problemset2

2 statement with z component only symmetry and with dv

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Unformatted text preview: r r r −ˆr� r i = − →� 2 |→ − − | r r B Follows from (A) immediately by substitution. Remember � is derived in terms of unprimed x, y, z . � does not affect x� , y � , z � . C → Φ(− ) = r → −� � V� → ρ(− � )dV � r →− 4πε0 |− − →� | r r C C ρ( r ) = charge density in m3 . We have λ in units of m . In this sense, ρ → ∞ at the ring. We can represent � → this in cylindrical coordinates by ρ(− ) = λ0 δ (z )δ (r − a). Then we can evaluate the triple integral r ��� λ0 δ (z )δ (r − a)rdrdφdz = −− 4π�0 |→ − →� | r r 2π � 0 λ0 adφ 4π�0 |r − r� | But, we can skip that unnecessary work by simply considering infinitesimal charges (adφ)λ0 around the ring. y x dφ a adφ Figure 2: A ring of line charge with infinitesimal charge elements dq = λ0 adφ. (Image by MIT OpenCourseWare.) We only care about z axis in this problem as well, so, by symmetry, there is no field in the x and y directions. 4 Problem Set 2 → Φ(− ) = r → Φ(− ) = r 6.641, Spring 2009 2π � λ0 (adφ) 4πε0 0 1 (a2 + z 2 ) 2 � �� � distance from the charge element λ0 adφ to the point z on the z -axis λ0 a 1 2ε0 (a2 + z 2 ) 2 on the z -axis. ⎛ 0 0 ⎞ ∂� ∂⎟ → − ⎜ ∂� − E = −�Φ(→) = − ⎝ˆx � + ˆy � + ˆz Φ⎠ r i Φi Φi ∂x ∂y ∂z � � ∂ → − E = −ˆz i ∂z � → −ˆ E = iz � aλ0 z λ0 a 1 2ε0 (a2 + z 2 ) 2 3 2ε0 (a2 + z 2 ) 2 → Using the equation from the Problem 2.2 Statement with z component only (symmetry) and with ρ(− � )dV � → r λ0 adφ � 2π λ0 adφ cos θ z Ez (z ) = , cos θ = 1 4πε0 (z 2 + a2 ) (a2 + z 2 ) 2 0 � 2π = 0 = λ0 az (a2 + 3 z2) 2 dφ 4πε0 λ0 az 3 2ε0 (a2 + z 2 ) 2 Limit |z | → ∞ � a2 + z 2 → |z | Φ(z ) ≈ λ0 a 2ε0 (a2 + 1 z2) 2 ≈ 2πλ0 a Q ≈ 4πε0 |z | 4πε0 |z | Q = 2πλ0 a (total charge on loop). Φ(z ) looks like potential from point charge in far field. �Q z>0...
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