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1 MIT OpenouseWe 6.1/ESD.1J Electomgnetics nd pplictions, Fll 25 Plese use the following cittion fomt: Mkus Zhn, Eich Ippen, nd Dvid Stelin, 6.1/ESD.1J Electomgnetics nd pplictions, Fll 25. (Msschusetts Institute of Technology: MIT OpenouseWe. (ccessed MM DD, YYYY. License: etive ommons ttibution- Noncommecil-She like. Note: Plese use the ctul dte you ccessed this mteil in you cittion. Fo moe infomtion bout citing these mteils o ou Tems of Use, visit:

2 6.1/ESD.1J Electomgnetics nd pplictions Fll 25 Poblem Set 4 - Solutions Pof. Mkus Zhn MIT OpenouseWe Poblem 4.1 y Guss lw, D = (εe = ρ = fo < < b. In cylindicl coodintes, so 1 (εe = (εe =, 1 (t (εe = = (εe = = E ( =. (t only depends on time nd not on dius. so b b ( (t b v(t = E ( d = d = (t ln, v(t v(t (t = ( b = E ( = ln (. b ln Resistnce R = V/i, J = σe = J = σe = σv(t ln ( b ( b σv(t 2πlσv(t v ln i = J d = 2πl = = R = = ln ( b ln ( b i 2πlσ pcitnce = Q/V. y Guss lw boundy condition, the sufce chge density is D σ s = [ εe ( = + εe ( = ] = εe + ( = = 2πlεv(t 2πlε Q = 2πlσ s = ( b = = ( b ln ln y mpee s lw, H =. In cylindicl coodintes, εv(t ( b ln 1 H φ 1 H φ H = (H φê z ê = = (H φ =, = z z 1 (t (H φ = = (H φ = = H φ ( = 1

3 Poblem Set 4 6.1, Fll 25 E Use mpee s lw: (t I(t I(t H ds = I = 2π = I(t = (t = = H(, t = ê φ 2π 2π F Inductnce L = Φ/I b ( ( I(t µi(tl b µl b Φ = ds = l µ d = ln = L = ln 2π 2π 2π S G ( b ln 2πlε ε R = =, the sme R s pllel pltes. 2πlσ ln ( b σ H ( µl b 2πlε L = ln = µεl 2, the sme L s pllel pltes of depth l. 2π ln ( b The speed of light in the mteil is c m = 1/ µε, so L = l 2 /c 2 m. Poblem 4.2 ρ J + = = J = in D stedy stte. t J x J = = J x = J constnt. x σ(xe(x = J = E(x = σ e x/s s s s J x/s Js E(x dx = e dx = e x/s σ σ = V J s V σ (1 e 1 = V = J = σ s(1 e 1 V E(x = (1 e 1 se x/s J V V s(1 e 1 R = = = i J ld ldσ de ρ f = ε = εv e x/s dx (1 e 1 s 2 εv σ f (x = = εe(x = = (1 e 1 s εv σ f (x = s = εe(x = s = (e 1 s 2

4 Poblem Set 4 6.1, Fll 25 s εv e x/s s εv x/s ldεv Q v = ld dx = ld e dx = (1 e 1 s 2 (1 e 1 s 2 s εv Q s (x = = (1 e 1 s ld εv Q s (x = s = ld (e 1s ldεv εv εv Q totl = Q v + Q s (x = + Q s (x = s = + ld + ld = s (1 e 1 s (e 1s Poblem 4. x ρ t/τ f (t = ρ e, τ = s ε σ E x ρ f (t x 2 E = = = E x = ρ e t/τ + (t x ε 2εs s 2 E x dx = ρ e + (ts = = (t = ρ e 6ε 6ε ( 2 2 x t/τ e + ρ 2εs 6 sε e t/τ 1 t/τ 2 s E x = ρ = ρ e x 2εs σ f (x = = εe(x = = ρ e 6 σ f (x = s = εe(x = s = ρ e D i(t E x σ 1 = σe x (x = s + ε (x = s = ρ t ε e τ ρ e = Poblem Φ 2 Φ 2 Φ(x, y = x 2 + y 2 =, d 2 X(x d 2 Y (y Y (y dx 2 + X(x dy 2 = Φ = X(xY (y 1 d 2 X(x 1 d 2 Y (y Renging + = X(x dx 2 Y (y dy }{{}}{{ 2 } function of x only function of y only

5 Poblem Set 4 6.1, Fll 25 The only wy the two tems cn dd to zeo fo evey x nd y vlue is if 1 d 2 X(x 1 d 2 Y (y X(x dx 2 = k 2, Y (y dy 2 = k 2 Fo k 2 =, we hve d 2 X(x d 2 Y (y =, nd = dx 2 dy 2 Theefoe, { X(x = x + b Y (y = cy + d = Φ = xy + x + y + D, whee we hve used Φ = X(xY (y, nd, b, c, d,,,, D e bity constnts. oundy onditions: ; x = (1 Φ(x, y = ; y = (2 V ; xy = b ( Φ(x, y = xy + x + y + D (we know Φ(x, y is of this fom oundy condition (1 Φ(x =, y = = y + D =. This hs to hold fo evey vlue of y. This mens tht = nd D =. oundy condition (2 Φ(x, y = = = x + D =. We ledy know tht D =, so x =. This hs to hold fo evey vlue of x, so =. oundy condition ( Φ(x, y such tht xy = b = V. We know D =, =, =, so Φ(x, y = xy on the boundy xy = b. Φ(x, y = b = V = = V b V Φ(x, y = xy b D Φ Φ Φ V V E 1 = Φ = xˆ ŷ ẑ = y xˆ x ŷ x y z b b We use the boundy condition nˆ [E 1 E 2 ] = σ s on the x = plne nd the noml nˆ = xˆ. ε V xˆ [ε 1 E 1 ε 2 E 2 ] = σ s = σ s = +ε 1 E 1,x = y }{{} b b/c pefect conducto On the y = plne, nˆ = ŷ nd E ε V ŷ [ε 1 E 1 ε 2 2 ] = x = σ s b 4

6 Poblem Set 4 6.1, Fll 25 E F dy x = = y dy = x dx = dx y 2 y = x = 2 (y x = y x = (y x 1 1 = +k 2 nd = k 2 d 2 X(x d 2 Y (y X(x dx Y (y 2 dy 2 The solution is X(x = e + e Y (y = sin(ky + D cos(ky whee,,,, nd D e bity constnts. Φ(x, y = X(xY (y = [ sin(ky + b cos(ky]e + [c sin(ky + d cos(ky]e whee, b, c, nd d e bity constnts. G Φ 1 (x, y = [ 1 sin(ky + b 1 cos(ky]e + [c 1 sin(ky + d 1 cos(ky]e Φ 2 (x, y = [ 2 sin(ky + b 2 cos(ky]e + [c 2 sin(ky + d 2 cos(ky]e Region (1 is fo x nd Region (2 is fo x. oundy onditions: (1 Φ 1 (x, y = ; x (2 Φ 2 (x, y = ; x ( Φ 1 (x, y x= = Φ 2 (x, y x= = V sin(y oundy condition (1 = no e tems fo Φ 1 (x, y becuse they blow up s x, so Φ 1 (x, y = [c 1 sin(ky + d 1 cos(ky]e. oundy condition (2 = no e tems fo Φ 2 (x, y becuse they blow up s x, so Φ 2 (x, y = [c 2 sin(ky + d 2 cos(ky]e. oundy condition ( = c 1 sin(ky + d 1 cos(ky = c 2 sin(ky + d 2 cos(ky = V sin(y. lely c 1 = c 2 nd d 1 = d 2 becuse sine nd cosine e independent (you cn t mke sine equl cosine fo ll y. Tht sid, c 1 = c 2 = V, d 1 = d 2 =, k = Φ 1 = V sin(ye x ; x Φ 2 = V sin(yex ; x 5

7 Poblem Set 4 6.1, Fll 25 H ( ( ( Φ Φ Φ E 1 = Φ = xˆ + ŷ + x y ẑ x x E 1 = V sin(ye x xˆ V cos(ye x x E 2 = V sin(ye xˆ V cos(ye ŷ ŷ To find the sufce chge we need to use the condition nˆ [ε 1 E 1 ε 2 E 2 ] = σ s t x =, whee nˆ = xˆ = ε 1 E 1,x x= ε 2 E 2,x x= = σ s. E 1,x x= = V sin(y, E 2,x x= = V sin(y Fo x >, σ s = V sin(y(ε 1 + ε 2 E 1 = V e x [sin(y xˆ cos(y ŷ] dy cos(y sin(y = = cot(y = dx = dy dx sin(y cos(y Let u = cos(y so tht du = sin(y dy : 1 du 1 1 dx = + = x = + ln(u + = + ln(cos(y + u 1 = x ln(cos(y ( 1 cos(y (x x = + ln cos(y 6

General Physics II. number of field lines/area. for whole surface: for continuous surface is a whole surface

General Physics II. number of field lines/area. for whole surface: for continuous surface is a whole surface Genel Physics II Chpte 3: Guss w We now wnt to quickly discuss one of the moe useful tools fo clculting the electic field, nmely Guss lw. In ode to undestnd Guss s lw, it seems we need to know the concept