PHYS1231 Higher Physics 1B Solutions Tutorial 2

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1 PHYS3 Higher Physics Solutions Tutoril sic info: lthough the term voltge is use every y, in physics it is mesure of firly bstrct quntity clle Electric Potentil. It s importnt to istinguish electric potentil from electric potentil energy (U). They re similr in tht they re both sclrs but they re not the sme. You cn think of electric potentil s the potentil for moving positive chrge t some point in spce. So, points closer to positive chrge hve greter electric potentil. However, more often we tlk bout the potentil ifference between two points:!potentil(!v) = Work Q or!v =!U Q We cn lso efine the electric potentil ifference between two points ( n ) in n electric fiel E:!V = # E.s Where the integrl is long the pth from point n point. Remember lso tht with electric potentil the principle of superposition pplies. Questions: In ll of the following solutions k = 4! 0. () Electrons ten to be ttrcte towrs regions of high potentil. (b) Equipotentil surfces o NOT intersect. (c) No. t position equiistnt between two equl positive point chrges the totl electric fiel cn equl zero becuse the electric fiels will hve equl mgnitues but exctly opposite irections. However, the electric potentil ue to either of these chrges hs no irection n, in fct, the electric potentil from both chrges will together. () No. You nee to know the integrl of E.s long pth from point of known voltge. (e) From the mthemticl reltionship between V n E, -V/s = E, where V is constnt, therefore E = 0.. (refer to the igrm bove) For the first chrge to be e to this configurtion no work is one becuse there is no pre-existing electric fiel. For the rest of the question we recognise tht work one = ΔU n tht ΔU = QΔV. For the secon (negtive) chrge to be brought within of the first we ssume tht the secon chrge origintes t n infinite istnce wy, where the electric potentil = 0V (this ssumption is repete with ll the subsequent itions). Then, for the secon point chrge, the work one is:

2 kq q!v = For the thir we consier its proximity to both the first n secon chrges n this to the bove: n for the fourth q!v = kq kq # kq q!v = % kq kq & ' kq kq kq kq = ( 4 ) kq q = () The electric potentil t istnce r from point chrge is given by V = kq/r. Where V = 30V, k = 8.99 X 0 9 Vm/C n Q =.5 X0-8 C, r must = 4.495m, or, to two significnt figures r = 4.5m. (b) Potentil is hyperbolic function of r (ie. V α /r). For equipotentil surfces to be evenly seprte you woul hve to hve V α r. 3. The potentil t istnce r wy from point chrge is given bove. Then, for position, V = kq/ 3kq/(). Similrly, for point, V = kq/() 3kq/. Then V - V = [kq/ 3kq/()] [kq/() 3kq/] or, n expning this we get: V! V = V! V = 4kq & 4 # 0 % V! V =! 4kq ( ) 4q( ) ( )! 4 # 0 4q ' ( ) ( & 4q ' % ( ) ( Clerly, when = 0 the bove expression = 0 n simultneously V = V. 4. () If q n q re point chrges, V = kq /(0.5) kq /(0.05). Substituting in the vlues for both chrges n k = 8.99 X0 9 Vm/C, V = 0.6 X0 5 V. (b) First we nee to etermine the potentil t, V which cn be one in similr fshion to prt (). V = kq /(0.05) kq /(0.5) = -7.8 X0 5 V. Then the potentil ifference V - V = 8.4 X0 5 V, n from:!v = Work Q then Work = Q!V. Using the vlue of Q = 3.0 X0-6 C, the work one in moving tht chrge from to =.5J. Externl work is one to move the positive chrge to higher electric potentil so the system of chrges hs higher potentil energy. 5. () For given section of the chrge line r, there is chrge λr. The totl electric potentil t P is foun by summing the contribution to the potentil V P ue to ech of these sections from r = 0 close to P, to r = L furthest from P. or L k!r V = which gives V = k! ln(l y) k! ln(y) 0 (y r) V = k! ln L y % # y &

3 (b) Now we use y s vrible quntity in E = -V/y in the y-irection. This prouces n expression for E: then # E =!k% (L y)! & y' L E = k! % # y(l y) & (c) Now if we efine the irection to the stright line with the isplcement vrible x, we notice tht, t P, the electric potentil epens only on y. Therefore E x = -V/x = () V = kq/(.06) = 506V n V = kq/(.7) = 893V. Then the potentil ifference V V = kv. (b) The electric potentils clculte in prt () epene on the istnces to n not the isplcements. Direction plys no prt in the clcultion, so the potentil ifference remins the sme. 7. Firstly we must consier the wire to be goo conuctor n ssume tht it hs negligible resistnce. With n externl electric fiel pplie to ny neutrl conuctor, the mobile chrges within tht conuctor will move until the net fiel t every point insie the conuctor is zero. In this cse negtive chrge -Q will ccumulte t the sphere closest to the q chrge. If the rius of these spheres r << R, then we cn consier them s point chrges. The potentil ue to point chrge q t istnce r is Kq/r. We lso ssume tht no chrge ccumulting on the wire (though in prctice this ssumption my not be vli). Then, potentil t centre of sphere Q is kq kq kq! r L R L Potentil t centre of sphere Q is kq kq kq! L r R Since the spheres must be t the sme potentils, we hve: Q Q q Q Q q! =! r L R L L r R & # & # i.e. Q '! = q ' % r L % R R L! Q( L ' r) q & L ( )! # ( = rl % R R L Q = R ql r ( R L)( L! r) Since L>>R>>r, this becomes qr Q! R

4 CPCITORS ND DIELECTRICS sic info: cpcitor is n electronic evice for storing chrge (Q) given supplie voltge (V). s shown symboliclly in circuit igrms, n often in relity, cpcitor cn be consiere s two chrge metl pltes seprte by non-conucting meium (eg. ir), The efinition of cpcitnce, C: C = Q V or in terms of the re (), the seprtion () of the pltes n the ielectric constnt of the intervening meium ε m. C =! 0! m Totl cpcitnce, of severl cpcitors connecte in series: C = T C C C... 3 connecte in prllel: The energy store in cpcitor U: CT =C C C 3... U = CV Questions: 9. Using the following: C = 0!! m where = π(0.08) m, = 0.00m n ε 0 = 8.85 X 0 - F/m n ssuming vcuum between the pltes ε m =, then C =.78 X 0-0 F. Then CV = Q n Q =.8 X 0-8 C. 0. First combine C n C in series, so C = (/C /C) - = 3.7 X0-6 F. Then combine C n C3 in prllel: 3.7 X X0-6 F =7.7 X0-6 F. () The over-riing rule here will be the conservtion of chrge. First we nee to fin exctly how much chrge is store n where. With 96.6 V cross C there must be Q = CV =. X 0-4 C on C. y the sme formul the chrge on C is 3. X 0-4 C. When the switches re close the effective cpcitnce will be = C C = 4.38 µf n the net chrge cross this will be = 3. X 0-4 C. X 0-4 C =.99 X 0-4 C. Therefore the voltge cn be foun by V = Q/C = 45.4V (b) Now, given the voltge cross both cpcitors, we cn fin their respective chrges by Q = CV. For C, Q = 5.7 X 0-6 C n (c) for C, Q = 46 X 0-6 C.. () With cpcitors rrnge in series the chrge cross ech cpcitor is the sme even if the voltge isn t. Here we begin by clculting the net totl cpcitnce =[/(µf) /(8µF)] - =.6 µf n, given the voltge cross this equivlent cpcitor is 300V, the chrge ech of the cpcitor components will be Q = CV = 4.8 X 0-4 C. Now, knowing the chrge cross ech cpcitor the voltge cross the µf, V = Q/µF = 40 V n cross the 8µF, V8 = Q/8µF = 60V. (b) Now the cpcitors re effectively connecte in prllel with combine cpcitnce of 0µF n cross cpcitors in prllel the voltge is the sme. Hence we cn sy, through conservtion of chrge, tht the chrge on this equivlent cpcitor is Q = C n from V=Q/0µF, the potentil ifference cross ech of the component cpcitors is = 96V. However the chrge on ech cpcitor will be

5 reistribute: C hs Q = 96V X µf =.9 X 0-4 C; C8 hs Q8 = 96V X 8µF = 7.68 X 0-4 C. (c) With equl chrge on ech cpcitor, reconnecting them in reverse will result in net chrge = 0 everywhere. With zero chrge there is lso zero potentil ifference. 3. Wht is being chnge here is the cpcitnce properties of the cpcitor. fter the bttery hs been isconnecte there is fixe mount of chrge on the cpcitor but, more thn likely, the ielectric will increse the cpcitnce (ε m > ) which mens (by C = Q/V) the voltge cross the cpcitor shoul ecrese. If the voltge ecreses then so must the electric fiel n the store energy = / Q /C 4. The ielectric mteril will brek own when there is n electric fiel strength of 8 X0 6 V/m cross it. The electric fiel cross cpcitor is given by E = σ/ε 0 ε m (see lso Guss s Lw pplie to chrge sheet in Tutoril 0) where σ is the surfce chrge ensity = Q/. OK, so to prevent brek own, Q/ε 0 ε m < 8 X0 6 V/m. Now, we cn work out the mximum chrge, Q = CV =.8 X0-4 C, n, with ε m =.8 sy: >.8 X0-4 C/( X.8ε 0 X 8 X0 6 V/m) Therefore the re of one plte = 0.34m. Or, for both pltes = 0.63m 5. If we consier the rrngement s two prllel cpcitors of equl re = / then we fin: C Totl = 0 0 from which we get Two limiting cses. Consier C Totl = 0 # & % ( ' = 0 then C totl = 0 i.e., cpcitor of re / n = then C totl = 0 i.e., cpcitor of re. 6. Totl cpcitnce = 6µF n the store energy = / X CV. Then substituting in V = 300V the store energy = 0.7J