problemset2

2 statement with z component only symmetry and with dv

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 03/19/2014 for the course EECS 6.641 at MIT.

Ask a homework question - tutors are online