2 This C, is in Now The system becomes Define capacitance of a capacitor. What is the ratio of electric field intensities at any two points between the plates of a parallel plate capacitor? The graph [Fig. 2 135] shows the variation of voltage V’ the plates of two capacitors A and B versus increase of cl 2 stored on them. Which of the tw pacitors has “apacitance? Give reasons for your answer.sto show how charge Q given to eapacitor of espachy) : are the dimensions of a capacitor? je fick! between two metal plates 3 mm apart, tc capacitor, the capacitance increases from 4 pF to wn hetween the plates. What is the dielectric const 5, What 6. Find the ele 7. Ina parallel pl a diclectrie me 8. Define the term ‘dielectri 9, Fora given potential difference, does a capacitor store more diclectrie than it does without a dielectric? 10. A parallel plate capacitor with ar between the plates has a capaci between the plates is now reduced by half'and the space between: of dielectric constant 5. Calculate the new value of capacitance, shown Fig. 2.136, which has lower capacitance and v v |. In the graph Fig. 2.136 12. In the graph shown in Fig. 2.137, which has higher capacitance 2 Fig. 2137 13. In a parallel plate capacitor, the potential difference of 10° V is maintained between the plates 2136) What willbe the electric field at points A and 14. Two protons A and Bare placed between plates’ the t Vas shown in figure 2.139. a i : i Fig. 2139 Will these protons experience equal or unequal force? “aENERGY No 45, Deseribe any to Uses of a capacitor Say IG Which of the following ar polar molecules: 1. ¢ Hic Hy ifthe plates of charged capacitors be suddeniy ooo? Cs Ny CO,2 Tam 7 Somnected 1 cach other by wie, wp Bf; Onihat factors docs the capacitance o capacitor fact ‘apacitor depend? ‘Acapaitoris charged through a potential ditereneeo 100V, init. How much energy will be released iti dichargeg? =" 2¥Cears sre Substances? Which of the follo wing is a dielectric: silicon, mica dielectric strength the help of an example show that farad is ay many picofarads are there in one farad? electric field induced in a dielectric when field. Calculate the re ery big unit for capacitance. placed in an external itis times the Permittivty ofthe dielectri, ; xI0 me $e of a capacitor is the magnitude of charge on the conductors (plates) when ference between the conductors (plates) i unity intensity between the plates of a charged. V Therefore, the ratio of electric field two points between the plates of a As Cis constant, therefore, Q « ¥. (See Fig, 2141) ial difference SIM L? 7442) v=Exd é - 12= £x3x1092, B=4x10 Vor 7. We know that C= KC, . : Where C’= capacitance of the parallel plate capacitor wit dielectric medium it ic medi | 4nd _C, = capacitance of the parallel plate capacitor without dielectric medium oe rer) Fig. 2.141 medium is the ratio of the capacitance ofa capacitor with dielectric acitor without dielectric medium between its plates,La have their centres separated by a g rest electrostatic repulsion ifthe charge yeep gible compared tothe distance of separa, me of 506m eo 44088 is charged double the above amouny hh sphet spulsion if and is halved? ANSWERS ofa single capacitor be C 3 ; Fj a \ Fig. 2.146 ih ak lly in one capacitor =~ CI" = ( stored finally wth two capacitors= {5 ) {The pot al will be divided equally je medium between the plates of a parallel plate capacitor the s decreases. This leads to dec the potential difference decrea es, more charge can be store 941075 1 + & x 470 x 10° x (20 Bp 470 = 10™® x (20) cd 1x10? B85 x10 The size of the plate is too big to According to Ciauss's 18%,inside the Gaussian surface my seonductor. But charged rewde an 4, The potential a the surface of 4 ig ie 4ne, r, and the potential at the surface of yj be zero, LY on the ote Surface 1 y, 4 "ane, 4, ioce Woy, Therefore electrons Nows from A tof and curren and current flows from Bt 4 Q-cv y=2 c the capacitors are in series, (is s ec n series, is same on both the capacitors larger potential difference will be across 3 uF cap: ee spacitor. 20+20+C 40C = 1000 + 25¢ C= 66.67 uF lapacitance is more in case of parallel arrangement as compared o serie total charge, the capacitors should be connected in parallel 0-C,,¥. ‘total energy, the capacitors should be connected in parallel 1 mu=5C., fhe placed in an external clecric fc, the induced charges appearin fic produce an extra electric ficld which is opposite to the direct ‘Asa result, the electric field decreases. jpermanent electric dipole moment because ofthe unsymmetrical Placement of the atoms constituting them. Its magnitude is about 0.6 « 10 ~ Cm, which is ‘bout ten thousand times more than the induced dipole moment acquired by the molecule Because of large value of the intrinsic dipole moment of the molecules of water, the dielectric ‘constant of water is large. te Reeto art 2151. oe , The maximum charge given to metallic sphere of radius R does not depend on whether the sphere is hollow eolid. This is because the charge resides on the surface of sphere | Therefore, the two cases are identical inthis respect: of three identical capacitors in series i given by n The equivalent capacitance Cy are in series.) Therefore, c 3uF-5 > sntical capacitors in parallel is given by equivalent capacitance of three We‘ 2, wlo ed in series. The equivalent capacitance is 2 Bsn 5c. = 16F=6% 1.521 = 24,336 x 107 = 0.24N SHORT ANSWER T . itor is charged toa potential difference VB om the source. If the distance betweenELECTROSTATIC Pi or ENTIAL, POTENTIAL ENERGY aND CAPACITANC € 4 7 rcapacitor of capacitance Chas distance between plate ana 3 Z.Avery thin metallic mesh wire is placed as Ms ‘nig. 2148. Calculate the new Sy a Be | ive an expression for the energy store: 4, Derive a” rBy stored in a charg. q a reed oo capacitor with air as the medium between ‘What is oe by the polarisation of a dielectric? What jsits effect? iA parallel plate capacitor is charged by a battery. After Deano] ‘some time a dielectric slab of dielectric constant K equal ‘othe gap between the plates is inserted between the plates the battery connected. How will the () capacitance (i) potential derence been e plates (ii) the energy stored in the capacitor get affected? elplate capacitors charged bya battery. After sometime the battery is disconnect fa dielectric slab with its thickness equal to the plate separation is inserted b: “How will () the capacitance of the capacitor (i) potential difference b fev and Gil the energy stored in the capacitor be affected? Justify your answer series, as shown in , and C, having plate area each are connected ins Meal Partin c eof this combination with the capacitor Hofarea A each, but made up as shown 12 across the plates of the jal difference are 2.150. 8. Determine the pot capacitor C, of the ‘network shown in Fig. Given E, > E,) 9. Fig. 2. 1st ea a sheet ofaluminium Pilaf meats peace m - How wil i placed between the plates ofa capacioe be affected if (a the fil is electrically insulted? (6) the foil is connected to the upper wire? 10. When a dielectric is inserted betw rie, fhe intervening Tes Psat plate apaiog cco ere medium affect the net electric fcld? For linear tric show that the intl a rteristi of the dielectric. ich se characte 1, are connected () in series (ii in parallel. Derive ce Xp + for each of these combinations ‘capacitance Fane of parallel plate capacitor using plates of surface plate with a conducting cen the plates of a chargedve of ont puysice-xi ee ppacitance of a parallel plate capacitor using plat with a diclectric slab of thickness "Fang sn expression for the ea ‘area ‘4° and kept at a distance ‘d’ apart constant ‘K* present between the plates 14 Distinguish with the help ofa suitable diagram, the diference in the behavienr of ‘and a dielectric placed in an external electric field. How does polarised diet sey etre menie the original external ficld? ietenn ANSWERS 1. Let Q, E, C, Vand Vbe the charge, electric field, capacitance, potential ener, difference associated with the capacitor. When the battery is ciel tae between te plate is doubled let the related parameters be QC: ase Thon Q'=Q beonuse no chatge leaves or comes into the plates ofthe capacitor 1-7 tre the surfce charge denitie in the final and initial situations and A the ares of on” OF crons. which is same as the capacitance of the initial eap Refer to Art. 2.15. Refer to Art, 2.10.2. Before V (Because the battery is a= CV=Kq w= tovELECTROSTATIC POTENTIAL, POTENTIAL ENERGY AND CAPACITANCE & 2115 After insertion of plate Ay Capacitance, C= = toy 2 K Ae, K,e,A C= 2 and c,= 27> “The equivalent capacitance of C, and C, is K,4e) | Kite 5 ; menage - 4 _K Bess 7 15+K, a d actor C, can be assumed to be made up of two capacitors Cand Cy each of plate ‘separation d connected in series. cs mcr” Cs 4. Let V, and P, be the potential diflerence across the plates of Cy and C,. Then a=ch, eg =CY, Also V+; -B-F CC, - £1 =F ee (4 -F:) Guay iy C, oe Sees) we G+G, When a sheet of aluminium fol is placed, the set up converts into two identical capacitors 2. (@) connected in series. The capacitance of each capacitor is was not placed.2.116 = PHysics-xi 10, 13. 4 eo ica’ Cc” G=ac (b) If the foil is connected to the upper plate with a conductin; g wire then the tosh fl BRM Retail be only one capacitor formed betveealeaiantan foil and the lower plate and its capacitance will be C’ = 2C. aa ‘The polarisation of the dielectric medium decreases the electri s ic field present plates of the capacitor. The new electric field Prcoeupuerizen te E E = 7% where E, is the electric fcld in the absence of dielectric, Now V=Ed the absence of dielectri] New capacitance Refer to Art. 2.14, Refer to Art. 2.12. Refer to Art. 213. When a conductors placed in an external is zero, This happens because a conductor has' under the influence of external electric field tll the zero. When a dielectric is placed in an external - uniform electric field Ey, induced charges are developed and molecules convert into dipoles (if not originally a dipole). The electric field Fy exerts torque on these dipole and tend to orient the dipoles such that p'and £ have same reates a net negative and a net directions. This ¢ positive charges on the edges of dielectric and therefore an electric field inside the dielectric called electric field due to polarisation E., The direction of is opposite 1° Ey Asa reeult the electric field inside the dielectric is 5ELECTROSTATIC POTENTIAL, POTENTI sY AND CAPACIT }OSTATIC POTENTIAL, POTENTIAL ENERGY AND TANCE. Meet a, le a ie Ty I 27 30 60: @ o Find the resultant electric force on a charge Q. 2 4 ent capacitance of the network. Fora 300 V supply. de joross each capacitor. (See Fig. 2.156) Fig. 2.156 plate of a parallel plate cap ge on the capacitor, and E ged fully by connecting it toa battery of emf £ > he separation between the plates of the capacitor is ae (@ charge stored by the cap citor, (i field strength between the plates: a (iii) energy stored by the capacitor. 1 answer in each ©88@: . capacitors A and B are connected th the switch S closed. The switch is now opened and the free space between te plates ofthe Gapacitors is filled with a dielectric of dielectric constant Prind the () ratio of the total electrostatic energy stored tors before and after the introduction of the ‘charge in A and B after the introduction Justify you! ‘Two identical parallel plat to a battery of V volts witCe + consists of two concentric spherical aansors, held in positon by suitable insulating supports (see Fig 2158), Find the capacitance of the spherical capacitor (@) Show thatthe potential energy stored in a charged parallel plate E? where E is the electric field between the 5. (@ A spherical capacitor capacitor is Hf platen ICT} neraies 6 @ A parallel plate capacitor has a dielectric slab of : dielectric constant K between its: plates that covers 1/3 of the area ofits plates, as shown in the Fig. 2.159. 2 The total capacitance ofthe capacitor is C while that ofthe portion with dielectric in between is C,. When the capacitor is charged, the plate area covered by the dielectric gets charge Q, and the rest of the area F Ge gets charge Q,. The eletre field inthe dielectric is E, and that in the other portion is E,. Find CIC,, qq, and EVE. ‘ a" o Six pol cae apes toes ace exagon of side L. Find the electri field and electric potential at point Q. (1) $ of magnitude ‘g’on it plates Fig. 2464 (0 the final charges of the capacitors of potential energy during the process: (ii) the loss a A cuboid is placed with its faces = and xz planes as 162, An electric O) parallel to xy, shown in the Fig. 2.1 field exist in the region given by E=2yj NC Find the electric flux passing through the cuboid and charge enclosed in it. 2ELECTROSTATIC POTENTIAL, POTENTIAL ENERGY AND CAPAC; TANCE, 249 potential difference across the plates of capacitor (2c) = 82 = 32 2CREC 30 c 60 ence on the left itor= © capacitor= 30 = ‘on the right capacitor 2 d, charge moves from plate of right capacitor (at higher potential) to that apacitor (at ow potential). Therefore charge on left capacitor increases. Jon Q due to charges q and q. The resultant of these forces in equal to each other is F'=V/2F PARALLEL —~a, —/ =| I G sv G Gy,C, and C,, are in series. Therefore ‘equivalent capacitance isdifference will get divided applied potent eh ia hil ane "3 5 equally ag Coe Gand Ga Te HUEY Tal ifferenee across Cand C;,is 100 V and there, er ae cy we = 20,000 pC x 10°C. entiat difference remains same. Therefore potential difference ary to al rip be 100 ‘V and therefore charge will be 1 * 18°C. ad o ‘The electric field due to the aa plate at point P is 9 : ~ @ charge is placed in the above field. The force on the plate + i carrying — Q charge is ; E.. Fee i) i atthe electric field Ebetween theplates uetochargesonboth | | | . the plates) is i s i ; a m Go) () From equations (i), (ii) and (fii), we have Fig. 2165 E F=0xe aie The origin of the factor 1/2 i that electric field produced by a plate (having charge -Q) does not apply a force on itself. Before After Ag, @ G==* 2 %=Co%o is disconnected) o_ & Ota aa @ AU, 2G (i) When S is closed Potential difference across each capacitors is 4 Cis the capacitance of each capacitor. 1 ils Total energy store = 300 co 1 When ‘S” is opened and dielectric is introduced Total energy stored = 0 2 aeT G(R +t 2 K K _ sharge.on B’ will not change after introduction ofthe dielectric slab (KOV. ov or @ {1 _1]_ 6-7) Arnel 4] Anon Q Ie gTi?2 eet 2 Le =—e,E" 2 2°! in parallel. Therefore 6. (@) This is a combinatic %G=C-C, C=G+G KA 24 FA and C-G= ee where G=3¢ 9 1 Fed(Let; be the electric field at 0 due to charges at F and C. Its directed fp m0 Les Fe nlak s| Again E isthe electric field at O due to charges at Band F. ‘The resultant ofthese electric fields is Ep = “+E? +2£,E, 008120° = B, directed from O to D Further Eis the electric field at O due to charges Aand D and is directed from O to D. py ies aa a “| ‘Therefore the total electric field at ‘O” is oO Ey = E,tE, 4 1g 1_2q | log, 2ax|_4|42|_1 29 eee ‘laste at Directed from O to D. Electric potential at “O° will be zero equal and opposite charges kept equi 0. 7. (@) Letq, and. gy be the find charges on: be the final potential difference across b g=4+q,and 06, Fig. 2.166 respectively andl" . Then Aca collects a dielecti The diel ws @25 2. Two cap connect potential from the (@ 50 nC (1800 3. The is C wil space b dielectr slab. T become c ws (© zr0 4. In the 20 differeELECTROSTATIC (b) Flux through face AHGC is , =EA cos 0=0 Flux through face AEFB is , = EA cos 0 = ax (2ax a) x cos 0° 2= 4a" Flux through all other faces are zero as direction of electric field is along their faces. Total flux = 4a? Charge enclosed = $ ¢ = 4a°e, OBJECTIVE TYPE QUESTIONS uw ipte Choice Questions 1. A capacitor connected to a 10V battery collects a charge of 40 uC with air as dielectric, and 100 1C with oil as dielectric. | i The dielectric constant of oil is @c (6) 2c , d (ys (© 3C @ = 25 A parallel plate capacitor is first charged “a be S es ey. Thee between