PX3008 Problem Sheet 1

Transcription

1 PX38 Poblem Sheet 1 1) A sphee of dius (m) contins chge of unifom density ρ (Cm -3 ). Using Guss' theoem, obtin expessions fo the mgnitude of the electic field (t distnce fom the cente of the sphee) in the egions < nd >. ) () Using Guss' theoem, obtin n expession fo the electic field ne to plne metl sufce which hs sufce chge of σ Cm -. [Use the fct tht the field within the metl is zeo.] (b) Obtin n expession fo the electic field between two oppositely chged metl pltes being sufce chge densities ±σ Cm - on the fcing sufces. [pllel plte cpcito] [Answe: E = σ/ε o ] (c) The electic field between the pltes of pllel plte cpcito is 1kV/mm [1 Vm 1 = 1 NC 1 ]. If the pltes e mde of coppe clculte the popotion of coppe toms t the sufce of the positive plte tht e Cu + ions. [Assume tht the sufce lye of coppe toms is sque y with tomic seption.36nm.] 3) (i) Find the electic field E i between, nd the field E o outside of, pi of coxil cylindes of dii 1 nd being line chge densities ±λ Cm -1 espectively. (ii) Evlute E i t = 1cm (ssumed to be in the gp) when λ =.56 μc m -1. [Ans: 1 6 NC -1 ] 4) Find the flux of E though ech of the six fces of cube of side 1cm when chge of 3 μc is locted: (i) t the cente of the cube. (ii) t cone of the cube. PX38pob(1) 1/1

2 PX38 Solutions to poblem sheet 1 1) Fo the dotted sphee of dius <, the electic field is noml to the sufce (by symmety) nd the flux of E is E() 4π. The chge enclosed by the sufce is 4 3 π3 ρ. So, Guss theoem gives E() 4π = 4 3 π3 ρ/ε ο. So, fo < E() = ρ 3ε Fo > E() 4π = 4 3 π3 ρ/ε ο, i.e. E() = Note tht both expessions give the sme vlue t =, s they must. The vition of E is s shown. o 3 ρ 3ε o ) () Close to the sufce of the metl the E field is noml to the sufce. Fo the imgined Gussin cylinde: Thee is no flux of E though the cuved sufce. Thee is no flux though the flt sufce inside the metl. The totl flux of E is theefoe E P A, whee E P is the mgnitude of E t P. The totl chge inside the Gussin sufce is tht which lies on the bit of the metl sufce within the cylinde. This is σ A. Theefoe, by Guss theoem: E P A = σ A/ε o, so E P = σ/ε o. (b) The gument in pt () cn be epeted exctly. So E P = σ/ε o, s befoe. Note tht E P is independent of the distnce of P fom the sufce, nd so is constnt between the pltes. (c) E = σ/ε ne chged conducting sufce gives: σ = ε E = Cm - = poton chges/m. PX38soln(1) 1/

3 The sufce density of coppe toms is 1/( ) = m -. So theε is one positive coppe ion fo evey coppe toms. 3) (i) Dw cylindicl Gussin sufce betwen the two conducting cylindes By the usul guments: E() πl = λl/ε o So E() = λ πε o If the Gussin cylinde hs dius >, then: totl chge enclosed =. So: E() = fo ll points outside the coxil conducto. (ii) Fo = 1 cm (in gp) nd λ =.56 Cm 1, E = π NC 1. 4) Accoding to Guss' theoem the flux of E out of the closed sufce is equl to (the totl chge enclosed)/ε. (i) Since the chge is t the cente, ll fces e equivlent nd the flux though ech fce is 1/6th of the totl = Vm. [The e of the fce is ielevnt.] (ii) The flux though the thee fces, on which the chge lies, is zeo. The flux though the othe thee fces is ¼ of esult (i), i.e Vm. PX38soln(1) /

4 PX38 Poblem Sheet 1) Two flt, pllel, closely spced metl pltes of e.8m cy totl chges of -.1μC nd +3.8μC. Find the sufce chge densities on the inne nd oute sufces of ech plte. Assume tht the extenl electic fields hve equl mgnitudes to the left nd ight sides of the pltes. ) A sphee of dius R nd unifom chge density +ρ (Cm -3 ) contins spheicl cvity of dius R/ s shown. Find the electic field in the cvity. [This is poblem in supeposition. Add the fields due to complete sphee (R), of chge density +ρ, nd futhe sphee (R/), of chge density -ρ. The E field is constnt in the cvity. If you cnnot stisfy youself of this, then wok out the field t chosen point in the cvity. Answe: E = ρ R in 6ε the diection shown] 3) The electic field E() t position eltive to point chge q is q E () = ˆ 4 πε whee = nd ˆ =. Show tht E() stisfies.e() = t ll points othe thn =. 4) Show tht the electic field between the conductos of coxil cble stisfies.e() =. 5) The digm shows pllel plte cpcito with chge densities ±σ (Cm - ) on the metl fces. Dw sketch gph of the vition of.e long the imginy line GH (tke this to be the x-diection). PX38pob() 1/1

5 PX38 Solutions to poblem sheet 1) σ 1 + σ = α whee α = (-.1μC/.8m ) σ 3 + σ 4 = β whee β = (+3.8μC/.8m ) Fo Gussin sufce X thee is no flux so: σ + σ 3 = So we hve 3 equtions in the 4 unknowns. The finl eqution comes fom the fct tht the extenl field E A hs the sme mgnitude s, nd opposite diection to E B. Gussin sufces Y nd Z give E A = σ 4 /ε nd E B = σ 1 /ε nd so σ 1 = σ 4. Solving the 4 equtions gives: σ 1 = 1.6μCm -, σ = -36.9μCm -, σ 3 = 36.9μCm - nd σ 4 = 1.6μCm -. ) Conside the poblem s supeposition of two sphees of chge, the lge with chge density ρ nd the smlle with -ρ. The electic field inside unifomly chged sphee t distnce fom the cente is dil nd hs mgnitude ρ [pove this] 3ε A is the cente of the lge sphee, so the E field t B due to this complete sphee is epesented by vecto popotionl to AB (pointing wy fom A fo ρ>). C is the cente of the smll sphee, so the E field t B due to this complete sphee is epesented by vecto popotionl to BC (pointing towds C fo ρ<). The esultnt E field t ny point B within the smll sphee is lwys epesented by vecto popotionl to AB + BC = AC. Since AC is fixed vecto the esultnt field E is constnt. PX38soln() 1/

6 3) E =.E = q q q ˆ = = (xi+ yj + z k ) 4πε 4πε 4πε 3 3 o o o E E x y E + + x y z z q x Ex q 1 x 1 E x = so = x 4πε o x 4 o x x πε x 1 3 Now = 1 nd x x 3 = 4 x nd since = x + y + z then = x nd so = x x Putting the foegoing togethe, we get Ex q 1 3x = x πε q 1 3x 1 3y 1 3z So.E = πε q 3 3(x + y + z ) i.e..e = = πε x λ 4) We hve seen tht the E field hs mgnitude, whee is the distnce fom the πεo xis of the cylindicl geomety. Also the diection of E is pependicul to the cylindicl xis. A unit vecto in this diection is xi + yj ˆ = x + y So, in vecto tems: λ λ 1 x y E = ˆ i + j = πε o πε o x + y x + y Giving λ x E x = πε o x + y E Fom this obtin x E y nd wite down x y (by symmety). E Since, E z =,.E = x E y + x y which tuns out, supise! supise!, to be zeo. 5).E = ρ/ε, so the gph of.e is the sme s tht of ρ (pt fom multiplying constnt). It hs positive spike whee GH cosses the fce of the positive plte nd convesely fo the negtive plte. PX38soln() /

7 PX38 Poblem Sheet 3 1) An electic dipole of moment p = q lies long the x-xis with its cente t the oigin. Using supeposition, deive expessions fo the electosttic potentil V(x) nd the electic field E(x) t points on the x-xis fo which x >>. dv(x) p Confim tht E(x) = =. 3 dx πε x q ) The potentil due to point chge q is V() = (zeo t infinity). Obtin 4πε the coesponding expession fo the electic field by computing E() = V(). Q 3) Pove the esult E S = fo the enegy stoed in cpcito, chged to ±Q, by C consideing the wok equied to move chge dq fom the negtive plte to the positive plte when the pltes cy chge q. [Integte fom q = to Q ] 4) The digm below shows 3-petue electon lens (such s is used to focus the electons in the gun of n oscilloscope tube). The lens system hs cylindicl symmety bout the hoizontl xis nd the electode potentils e s shown. Copy the digm nd sketch the equipotentils nd electic field lines fo the ngement, giving esons fo you nswe. 5) Clculte the mximum voltge tht cn be pplied between the inne conducto (oute dim. 1mm) nd the oute conducto (inne dim. 1mm) of n i spced coxil cble. The bekdown field of dy i t tmospheic pessue is bout 3kV/mm. PX38poblems(3) 1/1

8 PX38 Solutions to poblem sheet 3 1) Dipole moment p = q Clculting the E field, t P(x,,), diectly, the contibutions fom the two chges e both diected long the x-xis, so E only hs n x component. ( ) ( ) ( x+ ) ( x ) ( ) ( ) 1 q q q p E (x) = = = fo x x 3 4πε x x+ 4πε x x+ πε x Deiving E fom V: 1 q q p V(x) = = fo x 4πε ( x ) ( x ) + 4πεx V(x) p E = V so E (x) = = E (x) = E (x) = x 3 y z x πεx ) q V( ) = 4 πε q 1 q E () = V() = = i + j + k 4πε 4πε x y z 1 1 x = = 3 x x q q So : E ( ) = ( xi+ yj+ zk) = ˆ 4πε 4 3 πε 3) When the potentil diffeence between the pltes of cpcito is V, the wok equied to move dq fom the negtive to the positive plte is Vdq = (q/c)dq. The wok equied to chge the cpcito fom to Q is, theefoe, Q qdq Q W = = C C PX38soln(3) 1/

9 4) 5) If the chge pe unit length on the inne conducto is +σ Cm -1. (-σ on the oute). Then: σ E() =, nd so πε V = V() - V(b) = σ b n l πε So the eltion between the field nd the potentil diffeence is (eliminte σ): V E = b ln This field is getest t =. The mximum voltge diffeence between the conductos is: b Vmx = Emx l n 3.5kV PX38soln(3) /

10 PX38 Poblem Sheet 5 1) Wite down the dimensions of the following (in tems of m, kg, s nd C): electic field, ε, electic dipole moment, mgnetic field (tesl), μ, mgnetic dipole moment. ) Two identicl plne pllel cicul loops hve dius nd e septed by distnce long thei common xis. The loops cy the sme cuent I with opposite senses of cicultion. (i) Show tht the mgnetic field B on the xis t distnce y fom the cente point is given by μoi 1 1 B( y) = ( + ( + y) ) ( + ( -y) ) [Use the known esult fo the on-xis field fo one coil plus supeposition] Sketch the vition of B(y) long the xis between y = - nd y =. (ii) Show tht the coil ngement poduces mgnetic field tht chnges linely with y in the vicinity of the mid-point nd tht the field gdient is given by db dy 3μI = 5 (iii) This ngement of opposed coils is used to poduce constnt mgnetic field gdients in nucle mgnetic esonnce imging body scnnes. Wht cuent is equied to poduce field gdient of 1. mt/m if = cm? PX38pob(5) 1/

11 3) Find the B field between coxil cylindes of dii 1 nd cying opposite cuents, I, s shown. Evlute B t = 1 cm fo 1 =.5 cm, = 1.5 cm nd I = 1 A. 4) () A long stight wie, of cicul coss-section with dius, cies cuent I. Deive the fom of the mgnetic field vition s function of the distnce fom the xis (both fo > nd < ), when (i) the cuent density is unifom. (ii) the cuent flows only on the sufce of the wie. (b) Show tht, if the wie is ligned with the z-xis nd the cuent is flowing in the diection k, then the B field outside the wie hs the fom B(x,y) = μoi (yi xj) π Show tht this stisfies B =. (c) Show tht the B field inside the wie in cse ()(i) hs the fom B(x,y) = μoi π (yi xj) Show tht this stisfies B = μ o J. PX38pob(5) /

12 PX38 Solutions to poblem sheet 5 1) [E] kg m s - C -1, [ε o ] kg -1 m -3 s C [p] C m [B] kg s -1 C -1 [μ o ] kg m C - [μ] C s -1 m ) On the xis t distnce x fom the cente of the loop B x μi = ( + x ) 3 (the negtive sign is becuse of the sense of I). Fo two loops, s in the question nd mesuing y fom the mid-point, the totl field is μi μi 3 3 B(y) = B1 + B = + ( + ( + y) ) ( + ( y) ) The gdient of the field is: ne y= μ B( ) I [neglecting the smlle μ contibution,.1 I, fom loop ] B() = 3 3 db μi ( + y) ( y) = + dy + (+ y) + ( y) db dy [B 1 nd B cncel] μ B( + ) + I [neglecting the smlle contibution fom loop 1] The genel ppence of B(y) is s shown. 5 5 { y y}.53 3μ I 3μ I μ I = + + = = 5 5 ( ) The field gdient ne the cente is constnt. This is the bsis of method of poducing field gdients tht is used in mgnetic esonnce imging body scnnes. [Note; quick PX38soln(5) 1/3

13 μ guess t the gdient my be mde by sseting tht the field goes fom ~ I μ to ~ + I μi μi in distnce, giving gdient =.5 which is close to the bove.] (iii) I = 6 A 3) Mxwell's 4th eqution (Ampee's theoem) fo sttiony cuents, pplied to I cicul pth of dius [ 1 ] gives B ()= μ. π Numeicl nswe: 1-4 T. 4) () Outside eithe wie Bdl. =μi B() π =μ I μi B() = π > Inside the wie Cse (i) A cicul loop of dius encloses fction of the totl cuent given by the tio of the es π Ithough loop = I totl = I π Bdl. = B() π =μ μi B() = < π I Note: the solutions fo inside nd outside gee t =. Cse (ii) The Ampein loop now encloses ZERO cuent Bdl. = B() π =μ Ienclosed= B() = < PX38soln(5) /3

14 (b) A is the point (x,y). The field t A is pependicul to nd μ hs mgnitude B() = I fo >. π Fo ight hnded set of xes, the z-xis points out of the ppe. The cuent is given to be in the k diection nd so flows into the ppe. The ight hnd scew ule gives the diection of B s shown. B x = B() sin(θ) = B y = -B() cos(θ) = B = μoi π μ (y i x j) Iy π y sin( θ ) = μix x cos( θ ) = π B B z y B B x Bz y Bx B = i + j + y z z x k x y The i nd j components vnish since B z is zeo, nd B x nd B y do not chnge with z. Using i.e. x = x etc. μi x y B = k x y π μ I 1 ( ) x 1 x ( y) y = k π μi x y B = k 1 1 since x y + + = + = π (c) Fo unifom cuent density, the B field inside the wie hs the fom: But B = μoi π ( y i x j) = μoi π (y i x j) μi μi B = k ( x) ( y) = k x y π π I J = π k So B = μ o J PX38soln(5) 3/3