Download Dieux et héros de l'Antiquité - Toute la mythologie grecque by Odile Gandon PDF

By Odile Gandon

D'Achille à Zeus en passant par Héraclès, Jason, Médée ou bien encore Ulysse, ce dictionnaire présente les principales divinités grecques et latines par ordre alphabétique. Chaque become aware of se présente comme un véritable récit obtainable aux jeunes lecteurs : des renvois, des index et des cartes en facilitent los angeles session.

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The mathematical skills obtained in chapter 1 will be applied here to analyze different charge distributions in Cartesian, spherical, or cylindrical coordinates. 85 × 10−12 C2 ) Nm2 is the permittivity of free space. 2 Electric field In general, for a volume charge density ρ(r ⃗ ), the electric field at r ⃗ is given by JG 1 ρ(r ⃗′) rˆ dτ′ . 1088/978-1-6817-4429-2ch2 1 4πεo ∫S σr(r2⃗′) rˆ da′. 2-1 ª Morgan & Claypool Publishers 2016 Electromagnetism For a linear charge density λ(r ⃗ ), the electric field is given by JG 1 λ(r ⃗′) rˆ dl′ .

Therefore, JG E = ρ 4πεo 2π R h ∫ ∫ ∫ 0 0 0 (d + h − z )s zˆ ⎡ s 2 + (d + h − z )2 ⎤3/2 ⎣ ⎦ dz ds dϕ 2πρ ⎡ 2 R + d 2 − R2 + (d + h)2 + h⎤⎦zˆ 4πεo ⎣ JG ρ ⎡ 2 E = R + d 2 − R2 + (d + h)2 + h⎤⎦zˆ . 2εo ⎣ = 2-6 Electromagnetism Note if R ≫ d and R ≫ h, the field reduces to JG ρh E = zˆ , 2εo which is the field given by an infinite sheet of surface charge σ = ρh. 4. Given the bottom hemisphere of a spherical shell of radius R , thickness d , and volume charge density ρ, find the electric field z above the center (above the open part, ignoring edge effects).

1 Since 2 ∈ (1, 3) and f (x ) = 2x 2 − x + 4, we have 3 ∫ ( 2x 2 − x + 4)δ(x − 2)dx = f (2) = 2(2)2 − 2 + 4 = 10. 1 b) 1 ∫ ( x 2 + 4)δ(x − 2)dx. −1 Since 2 ∉ ( −1, 1), we have 1 ∫ ( x 2 + 4)δ(x − 2)dx = 0. −1 c) 6 ∫ 2 ⎛ 3x ⎞ sin ⎜ ⎟δ(x − π )dx . ⎝ 2 ⎠ 1-31 Electromagnetism 3x Since π ∈ (2, 6) and f (x ) = sin( 2 ), we have 6 ⎛ 3x ⎞ ⎛ 3π ⎞ sin ⎜ ⎟δ(x − π )dx = f (π ) = sin ⎜ ⎟ = −1. ⎝ 2 ⎠ ⎝2⎠ ∫ 2 d) 2 ∫ ( 2x3 + 1)δ(4x)dx. −2 Since 0 ∈ ( −2, 2) and f (x ) = 2x 3 + 1, we have 2 ∫ ( 2x3 + 1)δ(4x)dx = −2 e) 1 1 1 2(0)3 + 1) = .

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