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Unformatted text preview: 416. Circular ring of charge. (a) The potential at the origin is given by
Q2 02 '1’ 3 “3”" ““5 + 49ml? + 41mm) _f PIESW+ Q2 (h _1_) 0 4116ch +41reu R+ 2R
;(”’R+ﬂ)= 1 ( ”‘32)
Hm. R 2R «Tn ' where we have noted that p; = (El/(1R). Thus, (I! =_{i teqnirw Q; = —2Q1/3. (h) Noting that, due to symmetry, the halfring electric ﬁeld at the origin has no 3: or 3 components.
so that we can simply work with the 2 component of the ﬁeld. The 2 component ofthe elemental electric ﬁeld atthe origin duetoaehargeelemempidt’ alongthehaifdngie givenby
CH, 1' l'
'5“ sh13=—~1P‘ﬂsin(i) Iw‘hm 3 4mm 4neoR28 Fig. 4.7. Figure for Problem 416. which can be integrated to ﬁnd the total ﬁeld at the origin due to the halfring. i.e., ﬁ‘h‘G ) d“ anal/«r E f dE _
I 31h" m dﬂEoRzu 4111932
so thatthe total ﬁeld at the origin due to both the helf~ring and the two charges is =ZQI/r _ 02 _ _Q2_ 1 26.31 SQ:
we) “of?” ﬁg mot—MT! *omm [T‘T Setting E = 0 we ﬁnd Q; = 301/(5n). 418. Charge on a hemisphere. {:1} Considering the geometry shown, and using [4.281 widi 11"; = o,
r = 23 and R: r— 1"} a «la: +25 — ﬂeeces 3’, ﬂleelecttostetie potential atpointPis given by 1“ Lin we} I azsin 9w (pp: 9;sz =0 417—603 =(2ﬁ'aZJII4WE9) me w/a2+ 3? — Zozoas 6" where wehaveuoted thatps _ Q/(41m2) amide —. ”o. 2sin0'dﬂ’dqb’. The integml can heevaluated
by substitution of variables (e.g.. let u = £12 + 2'2 — Zazcosﬂ', in which case do =  sind'dﬂ")
or by looking up in a Table of Integrals or by other means (cg, using a madmnmtieai analysis
package such as MAII‘IIEMATICA). We ﬁnd Fig. 43. Figure ﬁJr thlun 44m (b) By symnmtry, we expeetlhe electric ﬁeld at point P to be in the z diredien, i.e., E1: = EEG).
WE: canﬁnd E(z) by taking the negative gradiolt of dip in the z direction, Le, 843(2)=_[1 5+3 1 Val+23 3(3) = _ HZ ”2 amid2 + 32 + M2 (c) We follow the same procedure, except that we now consider a volume charge density p ._
Sta/(2:17:33) and also mtcgrate over 1*. In other words, we have 32 _" '39—— __ 3am) m] f__ (r’)1 5111 B’dﬂ'dr’
(21753}{4ﬁ' £0) ran «)1 965 + 25 — 2a: c036"
Where the integration over e5“ 18 already dime. Looking up the integral from tables or using an
analysis package we ﬁnd “’9“ i T‘ 32 39 32 0.204 + 3;) (a? + 22F”
ﬁler$933 + The electric ﬁeld Ep = 313(3) and Eta) can onmagain he found as the negative gradient of @p in
the z direction. We ﬁnd [11
Wz)_ SQ [220,2 +M+W_ w] 3(3) = _ 33 _ 41:59:33 423. Spherical charge distribution. (a) The total charge Q can be mlculawd as Q= wp(r')d”’= [ﬁg] Mr’lr’z line’dr'dwe' r1301 rfsa n
=4“ '3' " at?
a3 0.3 811309113
=4 A w _ _
“m" [a 5 l 15 (b) Due to spherical symmetry. the electric ﬂux densiljr D. and hence E, are in the radial direction
and depend only on r. We consider a ephedtal Gaussian smfaee S of radius r altered at the
origin and apply Gauss“: law: Errpo’ra 317,0033
15 r<a f; eE  ds = sneeze“) = em = 3"}6 from which the alcctric ﬁeld failows as 2pm"
156.;
21:30:11.3
1559:"2 1'50, Er: 1")(1 (c) The. electric potential can be found from the electric ﬁeld as Mr) =— E all
an
Farr 1': a, this integral mulls in E
r
)0 3 3 r 3 r 3
ﬂu?) = — 2‘93“sz = _2ﬂnﬂ f (1—: = 2m“ [1] = 2m“
on 15601" 1560 w r 1550 r m 15:59:“ Forr («.mmteyalcanbeevaluawdas 3 race. 2 2 __ 2
ﬁﬂ=2pua _j Zﬂdr=2p0a _ﬂr a.
15604.; ﬂ 1560 1550 1550 2
_2ma‘+ﬁ_wi_ﬂﬁ=ﬂ_ﬂ=ﬂ “213]
15m 15m 1550 560 [Sea 559 3 {dTﬂleshelachesofbothErdeIPvursusrareshawninlTigmmlﬂ. h ________
a ___..___ Fig.4.10. Figuraﬁorl‘rahIm433. Erandlbmus r. ...
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