Practice Problem Set 3

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1 Prctice Problem Set 3 #1. A dipole nd chrge A dipole with chrges ±q nd seprtion 2 is locted distnce from point chrge Q, oriented s shown in Figure of the tetbook (reproduced here for convenience. For both nd b, consider the limit. Wht is the net torque (mgnitude nd direction on the dipole? We use the formul derived in the tetbook eressing the torque on dipole in terms of its electric dipole moment nd the electric field t its position: τ = p E. (1 In terms of the coordintes shown on the figure below, the field generted b the point chrge t the loction of the dipole is The dipole moment, on the other hnd, is so the torque is given b E = kq 2 i. (2 p = q(2 j, (3 τ = 2q j kq 2 i = 2Q 2 ( k, (4 where k = i j points out of the pge. This indictes tht the torque tends to rotte the dipole in the clockwise direction, in ccord with our intuition. b Wht is the net force (mgnitude nd direction on the dipole? We begin b not et ssuming, nd treting ech consituent of the dipole one t time. Ech constituent of the dipole is subjected to n electric force from the monopole s field, s shown in the djcent figure. Though the net force on the dipole in the direction cncels ectl, it does not in the direction. Insted, it is given b F = 2 Q 2 sin(θ. (5 + 2

2 But since we rrive t sin(θ = F = 2 + 2, (6 kq(2q ( (7 3/2 We cn now invoke the limit. In the present cse, the leding-order behviour (i.e. the most significnt contribution is given b the first term in the smll-u epnsion (1 + u α 1 + αu, so we onl retin tht term: ( /2 = 3 ( / }{{} higher-order term 3. (8 This cn be compred to #2 c, in which both terms in the epnsion must be retined, becuse the first is cncelled b some other term. Hence, we find F = kq(2q 3 j. (9 c How could ou hve ou hve predicted it bsed on other results from the chpter? This result could hve been forseen using Newton s third lw (which electromgnetism obes. Indeed, we know the field generted b dipole ( E = k p/r 3 on the perpendiculr bisector covered in the tetbook, nd we know tht the monopole eperiences force F = Q E. According to Newton s third lw, the dipole must feel nd equl nd opposite force due to the monopole. #2. A Chrged Cross, or Prctice With Integrls Consider thin rod of length 2 plced long the is nd centered t the origin (see Figure 1. Considering the rod hs uniforml distributed chrge q (where q > 0, find the electric field t the point P t distnce from the origin on the positive side of the is (ssume

3 P P P Figure 1 Figure 2 Figure 3 >. Does our result gree with our epecttions in the limit? B smmetr, the field t point P is directed long the is, nd clerl points w from the chrge distribution. Furthermore, in this cse, there re no sine or cosine fctors to include, becuse the entire chrge distribution is long this is. We set up the integrl: dq E (P = k r 2, (10 where r is the distnce bewteen P nd the infinitesiml chrge dq. We let dq = λd (where λ = q/2 is the line chrge densit, in which cse r =, nd the integrtion rnges from to : d E (P = kλ ( ( 2 1 = kλ ( 1 = kλ 1 + = kλ , (11 where we hve let 2λ = q. In the limit, we m let (which, s in #1 b, mounts to onl keeping the leding-order term in the Tlor epnsion, in which cse the field reduces to tht of point chrge q locted t the origin, s epected. b Consider thin rod, lso of length 2, plced long the is nd centered t the origin (see Figure 2. The rod hs uniforml distributed chrge q (where q > 0. Find the electric field t the point P t distnce from the origin on the positive side of the is. Does our result gree with our epecttions in the limit? Agin, b smmetr, the field t point P is directed long the is, though now there is trigonometric fctor to pick out the component of the field generted t point P. The integrl is s follows: E (P = k dq cos(θ. (12 r2

4 This time, the integrl is over between nd, the line chrge densit is λ = q/2, nd r is given b Pthgors theorem. Furthermore, cos(θ must be epressed in terms of the integrtion vrible: cos(θ = (13 The integrl hence becomes E (P = kλ = kλ d ( d. (14 ( /2 The integrl is evluted either vi trigonometric substitution or consulttion of n integrl tble: E (P = kλ = kλ ( ( = kλ = k( q 2 + 2, (15 where in the lst step we let 2λ = q. In the limit, once gin, we onl retin the ledingorder term in the Tlor epnsion: = }{{} higher-order term, (16 in which cse the field reduces to tht of point chrge q locted t the origin, s epected. c Now, suppose these two metl rods re positioned s shown in Figure 3 (ssume no chrge is echnged, with the negtivel chrged rod plced horizontll. Wht is the field t the point P t distnce from the origin, on the positive side of the is? Tr tking the limit. In this limit, with wht power of does the electric field dec? Is this behviour reminiscent of specific chrge distribution? We cn use the principle of superposition to combine the nswers from prts nd b nd

5 find the electric field t P in this sitution: E (P = k( q ( ( ( 2 2 ( ( = ( (17 Now we consider the limit. In the denomintor, s for the cses bove, keeping ledingorder behviour mounts to simpl letting 2 2 2, nd However, cre must be tken in the prentheses to the right, s this is n sitution. We use the pproimtion (1 + u α 1 + αu, vlid for smll u: E (P = ( 2 ( ( (18 Note how the lrger term from the epnsion, 2, disppers in cncelltion, so now the ledingorder term comes (in prt from the second term in the epnsion. We see tht in this limit, the field decs s 4 this is reminiscent of qudrupole, s ou will discover in this week s ssignment, nd indeed, squinting little mkes the cross in Figure 3 resemble qudrupole 1. 1 Disclimer: squinting is generll not good w to do phsics.

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