19-Capacitance

SECTION - I : STRAIGHT OBJECTIVE TYPE [k.M-I: lh/ks oLrqfu"B izdkj 19.1 In the figure shown the plates of a parallel plate capacitor have unequal charges. Its capacitance is 'C'. P is a point outside the capacitor and close to the plate of charge –Q. The distance between the plates is 'd', select incorrect alternative : fp=kesafn[k ;svuqlkjlekUrjiêla/k fj=kdhIysVksaijvlekuvkos'kgSAbldh/k fjrk'C'gSAPfcUnqla/k fj=kds ckgjfLFkrgSrFkkvkos'k–QokyhIysVdsiklgSAIysVksadschpdhnwjh'd'gS]vlR;dFkudkpqukodhft,& & (A) A point charge at point 'P' will experience electric force due to capacitor PfcUnqijfcUnqor~vkos'k]la/k fj=kdsdkj.kdksbZcyvuqHkodjsxkA 3Q (B) The potential difference between the plates will be 2C 3Q IysVksadschpfoHkokUrj 2C gksxkA (C) The energy stored in the electric field in the region between the plates is 9Q2 9Q2 8C IysVksadse/;fo|qr{ks=kesalafprÅtkZ 8C gSA Q2 (D*) The force on one plate due to the other plate is Q2 ,dIysV}kjknwljhIysVijvkjksfircy 2 0 d2 2 0 d2 gSA Sol. (D) E = 2Q 2A  2A  E = 3 Q 2A 3 Q 3 Q E = (ii) F = EQ/2 2 Cd  Ed = 2C = V  2Q  (Q) Q 2 F =  2A  × 1  A      Q 2 F = A  1 1  3Q 2 9 Q2 (iii) Energy ÅtkZ =  E2 Ad = 2 0   Ad = 2  2Cd 8 C 19.2 In the figure shown the equivalent capacitance between 'A' and 'B' is : fp=k esa n'kkZ;s vuqlkj 'A' rFkk 'B' ds chp rqY; /kkfjrk gS (A) 3.75 F (B*) 2 F (C) 21 F (D) 16 F Sol. (B) Equivalent circuit is rqY; ifjiFk gS   = 2 F M.Bank_Capa._2.15 x o y ds chp rqY; /kkfjrk : 5 (A) 6 F (B) 7 F (C*) 6 8 F (D) 1 F 3 Sol. (C) RearrangethecircuitifjiFk dks O;ofLFkrdjus ij Ceq = 8 F 3 19.4 In the figure initial status of capacitance and their connection is shown. Which of the following is incorrect about this circuit : fp=kesa/kfjrkrFk budsla;kstudhizkjfEHkdfLFkfrdsckjsesan'kZ;kx;kgSAifjiFkdsckjsesafuEuesadkSulk@ls dFku vlR; gSaA (A) Final charge on each capacitor will be zero izR;sdla/k fj=kijvfUrevkos'k'kwU;gSA (B) Final total electrical energy of the capacitors will be zero izR;sdla/k fj=kdhvfUredqyfo|qrÅtkZ'kwU;gSA (C) Total charge flown from A to D is 30µC A ls D dh rjQ cgus okyk dqy vkos'k 30µCgSA (D*) Total charge flown from A to D is – 30µC A ls Ddh rjQ cgus okyk dqy vkos'k – 30µCgSA Q1  Q2 Sol. (D) V = C1  C2 = 0 Final potential difference = zero Final charge = Zero Charge flow 30 c from A to D vfUrefoHkokUrj='kwU; vfUrevkos'k='kwU; 30 c vkos'k Als Drdizokfgr gksxkA 19.5 A parallel plate capacitor of capacitance C (without dielectrics) is filled by dielectric slabs as shown in the figure. Then the new capacitance of the capacitor is: C/k fjrk,dfcukijkoS/kqrokyslekUrjiêla/k fj=kdksfp=kuqlkjijkoS/kqrlsHkjktkrkgSla/kfj=kdhu;h/kfjrk gksxh% (A*) 3.9 C (B) 4 C (C) 2.4 C (D) 3 C Sol. (A) The equivalent capacitance rqY; /kkfjrkC = 2C 6C + 6C  4C = 3.9 C eq 2C  6C 6C  4C 19.6 A capacitor (without dielectric) is discharging through a resistor. At some instant a dielectric is inserted between the plates, then ,dla/kkfj=k(fcukijkoS/kqrds)dks,dizfrjks/k}kjkfujkosf'krfd;ktkrkgSfdlh{k.kIysVksadse/; ijkoS/kqrizos'kdjk;ktk;gS]rks (A) Just after the insertion of the dielectric, current will increase. ijkoS/kqrizos'kdjkusdsrqjUrckn]/k jkdkekuc> 1) (A) 20 In (B)  0 2In  (C*) 0 In (D) Information insufficient fp=kesan'kZ;svuqlkjnkslh/ksyacslekudkV{ks=kdsrkj,dnwljsdslekarjok;qesafu;ksftrgSAburkjksdhv{ksa dschpnwjhbudsdkV{ks=kdhf=kT;klsxq.k T;knkgSArkjksadhizfr,dkadyEckbZdh/k fjrkgksxh (>>1ysosaA ) (A) 20 In (B)  0 2In (C*) 0 In (D) tkudkjh vi;kZIr gSA Sol. Let us give equal and opposite charges to two wires so that they would have linear charge density as +  and – . ekukrkjksdkslekuifjek.kofoijhrizd`frdsvkos'kfn,x,gS]ftl sbudkjS[kh;vkos'k?kuRoØe'k% +  o – . Electric field at point P, fcUnq P ij fo|qr {ks=k a–a E =     dV  – E dx  –  E dx 2 0 x 20 (a– x) a where ais radius of wire tgka a rkj dh f=kT;k gS  C =   0 | V | In 19.24 A parallel plate capacitor is immersed in a liquid dielectric having dielectric constant  as shown in the figure. Find the force acting on a unit surface of the plate from the dielectric. ,dlekUrjIysVla/k fj=kijkoS|qrfu;rkaddsnzoesafp=k uqlkjMqck;kx;kgSAIysVds,dkad{ks=kijijkoS|qr inkFkZ}kjkvjksfircyKkrdfj,A (A) 0 V2 2d2 (B) 0 ( – 1)V2 2d2   (C) V2 2d2 (D*) ( – 1)0V2 2d2 Sol. The resultant force acting per unit area of each plate can be written as F = F0 – F’. where F0 is the force acting on unit area of plate due to other plate and F’ is the force acting on unit area of plate from the dielectric. izR;sdIysVds,dkad{ks=kijifj.k ehcyF=F0 –F’gSA tgkaF0,dIysVds,dkad{ks=kijnwljhIysV}kjkvkjksfircygSoF’IysVd,dkad{ks=kijijkoS|qr}kjkcygSA q2  0A 2 2 A  d V  1 0V2 Now vc, F = 0      F = 2d2 A Also F0 = F ×  F0 20A A  1   1   ( – 1)0V2 So vr% F’ = F0 –  = F0 1–    F1–    2d2     19.25 For the circuit shown in the figure, determine the charge of the capacitor in steady state. n'kZ;sfp=kesaLFkbZvoLFk esala/kfj=kijvkos'kKkrdfj,A (A) 4C (B*) 6C (C) 1C (D) Zero 'kwU; Sol. In steady state the capacitor is fully charged and is treated as open circuit, so no current flows through branch containing capacitor in steady state. So the circuit can be redrawn as : LFkbZvoLFk esala/kfj=kiw.kZvkosf'krgSo[kqysifjiFkdhHkafrO;ogkjdjsxkAvr%LFkbZvoLFk esala/kfj=kokyh 'k [k esa/k jkizokfgrughgksxhAvr%ifjiFkdks,slkekuktkldrkgSA Potential difference across the capacitor in steady state LFkbZvoLFk esala/kfj=kijfoHkokarj = V – 6 – V = – 6V (–ve sign signifies that left hand plate is of negative polarity) (–fpUgn'k ZrkgSfdcka;hIysV_.k RedgSA) Charge vkos'k = CV = 1 × 6 = 6 C 19.26 Two identical capacitors are charged to different potentials then they are connected to each other in such a way that the sum of charges of plates having positive polarity remains constant. Mark the correct statement. (A*) Sum of charges of plates having negative polarity remains constant. (B) Mean of individual final potentials is different from mean of individual initial potentials. (C) Total energy stored in two capacitors in final state may be equal to that in initial state. (D) Heat dissipation in the circuit could be zero. nkslekula/k fj+=kfHkUufoHkoijvkosf'krfd,x,gSobUgsblizdkjlaifdZrfd;kfd/kukRedIysVksijvkos'k sadk ;ksxfu;rjgrkgSAlghdFkufpfUgrdhft,A (A*)_.k RedIysVksdkvkos'k sadk;ksxfu;rjgrkgS (B)vafrefoHkoksadkvkSlrizkjafEHkdfoHkoksadsvkSlrlsfHkUugksxkA (C) vafrevoLFk esanksuksla/kfj=kasesalaxzfgrdqyÅtkZizkajfEHkdÅtkZvksdsrqY;gksldrhgSA (D) ifjiFkesaÅ"ek{k;'kwU;gksldrhgSA Sol. As the charge of isolated system remains conserved, so the sum of charges of plates having-ve polarity remains constant. As potential of two capacitors are different so some charge flows into the circuit till both acquire the same potential. foyfxrfudk;dkvkos'klajf{krjgrkgS]vr%_.k RedIysVksdsvkos'ksdk;ksxfu;rjgrkgSApqafdnksuksla/k fj=k s dkfoHkofHkUugSvr%lekufoHkoizkIrdjusrdifjiFkesavkos'kizokggksrkgSA As charge flows, H  0, and hence ui  Uƒ vkos'k izokg ij H  0, blfy, ui  Uƒ Let final common potential be V, then ekukmHk;fu"BfoHkoVgS]rc V = C1V1  C2 V2  V1  V2 [as pwafd C = C = C] C1  C2 2 1 2 SECTION - II : MULTIPLE CORRECT ANSWER TYPE [k.M- II: cgqy lgh fodYi izdkj 19.27 The figure shows, a graph of the current in a discharging circuit of a capacitor through a resistor of resistance 10 . 10vkseizfrjks/k}kjkla/k fj=kvukos'khdj.kifjiFkdsfy,/k jkoØn'k Z;kx;kgS& (A*) The initial potential difference across the capacitor is 100 volt. la/kfj=kijizkjfEHkdfoHkokUrj100oksYVgSA 1 (B*) The capacitance of the capacitor is 10 𝑙n 2 F. la/kkfj=k dh /kkfjrk 1 10 𝑙n 2 F gksxh 500 (C*) The total heat produced in the circuit will be 500 ifjiFkesamRiUudqyÅ"ek 𝑙n2 Jgksxh 𝑙n2 joules. 1 (D*) The thermal power in the resistor will decrease with a time constant 1 le;fu;rkad 2𝑙n2 lSd.MdslkFkizfrjks/kesaÅ"eh;'kfDr?kVsxhA Sol. (A,B,C,D) 2𝑙n2 second. V0 = 0 R = 10 × 10 = 100 volts (since, 0 = 10amp from figure) Hence (A) is correct vr% (A) lgh gSA (pwafd fp=k ls, 0 =10ampgS) Also rFkk :  =  e – t/RC Taking log y?kq xq.kd ysus ij ;    t n   =  C = t R𝑙n(0 / )   RC At ; t = 2 sec,  = 2.5 A t = 2 sec ij  = 2.5 A gS 2 C = 10𝑙n 10     2.5  2 2 1 C = 10𝑙n4 = 10  2𝑙n2 = 10𝑙n2 F Hence (B) is correct. vr% (B) lgh gSA 1 2 1  1  2 500 Heat produced mRiUu Å"ek = CV =   (100 ) = joules. Hence (C) is correct vr% (C) mÙkj lgh gS 2 2  10𝑙n2  𝑙n2 1 Thermal power in the resistor will decrease with a time constant 1 2𝑙n2 second. Hence (D) is correct izfrjks/kesaÅ"eh;'kfDrle;fu;rkad 2𝑙n2 dslkFk?kVrhgSAvr%(D)lghgSA 19.28 A parallel plate capacitor of capacitance 'C' has charges on its plates initially as shown in the figure. Now at t = 0, the switch 'S' is closed. Select the correct alternative(s) for this circuit diagram. fp=k esa 'C' /kkfjrk ds lekUrj iV~V la/kkfj=k dh IysVksa ij izkjfEHkd vkos'k n'kkZ;k x;k gSA vc t = 0, ij dqath 'S' cUn ¼pkyw½ djrs gSaA bl ifjiFk ds fy, lgh fodYi pqfu, - (A*) In steady state the charges on the outer surfaces of plates 'A' and 'B' will be same in magnitude and sign. (B) In steady state the charges on the outer surfaces of plates 'A' and 'B' will be same in magnitude and opposite in sign. (C*) In steady state the charges on the inner surfaces of the plates 'A' and 'B' will be same in magnitude and opposite in sign. (D*) The work done by the cell by the time steady state is reached is 5 2C . 2 (A*) LFk bZvoLFk (steadystate) esaIysV'A'rFkk'B'dhckgjhlrgksaijvkos'kifjek.k,oafpUgesalekugksxkA (B) LFk bZvoLFk (steadystate) esaIysV'A'rFk 'B'dhckgjhlrgksaijvkos'kifjek.kesaleku,oafpUgesafoijhr gksxkA (C*) LFk bZvoLFkk(steadystate) esaIysV'A'rFk 'B'dhvkUrfjdlrgksaijvkos'kifjek.kesaleku,oafpUgesa foijhrgksxkA 5 2C (D*) LFkkbZvoLFkk(steady state) esaigq¡pusrdlSy}kjkfd;kx;kdk;Z 2 Sol. [Tough] gksxkA Suppose charge flown through the battery is Q, then charge distribution will be as : A B The electric field in the region between A and B is = Q  2C 2A 0 – C  Q 2A 0 = 2Q  3C 2A 0 ekuk cSVªh ls Q vkos'k fudyrk gS rks vkos'k forj.k gksxkA A rFk Bdschpds{ks=kesafo|qr{ks=kgS = Q  2C 2A 0 C  Q – 2A 0 = 2Q  3C 2A 0  Potential differnce between the plates, 2Q  3C .d =   2Q  3C 1 =  2A 0 2 C  IysVksdse/;foHkokUrj 2Q  3C .d =  2Q  3C 1 =   2Q = 5 C  Q =  5C 2 2A 0 2 C 52C work done by battery = Q = 2  cSVªh }kjk fd;k x;k dk;Z = Q = 52C 2 19.29 The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now a dielectric of dielectric constant K is inserted to fill the whole space between the plates with voltage source remaining connected to the capacitor. (A*) the energy stored in the capacitor will become K-times (B) the electric field inside the capacitor will decrease K-times (C*) the force of attraction between the plates will become K2 -times (D*) the charge on the capacitor will become K-times. fcukfdlhijkoS|qrds,dlekUrjiV~Vla/k fj=kdhIysVkasdks,doksYVrkL=k srlstksM+kx;kgSAvcla/k fj=kdks oksYVrkL=ksrlstqM+kj[krsgq,bldhIysVksadschpdslEiw.kZHkxdksKijkoS|qrkadokysijkoS|qrlsHkjnsrsgSaSA (A*)la/k fj=kesalaxzfgrÅtkZKxqukgkstk;sxh (B)la/k fj=kdsvUnjfo|qr{ks=kKxquk?kVtk;sxk (C*)IysVksadschpvkd"kZ.k cyK2 xqukgkstk;sxk (D*)la/k fj=kijvkos'kKxqukgkstk;sxk 1 Sol. (A) E = CV2 2 As potential difference source between the plates is connected, p.d. remains constant. But capacitance C becomes KC hence energy stored is increased by factor K. V (B) Electric field d is not changed. (C) Charge on each plate is increased by factor K hence force between them increases by factor K2. For effect of the medium, they must completely lie in the medium. (D) Q = CV Hence charge becomes KQ as C becomes KC and V remain unchanged. 1 Sol. (A) E = CV2 2 ;fncSVjhla;ksftrgSrksIysVksadse/;foHkokUrjlekujgsxkysfdu/k fjrkKCgkstk,xhrFk la/k fj=kesalaxzfgr ÅtkZKxqukc<+tk,xhA V (B) fo|qr{ks=k d ifjofrZrughagksxkA (C) izR;sdIysVijvkos'kKxqukc<+tkrkgSblfy;smudse/;cyK2xqukc<+tkrkgSAek/;edsizHk odsfy;s budksaiw.kZr%ek/;eesaj[k gksukpkfg;sA (D) Q = CV vr%vkos'kKQrFk /k fjrkKCgkstk,xhrFk VvifjofrZrjgsxkA 19.30 A parallel plate capacitor of capacitance 10 F is connected to a cell of emf 10 Volt and fully charged. Now a dielectric slab (k = 3) of thickness equal to the gap between the plates, is very slowly inserted to completely fill in the gap, keeping the cell connected. During the filling process: (A*)the increase in charge on the capacitor is 200 C. (B*)the heat produced is zero. (C*)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab. (D*)energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab + heat produced. 10F/k fjrkdk,dlekukUrjIysVla/kfj=k10VoltdhcSVªhlstqM+kgqvkgSoiw.kZr;kvkosf'krgSAcSVªhdkstqM+kgqvk ghj[krsgq,]vcIysVdschpdsfjäLFk udscjkcjeksVkbZdh,dijkoS|qrIysV(k=3)dksIysVksadhchpiw.kZ:ils Hkjk tkrk gSA bl Hkjus ds çØe esa & (A*) la/kkfj=k ds vkos'k esa o`f) 200 C (B*)mRiUuÅ"ek'kwU;gSA (C*)cSVjh}kjkçnkudhxbZÅtkZ= la/kfj=kesalafprfLFkfrtÅtkZesao`f)+ijkoS|qrIysVHkjusokysO;fDrijfd;kx;k dk;ZA (D*) cSVjh }kjk çnku dh xbZ ÅtkZ = la/kkfj=k esa lafpr fLFkfrt ÅtkZ esa o`f) + ijkoS|qr IysV Hkjus okys O;fDr ij fd;k x;k dk;Z + mRiUu Å"ekA Sol. Charge on capacitor before insertion of dielectric slab = 100 µC Charge on capacitor after insertion of dielectric slab = 300 µC Increase in charge on the capacitor = 300 – 100 = 200 µC Heat produced = 0 Energy supplied by the cell = increase in stored potential energy + work done on the person who filling the dielectric slab + heat produced. Sol. ijkoS|qrIysVMkyuslsigysla/k fj=kijvkos'k=100µC ijkoS|qrIysVMkyusdscknla/k fj=kijvkos'k=300µC la/k fj=kdsvkos'kesao`f)=300–100=200µC mRiUu Å"ek= 0 cSVjh}kjkçnkudhxbZÅtkZ= la/k fj=kesalafprfLFkfrtÅtkZesao`f)+ijkoS|qrIysVHkjusokysO;fDrijfd;kx;k dk;Z+ mRiUuÅ"ekA 19.31 Capacitor C1 of the capacitance 1 microfarad and capacitor C2 of capacitance 2 microfarad are separately charged fully by a common battery. The two capacitors are then separately allowed to discharge through equal resistors at time t = 0. (A) the current in each of the two discharging circuits is zero at t = 0. (B*) the current in the two discharging circuits at t = 0 are equal but non zero. (C) the current in the two discharging circuits at t = 0 are unequal (D*) capacitor C1 loses 50% of its initial charge sooner than C2 loses 50% of its initial charge la/k fj=kC1dh/k fjrk1ekbØksQsjMrFk la/k fj=kC2dh/k fjrk2ekbØksQsjMgSAnksuksadksvyx&vyxlekucSVjh }kjklEiw.kZvkosf'krdjrsgSaAt=0ijnksuksala/k fj=k sadkslekuizfrjks/k}kjkvyx&vyxfujkosf'krfd;ktkrkgSA (A)izR;sdnksuksafujkosf'krifjiFksaesat=0ij/kjk'kwU;gSA (B*)nksuksafujkosf'krifjiFk saesat=0ijcjkcjijUrqv'kwU;/kjkgSA (C)nksuksafujkosf'krifjiFk saesat=0ij/k jkvlekugSA (D*)la/k fj=kC1izkjfEHkdvkos'kdk50%,la/k fj=kC2}kjkizkjfEHkdvkos'kdk50%R;kxusdhvis{k tYnhR;kxnsxk Sol. (B D) The instantaneous charge on the capacitor is q(t) = q0 [1 – e–t/R C] = CV [1 – e–t/R C] The instantaneous current la/kkfj=k ij rkR{kf.kd vkos'k q(t)=q [1–e–t/RC]=CV[1–e–t/RC] rkR{kf.kd/kjk dq  1  i = dt = CV  RC  e–t/R C    i = V e–t/R C or ;k i = R V ( t = 0 ) R Given that V and R are same for both capacitors, so the initial current in both condensers is same moreover this is not zero. During discharge, the instantaneous charge q is fn;kgSfdnksuksala/kfj=ksadsfy,VrFk RlekugSAvr%nksuksala/kfj=ksaesaizkjfEHkd/kjklekugSrFk ;g'kwU; ughagSA vukos'kudsle;]rkR{kf.kdvkos'k q(t) = q e–t/RC Let q' = q /2 at t = t, then q0 = q e–1t/R C 0 2 0 or t = RC loge 2 If t1 and t2 be the times in which the two capacitors lose 50% of their charge, then t1 t2 RC1 loge 2 = RC2 loge 2 t1 = t2/2 C1 1 = C2 = 2 This shows that C1 loses 50% charge sooner than C2 because it takes time t1 which is half of t2] ekuk q' = q /2, t = t ij, rks q0 = q e–1t/R C 0 2 0 ;k t = RC loge 2 ;fnt1rFk t2ogle;gStcnksuksala/k fj=kvius50%vkos'kR;kxrsgSarks t1 t2 RC1 loge 2 = RC2 loge 2 t1 = t2/2 C1 1 = C2 = 2 bllsiznf'kZrgksrkgSfdla/k fj=kC1vius50%vkos'kdksC2dhvis{k tYnhR;kxrkgSD;ksafd;gt1le;ysrk gStkst2dkvk/k gS] SECTION - III : ASSERTION AND REASON TYPE [k.M- III: dFku vkSj dkj.k izdkj 19.32 STATEMENT-1 : If the potential difference across a plane parallel plate capacitor is doubled then the potential energy of the capacitor becomes four times under all conditions. 1 STATEMENT-2 : The potential energy U stored in the capacitor is U = 2 CV2, where C and V have usual meaning. oDrO;-1: ;fnlekUrjIysVla/k fj=kijfoHkokUrjnqxukdjfn;ktkrkgSrkslHkhfLFkfr;ksadsvUrZxrla/k fj=kdh fLFkfrtÅtkZpkjxquhgkstk;sxhA 1 oDrO;-2: la/kkfj=kesalafprfLFkfrtÅtkZU= 2 CV2 gStgk¡CoVdklkekU;izpfyrvFkZgSA (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D*) Statement-1 is False, Statement-2 is True (A) oDrO;-1lR; gS] oDrO;-2lR; gS ;oDrO;-2,oDrO;-1 dk lgh Li"Vhdj.k gSA (B)oDrO;-1lR; gS] oDrO;-2lR; gS ; oDrO;-2,oDrO;-1 dk lgh Li"Vhdj.k ugha gSA (C) oDrO;-1lR; gS] oDrO;-2vlR; gS ; (D*) oDrO;-1vlR; gS] oDrO;-2lR; gS Sol. If potential difference across an isolated charged capacitor is doubled by doubling separation between plates, Q2 the energy stored is capacitor from U = 2C becomes double of previous value. Hence statement 1 is false. ;fn,dfoyfxrvkosf'krla/k fj=kdsfljksadkfoHkokUrjIysVksadschpdhnwjhnqxuhdjds]nqxukfd;ktkrkgS]rksU Q2 = 2C lsla/k fj=kesalafprÅtkZigysdsekudhnqxuhgkstkrhgSAblfy;soDrO;1vlR;gSA 19.33 STATEMENT-1 : A charged plane parallel plate capacitor has half interplanar region () filled with dielectric slab. The other half region  has air. Then the magnitude of net electric field in region  is less than that in region . oDrO; -1 : ,d vkosf'kr lekUrj IysV la/kkfj=k esa IysVksa ds chp ds vk/ ks LFkku () dks ijkoS|qr xqVds }kjk Hkjk x;k gSA nwljs vk/ks LFkku () esa ok;q gSA rks LFkku I esa ifj.kkeh fo|qr {ks=k dk ifjek.k LFkku  esa ifj.kkeh fo|qr {ks=k ds ifjek.k ls de gksxkA +Q +Q –Q –Q STATEMENT-2 : In a dielectric medium induced (or polarised) charges tend to reduce the electric field. oDrO;-2:,dijkoS|qrek/;eesaizsfjr¼;k/kzqfor½vkos'kfo|qr{ks=kdks?kVkusdhizo`fÙkj[krsgSaA A +Q C B –Q D A +Q C B –Q D (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 oDrO;-1lR; gS,oDrO;-2lR; gS;oDrO;-2oDrO;-1dk lgh Li"Vhdj.k gSA (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 oDrO;-1lR;gS,oDrO;-2lR; gS;oDrO;-2oDrO;-1dklgh Li"Vhdj.kugha gSA (C) Statement-1 is True, Statement-2 is False oDrO; -1lR; gS, oDrO;-2vlR; gSA (D*) Statement-1 is False, Statement-2 is True. oDrO; -1vlR; gS, oDrO;-2lR; gSA Sol. Let the electric field in region  and  be E and E . The potential difference across left half capacitor and right 1 2 half capacitor is same. Therefore E1d = E2d where d = inter planar gap.  E = E2 Hence statement 1 is false, statement 2 is correct by definition. ;fn{ks=kvkSj esafo|qr{ks=kE1vkSjE2gSAcka;svk/ksvkSjnka;svk/ksla/k fj=kdkfoHkokUrjlekugSAblfy;sE1d =E2dtgk¡d=ryksadschpnwjh  E = E2 blfy;soDrO;-1vlR;gSA oDrO;-2ifjHkk"kklslR;gSA 19.34 STATEMENT-1 : A dielectric is inserted between the plates of an isolated fully-charged capacitor. The dielectric completely fills the space between the plates. The magnitude of electrostatic force on either metal plate decreases, as it was before the insertion of dielectric medium. STATEMENT-2 : Due to insertion of dielectric slab in an isolated parallel plate capacitor (the dielectric completely fills the space between the plates), the electrostatic potential energy of the capacitor decreases. oDrO;-1 : ,d foyfxr :i ls j[ks gq, iw.kZ vkosf'kr la/kkfj=k dh IysVksa ds e/; ijkoS|qr izos'k djk;k tkrk gS ijkoS|qr] nksuksa IysVksa ds e/ ; iwjs LFkku rd Hkj tkrk gSA ijkoS|qr ek/;e Hkjus ls igys dh rqyuk esa ijkoS|qr ek/;e Hkjus ds ckn izR;sd IysV ij fLFkj oS|qr cy dk ifjek.k ?kVrk gSA oDrO;-2:fdlhfoyfxrlekUrjiêla/k fj=kesaijkoS|qrifêdkizos'kdjkusijla/k fj=kdhfLFkjfo|qrfLFkfrt ÅtkZ?kVtkrhgS¼;gk¡ijkoS|qr]IysVksdse/;iwjsLFk uesaHkjktkrkgS½ (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D*) Statement-1 is False, Statement-2 is True. (A) oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k gSA (B) oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k ugha gSA (C) oDrO; -1 lR; gS, oDrO;-2 vlR; gSA (D*) oDrO; -1 vlR; gS, oDrO;-2 lR; gSA Sol. The electrostatic force on metal of capacitor is 2 = pressure × area of plate =  = charge per unit area on plate. A 2 0 Since charge on metal plate of an "isolated" capacitor does not change, force on metal plate remains same. Electric field decreases due to induced charges in dielectric, but this does not effect the charge distribution on isolated metal plate. la/k fj=kdh/k fRodIysVijfLFkjfo|qrcy 2 = nkc × IysV dk {ks=kQy= 2 0 =vkos'kizfrbdkbZ{ks=kQy¼IysVij½ pwafdfoyfxrla/k fj=kdh/k fRodIysVijvkos'kifjofrZrughagksrkgSvr%/k fRodIysVijcylekujgrkgSA ijkoS|qresafo|qr{ks=kdkekuizsfjrvkos'k sdsdkj.k?kVtkrkgS]ijUrq;gfoyfxr/k fRodIysVdsÅijvkos'k forj.kdksizHk forughdjrkgSA 19.35 Statement I : During the charging of a capacitor using a battery, the electrons transferred from positive plate of capacitor to negative plate via dielectric medium in between the plates as shown. Statement II : The direction of electric field in between the capacitor plates is from positive plate to negative plate. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D*) Statement-1 is False, Statement-2 is True. oDrO;I:cSVªh}kjkla/k fj=kdsvkos'kudsnkSjku]la/k fj=kdh/kukRedIysVlsbysDVªkWu_.k RedIysVdhvksj ijkoS/kqrlsgksdjLFkukarfjrgksrsgSA(fp=kuqlkj) oDrO;II:la/k fj=kdhIysVksdse/;fo|qr{ks=kdhfn'k /kukRedIysVls_.k RedIysVdhvksjgksrhgSA (A) oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k gSA (B) oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k ugha gSA (C) oDrO; -1 lR; gS, oDrO;-2 vlR; gSA (D*) oDrO; -1 vlR; gS, oDrO;-2 lR; gSA Sol. The battery energizes the circuit and maintains the flow of electron from positive plate of capacitor to positive terminal of battery through wires and from wires to negative plate on other side. No transfer of charge takes place within the plates in spite of having the electric field in between the plates. cSVªhifjiFkdksÅtkZokucukrhgSobysDVªkuizokgdksla/k fj=kdh/kukRedIysVlscSVªhds/kukRedIysVdh vksj pkyd rkj esa cuk, j[krh gSA o nwljh rjQ rkj ls _.kkRed IysV dh vksj cuk;s j[krh gSA IysVksdse/;fo|qr{ks=kgksusdsckotwnIysVksdse/;vkos'kLFk ukUrfjrughgksrkgSA SECTION - IV : COMPREHENSION TYPE [k.M- IV:cks/ku izdkj COMPREHENSION # 1 The switch s has been closed for long time and the electric circuit shown carries a steady current. Let C1 = 3.0 F, C = 6.0 F, R = 4.0 k, and R = 7.0 k. The power dissipated in R is 2.8 W. fLop s yEcs le; ds fy, cUn gS vkSj fn[kk;s x;s fo|qr ifjiFk esa LFkk;h /kkjk gSA ekfu;s] C1 = 3.0 F, C2 = 6.0 F, R1 = 4.0 k, vkSj R2 = 7.0 k gSA R2 esa {k; 'kfä 2.8 W gSA 19.36 The power dissipated to the resistor R1 is izfrjks/kR1esa{k;'kfägS& (A) 2.8 W (B*) 1.6 W (C) 4.9 W (D) 0 Sol. (Moderate) i = 2 × 10–2 A P = i2 R = (2 × 10–2)2 × 4 × 103 = 1.6 W 19.37 The charge on capacitors C1 and C2 are respectively. la/k fj=kC1vkSjC2ijvkos'kØe'k%gS& (A) 940 C, 940 C (B) 440 C, 440 C (C*) 240 C, 840 C (D) 840 C, 240 C Sol. (Moderate) Q 1 V 1 × C1 = 80 × 3 × 10–6 = 240 C Q = VR × C2 = 140 × 6 × 10–6 = 840 C 19.38 Long time after switch is opened, the charge on C1 is : fLop[kksyusdsyEcsle;ckn]C1ijvkos'kgS& (A) Zero 'kwU; (B) 420 C (C) 240 C (D*) 660 C Sol. Q 1 = EC1 = 220 × 3 × 10–6 = 660 C 2 COMPREHENSION # 2 In the shown circuit involving a resistor of resistance R , capacitor of capacitance C farad and an ideal cell of emf E volts, the capacitor is initially uncharged and the key is in position 1. At t = 0 second the key is pushed to position 2 for t0 = RC seconds and then key is pushed back to position 1 for t0 = RC seconds. This process is repeated again and again. Assume the time taken to push key from position 1 to 2 and vice versa to be negligible. fp=kesafn[k ,x,ifjiFkesaizfrjks/k(R)]la/k fj=k/k fjrk(C)o,dvkn'kZlsy(fo0ok0cyEoksYV)gS]la/k fj=k izkjEHkesavukosf'krgSrFk dqathfLFkfr 1esagSAt=0le;ijdqathdksfLFkfr2ijt0 =RClsd.Mdsfy,/kdsyk tkrkgSrFk fQjdqathdksfLFkfr1ijt0 =RClsd.Mdsfy,fQjls/kdsyktkrkgSAblizØedksckj&ckjiqujkorZ fd;k tkrk gSA ;g ekfu, fd fLFkfr 1 ls fLFkfr 2 esa dqath dks /kdsyus esa ;k fLFkfr 2 ls fLFkfr 1 esa /kdsyus esa fy;k x;k le; ux.; gSA 19.39 The charge on capacitor at t = 2RC second is t=2RClsd.Mijla/k fj=kijvkos'kgksxk& 1 1   1  1  1 1  1  (A) CE (B) CE    e  (C*) CE   e2  (D) CE   e e2  19.40 The current through the resistance at t = 1.5 RC seconds is t=1.5RClsd.Mijizfrjks/kesaizokfgr/k jkgksxhA (A) E (1 1) (B) E (1 1 ) (C) E (1  1 ) (D*) (1 1 ) e2R e eR e R e e 19.41 Then the variation of charge on capacitor with time is best represented by la/k fj=kijle;dslkFkvkos'kdkifjorZufuEugksxkA q q q q (A) Sol. 39 to 41 39. & 40 (Moderate) 41. (Tough) (B) t (C*) t (D) t t For t = 0 to to = RC seconds, the circuit is of charging type. The charging equation for this time is  t q  CE(1 e RC ) Therefore the charge on capacitor at time t0 = RC is qo  CE(1 1 ) e For t = RC to t = 2RC seconds, the circuit is of discharging type. The charge and current equation for this time are q  qoe  tto RC and i  qo e RC  tto RC Hence charge at t = 2 RC and current at t = 1.5 RC are q q  qoe  2RCRC RC  qo  e 1 CE(1 1) e e t and i  qo e RC 1.5RCRC RC  qo  eRC E (1 eR 1 ) respectively e Since the capacitor gets more charged up from t = 2RC to t= 3RC than in the interval t=0 to t=RC, the graph representing the charge variation is as shown in figure Sol. 39 to 41 39. & 40 (Moderate) 41. (Tough) t=0lsto =RCle; ds fy,] ifjiFk vkos'khdj.k dh rjg dk gSA bl le;ds fy,vkos'khdj.k dhlehdj.k gksxh q  CE(1 e  t RC ) blfy, la/kkfj=k ij vkos'k dk eku t0 =RCijqo  CE(1 1 ) gSA t= RClst=2RClsd.M rd ifjiFk fujkosf'kdj.k rjg dk gSA vr% vkos'k o /kkjk dk lehdj.k bl le; ds fy,gksxk  tto vkSj i  qo e  tto RC q  qoe RC RC vr%t=2RCij vkos'k rFkk t=1.5RCij /kkjk Øe'k% gSA q q  qoe  2RCRC RC  qo  e 1 CE(1 1) e e t vkSj i  qo e RC 1.5RCRC RC  qo  eRC E (1 1 ) gSA pwafdla/kkfj=kt=2RClst=3RCle;vUrjkyesa]t=0lst=RCle;kUrjkydhrqyukesavf/kdvkos'kxzg.kfd;s gq,gksrkgSvr%vkos'kforj.kxzkQesafn[k ;kx;kgSA SECTION - V : MATRIX - MATCH TYPE [k.M- V:eSfVªDl&lqesy izdkj 19.42 Two identical capacitors are connected in series, and the combination is connected with a battery, as shown. Some changes in the capacitor 1 are now made independently after the steady state is achieved, listed in column-I. Some effects which may occur in new steady state due to these changes on the capacitor 2 are listed in column-II. Match the changes on capacitor 1 in column-I with corresponding effect on capacitor 2 in column-II. nks ,d tSls la/kkfj=kksa dks Js.khØe esa tksM+k tkrk gS rFkk bl la;kstu dks cSVjh ds lkFk fp=kkuqlkj tksM+k tkrk gSA LFkk;h voLFkk ds igq¡pus ds ckn] vc la/kkfj=k-1 esa LorU=k :ils dqN ifjorZu fd;s tkrs gS] tks LrEHk-I esa lwphc) fd;s x;s gSaA bu ifjorZuksa ds dkj.k la/kkfj=k-2 ij gksus okys dqN izHkko LrEHk-II esa lwphc) fd;s x;s gS] tks ubZ LFkk;h voLFkk esa gksrs gSA LrEHk-I esa la/kkfj=k-1 esa fd;s x;s ifjorZuksa dks LrEHk-II esa la/kkfj=k-2 ij laxr izHkkoksa ls lqesfyr dhft;sA Cap.1 Cap.2 Column I Column II (A) A dielectric slab is inserted. (p) Charge on the capacitor increases. (B) Separation between plates increased. (q) Charge on the capacitor decreases. (C) A metal plate is inserted connecting both plates (r) Energy stored in the capacitor increases. (D) The left plate is grounded. (s) Energy stored in capacitor is decreased (t) No change is occurred. LrEHk-I LrEHk-II (A) ,dijkoS|qriêhizos'kdjkbZtkrhgSA (p)la/k fj=kijvkos'kc<+rkgSA (B) IysVksa ds chp dh nwjh c<+k;h tkrh gSA (q) la/kkfj=k ij vkos'k ?kVrk gSA (C) nksuksa IysVks dks tksM+us okyh ,d /kkfRod ( r) l a / kkfj=k esa lafpr ÅtkZ c<+rh gSA IysV izos'k djkbZ tkrh gS (D) cka;hIysVdksHkwlEiZfdrfd;ktkrkgSA (s)la/k fj=keslaxzfgrÅtkZ?kVrhgSA (t) dksbZ ifjorZu ugh gksrk gSA Ans. (A) p, r (B) q; s (C) p, r (D) t Sol. (A) By inserting dielectric slab, capacitance of 1 increases there by increasing charge on capacitor 2 as more charge is flown through the battery. Energy stored in capacitor also increases. (B) By increasing separation between the plates, capacitor C1 decreases. Charge on C2 also decreases. (C) By shorting capacitor-1, only capacitor 2 remains in the circuit. Potential difference across C2 increases thereby increasing charge on 2 as well as energy stored. (D) By earthing plate of capacitor 1 potentials will change but there will be no potential difference change, making no overall change in the circuit. (A)ijkoS|qriV~Vhdksizosf'krdjkusls]la/kfj=k1dh/kfjrkc<+tkrhgSftlsla/kfj=k2 ijvkos'kc<+tkrkgSD;ksafd cSVjhlsvf/kdvkos'kizokfgrgksrkgSAla/kfj=kesalafprÅtkZHkhc<+rhgSA (B)IysVksadschpdhnwjhc<+kusls]la/kfj=k1dh/kfjrk?kVrhgSAla/kfj=k2ijvkos'kHkh?kVrkgSA (C)la/kfj=k-1,dksy?kqifFkrdjusij]ifjiFkesadsoyla/kfj=k2jgrkgSAC2dsfljksaijfoHkokUrjc<+rkgSftlsla/kfj=k 2ijvkos'kc<+rkgSlkFkghlkFklafprÅtkZHkhc<+rhgSA (D)la/kfj=k1dhIysVdkstehulstksM+usijfoHkoifjofrZrgksaxsysfdufoHkokUrjesadksbZifjorZuughagksxkftlsifjiFk esalexz:ilsdksbZifjorZuughgksrkgSA 19.43 In each situation of column-I some changes are made to a charged capacitor under conditions of constant potential difference or constant charge. Condition of constant potential difference means that a cell is connected across the capacitor and condition of constant charge means that the capacitor is isolated. Match the conditions in column-I with corresponding results in column-II LrEHk-IdhizR;sdfLFkfresafu;rfoHkokUrj;kfu;rvkos'kdsizfrcU/kdsvUrxZr,dvkosf'krla/k fj=kesadqNifjorZu fd;stkrsgSaAfu;rfoHkokUrjdkvFkZgSfdla/k fj=kdsfljksaij,dlsytksM+ktkrkgSrFk fu;rvkos'kdsizfrcU/kdk vFkZgSfdla/kfj=kfoyfxr(isolated)gSALrEHk-IesaizfrcU/ksadksLrEHk-IIesafn;sx;slaxrifj.keksalslqesfyrdhft,A Column Ι Column ΙΙ (A) For a capacitor maintained at constant (p) Then electric field inside the capacitor potential difference, the separation decreases in comparison to what it between plates is increased. was before the change. (B) For a capacitor maintained at constant (q) Then electric field inside the capacitor charge, the separation between the remains same. plates is increased (C) For a capacitor maintained at constant (r) Then potential energy stored in the potential difference, area of the both capacitor decreases in comparison to the plates is doubled. what it was before the change. (D) For a capacitor maintained at constant (s) The potential energy stored in the charge, area of both plates is doubled capacitor increases in comparison to what it was before the change. (t) Capacitance of capacitor decreases LrEHkI LrEHkII (A) fu;rfoHkokUrjijj[ksx;s,dla/kfj=kesaIysVksa (p)rksla/kfj=kdsvanjfo|qr{ks=k]ifjorZudsiwoZdh dschpdhnwjhc<+kbZtkrhgSA dhrqyukesa?kVrkgSA (B) fu;rvkos'kijj[ksx;s,dla/kfj=kesaIysVksadschp (q)rksla/kfj=kdsvUnjfo|qr{ks=kvifjofrZrjgrkgSA dhnwjhc<+kbZtkrhgSA (C) fu;rfoHkokUrjijj[ksx;s,dla/kfj=kesa] (r)rksla/kfj=kesalafprfLFkfrtÅtkZ]ifjorZu nksuksa IysVksa dk {ks=kQy nqxuk fd;k tkrk gSA d s iwoZ dh rqyuk esa ?kVrh gSA (D) fu;r vkos'k ij j[ks x;s ,d la/kkfj=k esa nksuksa IysVksa ( s ) la/kkfj=k esa lafpr fLFkfrt ÅtkZ] ifjorZu ds dk {ks=kQy nqxuk fd;k tkrk gSA iwoZ dh rqyuk esa c<+rh gSA (t) la/kkfj=k dh /kkfjrk ?kVrh gSA Ans. (A) p,r, t (B) q, s, t (C) q, s (D) p, r Sol. (A) At constant potential difference, when interplanar separation is increased, the capacitance decreases. From U = 1 CV2, the potential energy decreases. 2 V Also from E = d electric field decreases (B) At constant charge when interplanar separation is increased the capacitance decreases. Q2 From U = 2C , the potential energy increases Since charge density on plates is constant, electric field remains same. (C) At constant potential difference, when area of plate increases the capacitance increases. Hence from U = 1 2 V CV2, the potential energy increases Also from E = d , the electric field remains same. (D) At constant charge on increase in area of plates 1 Q2 From U = 2 C , the potential energy decreases and since charge density on plate decreases electric field decreases. (A) fu;rfoHkokUrjij]tcIysVksadschpdhnwjhc<+kbZtkrhgS]rks/k fjrk?kVrhgSA U= 1CV2ls,fLFkfrtÅtkZ ?kVrhgSA V E= d ls,fo|qr{ks=k?kVrkgSA (B) fu;rvkos'kij]tcIysVksadschpdhnwjhc<+kbZtkrhgSrks/k fjrk?kVrhgSA Q2 U= 2C ls,fLFkfrtÅtkZc<+rhgSA pwafdIysVksaijvkos'k?kuRofu;rgS]fo|qr{ks=klekujgrkgSA (C) fu;rfoHkokUrjij]tcIysVdk{ks=kQyc<+rkgSrks/k fjrkc<+rhgSA blfy;s U= 1CV2ls, fLFkfrt ÅtkZc<+rh gSA V E = d ls, fo|qr {ks=k leku jgrk gSA (D) fu;rvkos'kij]IysVksadk{ks=kQyc<+kusij 1Q2 U= 2 C ls,fLFkfrtÅtkZ?kVrhgSrFk pwafdIysVijvkos'k?kuRo?kVrkgS]blfy;sfo|qr{ks=k?kVrkgSA 19.44 The circuit involves two ideal cells connected to a 1 F capacitor via a key K. Initially the key K is in position 1 and the capacitor is charged fully by 2V cell. The key is pushed to position 2. Column  gives physical quantities involving the circuit after the key is pushed from position 1. Column II gives corresponding results. Match the statements in Column  with the corresponding values in Column  ifjiFk esa nks vkn'kZ lSy gSa tks dqath K }kjk 1 F la/kkfj=k ls tksM+s tk ldrs gSA izkjEHk esa dqath K fLFkfr 1 esa gS vkSj la/kkfj=k 2V lsy }kjk iw.kZr;k vkosf'kr gks tkrk gSA vc dqath dks fLFkfr 2 ij ys tk;k tkrk gSA LrEHk  esa dqath dks fLFkfr 1 ls /kdsyus ds ckn ifjiFk ls lEcfU/kr HkkSfrd jkf'k;k gSA LrEHk II esa mlds laxr ifj.kke fn;s x;s gSaA LrEHk  esa dFkuksa dks LrEHk  esa fn;s x;s laxr ekuksa ls lqesfyr dhft,A 1 2 Column Ι Column ΙΙ LrEHkI LrEHkII (A) The net charge crossing the 4 volt cell in C is (p) 2 (B) The magnitude of work done by 4 Volt cell in J is (q) 6 (C) The gain in potential energy of capacitor in J is (r) 8 (D) The net heat produced in circuit in J is (s) 16 (A) 4VcSVjhlsgksdjtkusokykdqyvkos'kCesagSA (p) 2 (B) 4VcSVjh}kjkfd;sx;sdk;Zdkifjek.kJesagSA (q) 6 (C) la/k fj=kdhfLFkfrtÅtkZesaykHkJesagSA (r) 8 (D) ifjiFkesamRiUudqyÅ"ekJesagSA (s) 16 Ans. (A) p (B) r (C) q (D) p Sol. (Moderate) The initial charge on capacitor = CVi = 1 × 2 C = 2 C The final charge on capacitor = CVf = 1 × 4 C = 4 C  Net charge crossing the cell of emf 4V is qf – qi = 4 – 2 = 2 C The magnitude of work done by cell of emf 4V is W = (qf – qi) 4 = 8 J The gain in potential energy of capacitor is U = 1 C(V 2  V 2 ) = 1 1 × [42 – 22] J = 6 J 2 f i 2 Net heat produced in circuit is H = W – U = 8 – 6 = 2 J (Moderate) la/kkfj=k ij izkjfEHkd vkos'k =CVi = 1 × 2 C= 2 C la/kkfj=k ij vafre vkos'k =CVf = 1× 4C= 4C 4V dhcSVjhlsxqtjusokykvkos'k qf – qi = 4 – 2 = 2 C 4V fo-ok-cydhcSVjh}kjkfd;sx;sdk;Zdkifjek.k W = (qf – qi) 4 = 8 J la/kfj=kdhfLFkfrtÅtkZesaykHk U = 1 C(V 2  V 2 ) = 1 1 × [42 – 22] J = 6 J 2 f i 2 ifjiFk esa mRiUu uSV Å"ek H = W – U = 8 – 6 = 2 J 19.45 In the given figure, the separation between the plates of C1 is slowly increased to double of its initial value then. Column-I Column-II (A) the potential difference across C1 (p) increases (B) the potential difference across C2 (q) decreases (C) the energy stored in C1 (r) increases by a factor of 6/5 (D) the energy stored in C2 (s) decreases by a factor of 18/25 (t) decreases by a factor of 9/25 fufnZ"Vfp=kesaC1dhIysVksdschpdhnwjh/khjs&/khjsizkjfEHkdekulsnqxquhdjnhxbZgSA dkWye -I dkWye -II (A) C1ijfoHkokUrj (p) c 10–3 m2  C = d k0 A d = C > 10–3 103 x C 103 x 50 x 1012 A > k0  A > (6) x  1 x109  = 300 mm2  36  LONG SUBJECTIVE fucU/kkRed iz'u 19.50 Two identical capacitor having plate separation d0 are connected parallel to each other across points A and B as shown in the figure. A charge Q is imparted to the system by connecting a battery across A and B and battery is removed. Now first plate of first capacitor and second plate of second capacitor starts moving with constant velocity u0 towards left. Find the magnitude of current flowing in the loop during this process. fp=kesan'k Z;svuqlkjnks,dlekula/kfj=kftudhIysVksadse/;vyxkod0gS]n'kZ;svuqlkjfcUnqvksaAoBij,dnwljs dslekUrjtqM+sgSA,dvkos'kQ,AoBij,dcSVjhdkstksM+djfudk;dksfn;ktkrkgSrFk cSVjhgVkyhtkrhgSA vcçFkela/k fj=kdhçFkeIysVof}rh;la/k fj=kdhf}rh;IysVfu;rosxu0lscka;hvksjpyukçkjEHkdjrhgSAbl çfØ;k ds nkSjku ywi esa çokfgr /kkjk dk ifjek.k Kkr djksA Sol. Let each plate moves a distance ‘x’ from its initial position. ekukizR;sdIysVizkjfEHkdfLFkfrls‘x’nwjhpyrhgSA Let q charge flows in the loop. using KVL ekukik'kesaqvkos'kizokfgrgksrkgSA  Q   2 q d  x  Q   2 q d  x       0 0 A Qx 0 A dq Q  dx   q = 2d ;  = dt = 2d  dt  ; Ans, Ι = 0 Q0 2d0 0   19.51 Given that CA = 1  F, C = 2 F and C = 2  F. Initially each capacitor was charged to potential differences of VA = 10V, VB = 40 V and VC = 60 V separately and are kept as shown in figure (a). Now they are connected as shown in figure (b). The + and - sign shown in figure (b) represent initial polarities. Find total amount of heat produced in µJ by the time steady state is reached. fn;k x;k gS CA = 1  F, CB = 2 F rFkk CC = 2  FA çkjEHk es rhuksa la/kkfj=k Øe'k% VA = 10 V, VB = 40 V rFkk VC = 60 V foHkokUrjksa rd vyx&vyx vkosf'kr fd;s x;s gaS rFkk fp=k (a) ds vuqlkj j[ks gq, gSaA vc bUgsa fp=k (b) ds vuqlkj tksM+ nsrs gaSA fp=k (b) es '+' o '–' fpUg çkjfEHkd /kzqork dks crkrs gSaA fLFkjoS+|qr voLFkk çkIr gksus rd mRiUu Å"ek dks dqy ek=kk µJ esa Kkr djksA Sol. Initial state Final state izkjfEHkdvoLFk vfUrevoLFk From conservationof charge vkos'k laj{k.k ls – q1 – q3 = 10 – 120 = – 110 µC (1) – q2 + q3 = – 80 + 120 = + 40 µC (2) Inthefinal state vfUre voLFkk esa q1 CA = q2 q3 CB + CC  q1 = q2 + q3 at q + q = 2q 1  q1 = 2 q2 + 2 2 3 1 q3 ; q + q = 2q ij 1 2 2 2 3 1 Solving weget gy djus ij ge izkIr djrs gSa q3=65µC. The charge on lower plate of capacitor CC changes from – 120 µC to – 65 µC. Hence the charge flowing through shown connecting wire is (120 – 65) = 55 µC. final charges q3 = 65µC ; q2 = 25µC ; q1 = 45µC Heat produced = Ui – Uf la/k fj=kCC dhfupyhIysVdkvkos'k–120µCls–65µCrdifjofrZrgksrkgS vr%fp=kuqlkjtqM+sgq,rkjesavkos'kizokg (120 – 65) = 55 µC. vafrevkos'k q3 = 65µC ; q2 = 25µC ; q1 = 45µC mRiUu Å"ek = Ui– Uf 120µC)2  (80C)2  (10C)2  65µC)2  (25C)2  (45C)2  =  2  2F 2  2F 2  1F  –  2  2F 2  2F 2  1F  = 3025 µJ Ans. 3025 19.52 In the figure shown the capacitor is initially uncharged. Find the current in R3 ( = R) at time ‘t’. fn[kk;s x;s fp=k esa izkjEHk esa la/kkfj=k vukosf'kr gSA le; 't' ij R3 ( = R) izfrjks/k esa /kkjk dk eku crkb;sA [4] Sol. Applying Kirchoff’slawin fdjpkWQ fu;e ls Loop1 ywi 1  – (i + i ) R – i R = 0 ...(1) 2 1 Loop 2 ywi 2 q – i2 R +  – C + i1 R = 0 (2) eliminating i1 from (1) and (2) lehdj.k (1)rFkk(2)ls i1gVkus ij  – q – i R +   i2 R = 0 or ;k 3 q 3 – – i R = 0 C 2 2 dq i2 = dt 2 C 2 2 q t  3C  2q = 3 R dq dq dt or;k = 2C 2 dt 3C  2q 0 3RC 0