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The electric field Submitted by: I.D. 040460479 The problem: An isolated electrical wire is charged uniformly with charge q and bent into a circular shape (radius R) with a small hole b R (where b is the arc length). What is the electrical field in the middle of the circle? The solution: The simple solution is to use superposition. The electric field in the middle of a complete ring is, of course, zero. Now we’ll sum up the field of a complete ring with the field of a very small wire of the same size and shape as the hole but with a negative charge. ~ = kλb x̂ Because b R the wire can be taken as a negative point charge and, therefore, the field is E R2 q (when we take the hole to be on the X axis and λ = 2πR ). It is also possible to calculate the field directly. Taking ~r = (0, 0, 0) and ~r0 = (R cos θ, R sin θ, 0) we have kdq (−R cos θ, −R sin θ, 0) R3 dq = λRdθ Z −α kλ kλRdθ0 ~ (−R cos θ0 , −R sin θ0 , 0) = (2 sin α, 0, 0) E = 3 R R α ~ = dE (1) (2) (3) where ±α are the angles of the edges of the bent wire (or, the upper and lower limits of the hole). Since b R we can approximate tan α ' sin α = b/2 R . Substituting into the expression for the field we obtain ~ = kλb x̂ E R2 (4) 1 BGU Physics Dept. Physics 2 for physics students Cylinder and dipole 1. Volume charge density of sphere- ρ = Q/πR2 L. The electric field on the z axis is 2π Z Ez = Z LZ R ρ(b − z)rdrdzdφ = (r2 + (b − z)2 )3/2 p p 2Q 2 + (b − L)2 − 2 + b2 L + R R πR2 L 0 = 0 (1) 0 (2) 2. The volume charge density is ρ = ρ0 cos φ. The straightforward way to solve this problem is to write down the expression of the field and expand it in a taylor series. However, we know that far away from the cylinder, b L, R, this charge distribution looks like a dipole in the x̂ direction. So there is no point in going through all the derivation again. We just need the dipole moment p~ and then field is ~ = p~ · r̂ r̂ − p~ E r3 r3 (3) The dipole moment is Z Z ρ(~r)~rd3~r = ρ0 cos φ(r cos φ, r sin φ, z)rdrdφdz = Z L Z Z 2π πR3 Lρ0 x̂ = x̂ρ0 cos2 φr2 drdφdz = 3 0 0R 0 p~ = (4) (5) We want the field at ~r = bẑ, so 3 ~ ≈ − πR Lρ0 x̂ E 3b3 (6) 1