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Gauss’ law - spherical symmetry Submitted by: I.D. 039023262 The problem: 1. Find the electric field of a point charge 2. Find the electric field from a spherical shell of a radius R charged with a charge Q uniformly. The solution: For a point source the charge density is ρ(~r) = Qδ(~r). Then I Z ~ · d~s = 4πk ρdv E s V Z 2 Er · 4πr = 4πk Qδ(~r)dV = 4πkQ Er = kQ r2 (1) (2) (3) ~ where R ~ = Rr̂. For a spherical shell the charge density is ρ(~r) = Qδ(~r − R) Z Z ρdV = V ( 0, ~ Qδ(~r − R)dV = 1, r<R r>R (4) Then ~ = E ( 0, kQ r̂, r2 r<R r>R (5) 1 Gauss’s law - plane symmetry The problem: Find the electric field for: 1. infinite uniformly charged plane (charge density σ) 2. infinite uniformly charged layer (charge density ρ and width h) 3. two adjacent infinite layers of width h charged ρ+ and ρ− uniformly 4. two infinite planes perpendicular to each other, both charged σ uniformly. The solution: H R ~ · d~s = 4πk ρdv, and knowing that for plane symmetry E ~ = Ez ẑ 1. According to Gauss’ law E we construct a Gauss shell enclosing a piece of the plane of area S: I Z ~ · d~s = 4πk ρdv E (1) 2Ez · S = 4πkσS ~ z = 2πkσ · sign(z)ẑ E (2) (3) 2. For infinite layer of the width h (0 < z < h), if z > h or z < 0 the solution is identical to the one for the infinite plane. In the case 0 < z < h we construct a shell of area S symmetrical relatively to the middle of the layer - its its bottom plate is at z, the top plate is at h − z and the height, therefore, is h − 2z. 2E · S = 4πk(h − 2z)Sρ (4) E = 2πkρ(h − 2z) (5) z>h 2πkρh, = 2πkρ(2z − h), 0 < z < h −2πkρh, z<0 (6) Then Ez 3. Two infinite layers of thickness h charged ρ+ and ρ− uniformly. We choose the ρ+ charged layer to be at 0 < z < h and the ρ− charged layer to be at −h < z < 0. Then using the result of for the one layer and the superposition principle we obtain: 2πk(ρ+ + ρ− )h, z>h 2πkρ (2z − h) + 2πkρ h, 0<z<h + − Ez = (7) −2πkρ h + 2πkρ (h + 2z), −h < z < 0 + − −2πk(ρ + ρ )h, z < −h + − 4. Two infinite plains perpendicular to each other, both charged σ uniformly. By using the principle of superposition, we simply calculate the field intensity of each plane, and ~ = 2πkσ(sign(z)ẑ + sign(y)ŷ) (assuming that the plains in combine them. For two planes we get E question are the XY and XZ). 1 The electric field Submitted by: I.D. 313947202 The problem: An electrical infinite straight wire with charge density λ is at the center of an infinite hollow cylinder with charge density σ and a radius a. What is the electrical field everywhere? The solution: The solution is superposition of two fields: one from wire and another one from cylinder. Both fields we will find with the Gauss’ law. Because of symmetry,the field are in the radial direction only. Field of the wire: E · 2πrl = 4πkQ = 4πλl 2λk E = r (1) (2) Field of the cylinder: inside: E · 2πrl = 4πkQ = 4π0 (3) E = 0 (4) outside: E · 2πrl = 4πkQ = 4πk(2πal)σ 4πkaσ E = r (5) (6) The field from the whole system: ( 2λk r<R r r̂, ~ = E 4πaσ 2λk r r̂ + r r̂, r > R (7) 1 Electric field Submitted by: I.D. 060645009 The problem: • Calculate directly the electric field flux through an infinite plane a distance d from a point charge. • Explain why your result does not depend on d. The solution: Let the plane to be at the x − y plane, so that any point on it is ~r = (r cos(ϕ), r sin(ϕ), 0) and the point charge is at ~r0 = (0, 0, d). By the Coloumb law kdq(~r − ~r0 ) |~r − ~r0 |3 ~ = dE (1) In our case ~r − ~r0 = (r cos(ϕ), r sin(ϕ), −d) 0 2 (2) 2 1/2 |~r − ~r | = (r + d ) (3) So that ~ = E (r2 kq (r cos(ϕ), r sin(ϕ), −d) + d2 )3/2 The flus is I Z ~ = ~ · dA E Ez dA Z 2π Z = dϕ 0 Z = 2πkqd 0 (4) (5) ∞ 0 ∞ rdr kqd 3 (r2 + d2 ) 2 rdr 3 = 2πkq (r2 + d2 ) 2 (6) (7) The total flux from a point charge if 4πkq. Since the plain is infinite, half of the electric field lines pass through this plane (not depending on d), that is half of the flux 2πkq. 1