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By Jerrold Franklin

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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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