P260sheet - Ch 21 1 Coulomb's law F = k qrq2 r k = 2 ^ Ch...

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Ch. 21 Coulomb’s law: F = k q 1 q 2 r 2 ˆ r , k = 1 4 π 0 E = F q , so dE = k ˆ rdq r 2 Given uniform charge density: dq = Q V dV = ρdV for a volume, σdA for a sheet, λdl for a line Dipoles For q at d and - q at 0: P qd . F net = 0 τ = r × F = P × E = PE sin θ U = - P · E Ch. 22 Φ = E · dA = q encl 0 . For a closed surface, A points outward. is the integral over a closed surface. E ( X ) = 1 0 X 0 ρ ( x )( x X ) D - 1 dx , where D is the dimen- sion of the symmetry (1 slab, 2 cylindrical, 3 spheri- cal). ring: E = kQx ( x 2 + a 2 ) 3 / 2 (a radius, x distance) line: 1 2 π 0 λ x x 2 /a 2 +1 (2a length, x distance) line: E = λ 2 π 0 r sheet: E = σ 2 0 . (Think one end (half) of a Gaus- sian cylinder.) disk: E = σ 2 0 (1 - 1 R 2 /x 2 +1 ) parallel plates: E outside = 0; E between = σ 0 insulating sphere: E inside = kQr R 3 A sphere = 4 πr 2 Conductors: A conductor is enclosed by an equipo- tential surface, so E = E at surface. E inside = q inside = 0. Ch. 23 F is a conservative force if W a a = a a F · dl = 0. W a b is path-independent. ≡ ∃ U : F ( r ) = -∇ U ( r ) U + K is constant Δ U = - W for a slow displacement from rest to rest. U ona = Σ U iona Δ U = - f i qE · dl Δ V Δ U q ; E ( r ) = -∇ V ( r ) ∇ × E = ∇ × F = 0 For V = 0 at : V pt ( r ) = kq r = Δ Er ; V dist ( r ) = charge kdq r ring: V = 1 4 π 0 Q x 2 + a 2 line: V = 1 4 π 0 Q 2 a ln( a 2 + x 2 + a a 2 + x 2 - a ) line: V = λ 2 π 0 ln( r 0 r )
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