EM-2. 1 Coulomb s law, electric field, potential field, superposition q. Electric field of a point charge (1)

Size: px

Start display at page:

Download "EM-2. 1 Coulomb s law, electric field, potential field, superposition q. Electric field of a point charge (1)"

Philip Brooks
5 years ago
Views:

1 EM- Coulomb s law, electic field, potential field, supeposition q ' Electic field of a point chage ( ') E( ) kq, whee k / 4 () ' Foce of q on a test chage e at position is ee( ) Electic potential O kq V ( ) (). ' The potential enegy of a test chage e at position is ev ( ) V ) V ( ) E dl ( () which is independent of the path. ' Supeposition of field and potential. Given chage distibution ( '), the potential at is ( ') ( ') V ( ) k dx' dy' dz' k dv' (4) ' ' And the electic field is ( ')( ') ( ')( ') E( ) k dx' dy' dz' k dv' (5) ' ' Uniqueness theoem: It is quite obvious fom Eqs. (4) and (5) that given ( ') then V ( ) and E() ae uniquely detemined. The evese, which is not obvious fom the equations, is also tue. This is quite impotant in some cases whee one can guess the chage distibution to poduce the given potential field. If the guess is coect, then it is the one and the only coect answe. Gauss s law Gauss s Law Q S E ds (6) Hee E is the total field, but Q is the chage within the space enclosed by S. Diffeential fom: E / (6a) Fo the electic potential, it is V / (6b)

2 This is called Poisson s Equation. In the space whee the chage density is zeo, we have the Laplace Equation V (6c) Example-: Veify Eq. (6) with a point chage. Take a spheical suface of adius centeed on the point chage at '. The electic field at kq any point on the suface is paallel to the suface nomal and its amplitude is. The total kq q suface of the sphee is 4, so E ds 4 4kq ans. Gauss s Law is useful in cases with high symmety. S Example-: Find the E-field and potential of an infinitely long staight wie caying unifom chage of pe unit length. Symmety analysis shows that the electic field is pointing outwads along the adial diection, and depends on (the adial distance) only. Take a closed Gauss suface as shown, the top/bottom sufaces will have L no flux. The flux though the size wall is LE. So E. The electic potential V() is likewise depending on only. Taking a line integal along the adial diection fom to, we have V ( ) V ( ) Ed O simply V ( ) ln( ) ans. Conductos d ln( ) L Conductos contain vitually infinite amount of fee chages, usually electons, and a fixed positive chage backgound. Theefoe, in equilibium, thee is no electic field inside a conducto, and the electic potential of the whole body of a conducto is equal. Thee is chage neutality eveywhee inside a conducto, so the net chage must all be at the suface. Example- Suface chage density and pessue on a chaged conducto suface. At a point vey close to the suface, the suface can be taken as flat and infinitely lage. Choose a small flat box with one of its lage suface in the conducto and the othe outside, use Gauss s law, it is staightfowad to show that the E-field is E / outside the conducto and inside. The diection is pependicula to the suface because any paallel E- field component would dive the suface chage to move aound. (The paallel component of the E-fields at both side of the bounday must be the same.)

3 Now let us find out the electic field foce upon a small patch of suface chage on a conducto. The foce is not σea, whee A is the aea of the patch, because the E-field contains contibution fom the chage on the patch, and that chage does not exet foce on itself. E / E / E / / E A patch of chage will poduce E-field E / on both sides with opposite diections. The est of the flat chage sheet poduces a field E / pointing to the ight. The combined field of the patch and the flat sheet with the hole is σ/ε pointing to the ight. So the foce of the flat sheet with hole on the patch is F A. The pessue is P. ans. The main poblem associated with conductos is that only the potential and/o total chage of each conducto is usually known. The exact chage distibutions on the conductos ae not known in advance, and have to be found as pat of the solution of the Laplace Equation o Poisson s Equation. In mathematical tems, the potential and/o total chage of each conducto is called the bounday conditions, which must be given in pio. Fo example, the suface of a sphee of adius R centeed at (x, y, z ) is( x x ) ( y y ) ( z z R. ) If it is a conducto held at potential V, then the potential field V ( ) must be equal to V when satisfy the above equation fo the sphee suface. The bounday conditions ae of the same in math as the initial conditions in mechanics. Conside fo example a paticle in unifom acceleation a along the X-diection. The equation is then dv x a dt. Solving it one gets v x at v, and v is a constant and detemined by the initial condition of v x at t =. The potential o the total chage of all conductos must be known in ode to detemine the potential field V ( ). Uniqueness theoem: Given the potential on the suface of a closed space and the chage distibution in the space, the potential (and electic field) in the space is uniquely detemined. Hee ae some examples of closed space. () The closed space is the est of the est of the space not occupied by the conductos, and the boundaies ae the sufaces of the conductos. S S S S

4 () Hee thee ae two closed spaces. One is the cavity, with its bounday being the inne spheical suface. The othe is the space outside the conducto, and its bounday is the oute suface of the conducto. These two spaces ae not connected. q Poof: Q Situation-: The potential of each and evey conducto is known. Conside the case as () above, assume that thee ae two potential fields, V ( ) and V ( ), both satisfy the Poisson s Equation and the bounday conditions that thei values on conducto-, -, - ae U, U, and U, espectively. Then, let V( ) V ( ) V ( ), then it is staightfowad to show that V ( ) satisfy the Laplace Equation (no chage density outside the conductos), and its value on all conductos is. This is the case whee thee is no chage eveywhee, sov( ). Situation-: The total chage on each conducto is known. (HW execise) Example-4 A point chage is in a cave inside a conducto held at potential V. The potential and E-field in the cave emain the same as long as the conducto is at potential V, egadless of how this is achieved, e. g., by putting chage on the conducto o by placing some chage Q nea the conducto. Likewise, the potential and E-field outside the conducto is uniquely detemined as long as the conducto emains at potential V, egadless of the inteio of the conducto, which can even be caved out till only a thin shell emains, completely filled, o the chage q emoved. ans. q Q The technique of image chage: Use a chage distibution to ceate the same potential distibution on the suface of a closed space to facilitate the finding of potential (field) in the space. Example-5 A point chage is placed at distance d fom an infinitely lage conducto plate at V =. Find the potential, E-field, the suface chage density on the plate, and the foce on the chage. The E-field in the space of x < is zeo. So is V. Fo the space x >, the bounday of it is the plate at x =. The bounday condition is V(x = ) =. This is maintained by the induced chage on the plate which is so popely distibuted and the point chage q at x = d. -q q x

5 Put an image chage q at x = d will poduce the same bounday condition. So fo the E- field in x > the effect of the induced chage on the plate is exactly the same as the image chage q while emoving the plate. The foce on q is the same as the foce of point chage q q on it, i. e., F, which is attactive. ans. 4 4d As an execise, find the E-field and the chage density at the plate suface. Note that the image chage q cannot be applied to the x < egion, because its pesence changes the condition thee (Oiginally thee is no chage in that egion). 4 Seveal conductos, capacitance, capacitos Suppose thee ae thee conductos (,, ) in a closed space. The electic potential of the conductos ae V i (i =,, ) and the total chage on each is Q i. Then it can be shown that V P Q i j whee P ij = P ji depend on the shapes and the positions of the conductos only. If thee ae n conductos, then i and j un fom to n. Note that P fo two conductos alone usually will change when the thid conducto is added. With only one conducto, V = PQ, and C = /P is the capacitance of the conducto. If conducto- is inside conducto-, then P emains the same egadless of the pesence of othe conductos (use uniqueness theoem to poof). They fom a capacito. The potentials due to chages on othe conductos will be the same fo both conducto- and -. So P j = P j, j >. Also, P = P. Let Q = Q, then V =, and V = (P P )Q. The capacitance of the capacito is C = /( P P ). In geneal, to find the capacitance of a pai of conductos that fom a capacito, we put opposite chages on the pai of conductos and find thei potential diffeence. ij j (7), Total enegy: N W Q i Vi i (8). Fo a capacito, CV W (8a). Example-6 A vey small conducto is placed at distance d fom the cente of a conducto sphee with adius R (< d). Find the potential of the sphee when the small conducto caies chage q. Solution. Let the sphee be the conducto- and the othe conducto-, V = P q. But P = P =V /q. So the poblem is conveted to finding the potential of the small conducto when the sphee caies chage q, the answe of which is eadily available: V = kq/d ans. Example-7

6 As shown in the figue, a thin conducting cicula ing is placed at distance a fom the cente of a conducto sphee of adius R. The line joining the cente of the sphee and that of the ing is pependicula to the ing plane. The ing caies total chage q >. (Let k / 4 ) R a () When the sphee caies no net chage, what is its voltage? () When the sphee is gounded, what is the total chage on it? () When the voltage of the sphee is V, what is the total chage on it? (4) Compae () with (), what is the amount and diection of the change of foce acting upon the ing by the sphee? (5) Compae () with (), what is the amount and diection of the change of foce acting upon the ing by the sphee? Accoding to P = P, the answe of () is equal to the voltage of the ing when the sphee is caying chage q and the ing caies none. The distance between the edge of the ing to the sphee cente is a. Note that the space occupied by the ing is an equal-potential one when the ing is absent. Theefoe the pesence of the ing caying no chage will not change the chage distibution of the sphee suface. This is only tue when the ing plane is pependicula to the line joining the two centes. So the answe to () is V= kq/ a. (A chage distibution ( ) on the sphee suface poduces an E-field that cancels pecisely that by the chaged ing, so the inteio of the sphee is zeo-field eveywhee.) Fo (), the chage q on the sphee must poduce q / R V kq/ a, so q krq / a. Note that the chage q is unifomly distibuted on the sphee suface. Fo (), V kq / R kq/ a, so q can be eadily found. Fo (4), the diffeence in the amount of chage is q q, which is unifom on the suface, so kq( q q) a kaq( q q) the foce is F / a a ( a ) Fo (5), the diffeence in the amount of chage is q, which is unifom on the suface, so kaqq the foce is F. ans. / ( a ) 5 Dielectics Media The micoscopic oigin of this effect can be undestood if we conside insulatos as composed of closed packed, non-pola atoms, with each atom composed of negative and positive chages centeed at the same point. When the insulato is put unde an electic field, the atom

will be polaized the positive chage is pushed away a little bit by the electic field and the negative chage is pulled towads the electic field a little bit (see figue).

The layes of induced chage poduce an electic field E ' that opposes the electic field E coming fom the capacito chage.

7 will be polaized the positive chage is pushed away a little bit by the electic field and the negative chage is pulled towads the electic field a little bit (see figue). Notice that the insulato as a whole is neutal except at the suface, whee a laye of induced negative (positive) chage q is found next to the capacito plate with positive (negative) chage q. The layes of induced chage poduce an electic field E ' that opposes the electic field E coming fom the capacito chage. The total electic field between the plates E tot E E' is smalle than E, esulting in an weake voltage between the plates. Some dielectics (like wate) have molecules with pemanent electic dipole moments. In the absence of extenal electic field the dipoles ae pointing in andom diections. When they ae put unde an electic field, the dipoles ae aligned with the extenal field (see figue below). The net effect is simila to non-pola neutal atoms except that induced effective chage q is usually lage, and the total electic field E tot E E' is smalle. E-field induces dipoles. Polaization P = dipole moment pe unit volume. 5. Bound Chage Conside the small volume of aea a and length dx, the net chage Px within the volume is a( Px ( x dx) Px ( x)) a dx, so the bound x x P chage density is x b. x In geneal, conside any closed suface, the net chage within is Q b P S ds, its diffeential fom is then (compae to Gauss s Law) b P ; suface chage density n P, whee n is the suface nomal diection. b Define electic displacement D E P x + dx

8 Gauss s law becomes: Q f S D ds ; D f The foce on a test chage is still ee( ), and the electic potential is still given by Eq. (). Linea media: P E. So D ( E E (9) ) ε is pemeability (dielectic constant) 5. Bounday conditions: ; (a) D D f // // E E (b). Genealized uniqueness theoem: In a space containing conductos (shapes, positions, and potentials given) and dielectics (shapes, positions, and ε s all given), and fee chage distibution, the potential V( ) thoughout the space is then uniquely detemined. Image chages can be used to poduce the same bounday conditions. Ohm s law: J E (), J v is the electic cuent density, the mobile chage density. σ is the conductivity of the medium. Both σ and ε ae the mateial paametes of a medium. Electic cuent: I J S ds. So in geneal one has to define the suface befoe talking about cuent. Unde steady conditions (nothing changes with time but things can move), J. So J () J at an inteface (bounday) between two media. Example-8 At the bounday between two media (ε, σ, ε, σ ), the electic displacement D in medium- is known. Find D and the fee suface chage density. // D Using Eqs. (9), (a) and (), we fist have D //. Using Eq. (), we then D D have. Then use Eq. (b) to find the fee suface chage density. The electic fields and the cuent densities in both media can be found using Eqs. (9) and (). Ans.

9 Example-9 Conside a paallel capacito made of two lage metal plates of L by L sepaated by distance d (<<L) with a neutal dielectic slab (thickness a, same aea as the metal plates). The potential diffeence between the two plates is V. Find the amount of chage on the plates and enegy stoed in (a) and (b). Metal L Dielectic (a) (a) Since thee is no fee chage in the space between the plates, D is the same between the plates, as one can check that such D satisfies the bounday condition fo D. The E-field inside the dielectic is E = D/εε, and in the ai gaps E = D/ε. Let the uppe ai gap thickness be x, and that of the lowe ai gap be x, then Metal x Dielectic (b) V = E (x + x ) + ae = D[(d a) /ε + a]/ε Then the suface chage density is D fo uppe/lowe plates. Total chage Q = σl. The capacitance C = Q/V = ε L /[(d a)/ε + a] Enegy W =.5QV (b) In the x potion the answe of (a) can be applied. Fo the est pat σ = ε V/d. Total chage is Q = ε xl/[(d a)/ε + a] + ε L(L x)/d. dw W = QV/ depends on x. So the foce of the plates on the dielectic is F. F being dx positive means W inceases with deceasing x, i. e., the foce is pushing the slab out. ans. Example- As shown, half the conducto sphee (adius R) is buied in a dielectic medium. The sphee is held at potential V. Find V, E, fee and bound chage distibutions. R The potential is V ( ) V, i. e., as if the dielectic was not thee! Let us examine whethe all the bounday conditions ae met, which ae (i) V( R) V ; (ii) Eq. (a) at the dielectic/ai bounday, and (iii) D D at the dielectic/ai bounday. (i) is obvious. Note that with the potential given above, the E-field is paallel to the bounday, so (ii) and (iii) do hold. RV The E-field is then E( ).

10 The fee chage density of the uppe hemisphee is V, which is also equal to the total R chage density at the lowe hemisphee suface. RV The electic displacement D( ) in the lowe half of the space, and the fee chage density at the lowe hemisphee suface is V. R ( ) The bound chage density is V ans. R Example- A point chage q is placed at x = d in font of an infinitely lage dielectic medium filling the space of x <. Find eveything also. Patial solution: Again we use image chages. Fo E-field in x >, we put a chage q at x = d. Fo E-field in x <, put a chage q + q at x = d. The bounday conditions at x = ae again Eqs. (a) and (b) with zeo fee chage at the inteface. Thee bounday conditions detemine q in tems of q and ε (pemeability).

Flux. Area Vector. Flux of Electric Field. Gauss s Law

Flux. Area Vector. Flux of Electric Field. Gauss s Law Gauss s Law Flux Flux in Physics is used to two distinct ways. The fist meaning is the ate of flow, such as the amount of wate flowing in a ive, i.e. volume pe unit aea pe unit time. O, fo light, it is