half of the sphee tansmit a tension that is paallel to the field. This is a stess pulling downwad on the chage fom below. The field lines daped ove the top of the imaginay sphee tansmit a pessue pependicula to themselves. This is a stess pushing down on the chage fom above. The total effect of these stesses is a net downwad foce on the chage. 5
Figue 4.6.3 An electic chage in a constant downwad electic field. We suound the chage by an imaginay sphee in ode to discuss the stesses tansmitted acoss the suface of that sphee by the electic field. Viewing the animation of Figue 4.6. geatly enhances Faaday s intepetation of the stesses in the static image. As the chage moves upwad, it is appaent in the animation that the electic field lines ae geneally compessed above the chage and stetched below the chage. This field configuation enables the tansmission of a downwad foce to the moving chage we can see as well as an upwad foce to the chages that poduce the constant field, which we cannot see. The oveall appeaance of the upwad motion of the chage though the electic field is that of a point being foced into a esisting m edium, with stesses aising in that medium as a esult of that encoachment. 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 4.6.3. 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 6
π t kˆ T 4 = sin (4.6.1) (a) (b) Figue 4.6.4 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 4.6.4 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 4.6.3 above, we intepet the field configuation in Figue 4.6.4(b) as indicating a net downwad foce on the stationay chage. The animation of Figue 4.6.4 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, as 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 7
acoss the suface of this sphee suounding the chage ae plausibly of ode ε /, as we claimed in q. (4.4.8). 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 4.6.5 Two pendulums fom which ae suspended chages of the same sign. Figue 4.6.5 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 4.6.6 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 4.6.6 Two pendulums with suspended chages of opposite sign. 8
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: Φ = d A = 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. 9
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 3
4.8 Solved Poblems 4.8.1 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 4.8.1. Find the electic field eveywhee in space. Solution: Figue 4.8.1 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. 31
Figue 4.8.3 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 /. 4.8. 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 4.8.4. Figue 4.8.4 lectic flux though a squae suface 3
(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 4.8.5 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)j ˆ. Since ˆˆ i j= ˆj kˆ = and ˆˆ j j= 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 = 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: 33
dx x = ( x + a ) a ( x + a ) 3/ 1/ dx 1 b a ( x + a )( x + b ) a( b a ) a ( x + b ) 1 = tan, b > a 1/ 1/ (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. 4.8.3 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! 4.8.4 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: 34
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 4.8.6 (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 4.8.7 lectic potential due to a unifomly chaged sphee as a function of. 35
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 4.1.1 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 4.1.1 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 36
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. The point P 1. 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. 4.1.3 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. 4.1.4 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.] 37
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. 4.1.6 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 38