Good quality notes on Physics - Capacitance. Very helpful for students, Study notes of Physics

Download Good quality notes on Physics - Capacitance. Very helpful for students and more Study notes Physics in PDF only on Docsity!   Digital Pvt. Ltd. [1] Capacitor Definition A capacitor or condenser consists of two conductors separated by an insulator or dielectric. Having opposite charges on which sufficient quantity of charge may be accommodated. It is a device which is used to store energy in form of Electric field by storing charge. Conductors are used to form capacitors. It is the maximum amount of charge that a capacitor can have Theoretically, infinite amount of charge can be given to any conductor Practically, only a certain amount of charge can be given because of dielectric strength of air Dielectric strength The maximum value of electric field which can be tolerated by a medium without breakdown Dielectric strength of air E = 3 × 106 N/C Corona discharge: There is more possibility of leakage of charge from sharp corners. Electrical Capacitance It shows the capacity of a conductor to store electric energy in the form of electric field. If charge(Q) is given to an isolated conducting body and it's potential increases by V, then Q∝𝑉 Q=𝐶 𝑉 Q C V =  constant [electrical capacitance of a conductor] Capacitance of conductor depends upon shape, size, presence of medium and nearness of other conductor. Capacitance TG: @Chalnaayaaar   Digital Pvt. Ltd. [2] Capacitance Note: Capacitance of a conductor does not depend upon charge (Q) and potential difference (V). Graph between Q and V: Electrical capacitance is a Scalar quantity. 9 111C 3 10 stat C 1F 9 10 stat F 11V stat volt 300  − = = =  − − 1F = 9 × 1011 state – F Dimensional formula of capacitance 1 2 4 2M L T A− −    Capacitor When 2 conductors which carry equal and opposite charges are separated by some distance having some medium between them, then this arrangement is known as capacitor. Capacitor is electrically neutral. Q C V = Q → Charge transferred Or Charge on (+)ve plate V → Potential difference b/w plates Principle of Capacitor When uncharged conductor is placed nearer to the charged conductor and uncharged conductor is connected to earth, then capacitance of charged conductor is increased. TG: @Chalnaayaaar   Digital Pvt. Ltd. [5] Capacitance If b >> a then C = 40a (Like isolated spherical capacitor) If dielectric mediums are filled as shown then: C = 4𝜋∈0∈𝑟𝑎𝑏 𝑏−𝑎 Illustration 2: The stratosphere acts as a conducting layer for the earth. If the stratosphere extends beyond 50 km from the surface of earth, then calculate the capacitance of the spherical capacitor formed between stratosphere and earth's surface. Take radius of earth as 6400 km. Solution: The capacitance of a spherical capacitor is C = 40 ( 𝑎𝑏 𝑏−𝑎 ) b = radius of the top of stratosphere layer = 6400 km + 50 km = 6450 km = 6.45 × 106 m a = radius of earth = 6400 km = 6.4 × 106 m  𝐶 = 1 9×109 × 6.45×106×6.4×106 6.45×106 – 6.4×106 = 0.092 F (b) Inner sphere is earthed Here the system is equivalent to a spherical capacitor of inner and outer radii a and b respectively and a spherical conductor of radius b in parallel. This is because charge Q given to outer sphere distributes in such a way that for the outer sphere r r r a b 3 2 1 TG: @Chalnaayaaar   Digital Pvt. Ltd. [6] Capacitance charge on the inner side is 𝑎 𝑏 Q and charge on the outer side is Q – 𝑎 𝑏 Q = (𝑏−𝑎) 𝑏 Q So total capacity of the system. C = 4  0 𝑎𝑏 𝑏−𝑎 + 4  0 b 𝐶 = 4𝜋𝜀0𝑏 2 𝑏 − 𝑎 Cylindrical Capacitor There are two co-axial conducting cylindrical surfaces where >> a and >> b, where a and b are radius of cylinders. When a charge Q is given to inner cylinder it is uniformly distributed on its surface. A charge –Q is induced on inner surface of outer cylinder. The charge +Q induced on outer surface of outer cylinder flows to earth as it is grounded Electrical field between cylinders E = λ 2πε0r = Q/ℓ 2πε0r Potential difference between plates V = ∫ Q 2πε0rℓ dr = Q 2πε0ℓ ℓn ( b a ) b a Capacitance per unit length C = λ V = λ 2Kλℓn b a = 4π∈0 2ℓn b a = 2π∈0 ℓn b a Capacitance per unit length = 2π∈0 ℓn b a F/m TG: @Chalnaayaaar   Digital Pvt. Ltd. [7] Capacitance Illustration 1: A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5 µC. Determine the capacitance of the system and the potential of the inner cylinder. Solution:  = 15 cm = 15 × 10–2 m;  a = 1.4cm = 1.4 × 10–2 m; b =1.5 cm = 1.5 × 10–2m; q = 3.5 µC = 3.5 × 10–6C Capacitance C = 2πε0ℓ 2.303 log10( b a ) = 2π×8.854×10–12×15×10– 2 2.303 log10 1.5×10–2 1.4×10–2 = 1.21 × 10–10 F Since the outer cylinder is earthed, the potential of the inner cylinder will be equal to the potential difference between them. Potential of inner cylinder, is V = q C = 3.5×10–6 1.2×10–10 = 2.89 × 104V Work done by external agent to charge a conductor: Work done by external agent to bring small charged element dq from infinity to surface of conductor. dw = V dq dw V.dq=  V= q C Work done by external agent to charge the conductor from q1 to q2 is 2 22 1 1 1 q qq 2 2 q q q q 1 q 1 q W .dq W W C C 2 C 2     =  =  =         As q1 = CV2, q2 = CV2 work done by external agent to increase the potential of conductor from V1 to V2 ( )2 2 2 1 1 W C V V 2 = − Illustration 1: The work done against electric forces in increasing the potential difference of a condenser from 20 V to 40 V is W. The work done in increasing its potential difference from 40 V to 50 V will be TG: @Chalnaayaaar   Digital Pvt. Ltd. [10] Capacitance Now, work done = 𝜀0𝑉 2𝐴 2𝑑 Parallel Plate Capacitor It consists of two large plane parallel conducting plates separated by a small distance. Capacitance of Parallel Plate Capacitor Surface charge density Q / A = Electric field intensity between plates 0 0 0 E 2 2   = +   0 0 0 0 Q E E A  =  =   Potential difference between the plates 0V E d=  0 Qd V A =  Capacitance of parallel plate capacitor 0 0 0 Q Q C C V Qd A =  =       0 0 A C d   = Illustration 1: One plate of parallel plate capacitor is smaller than other, then charge on smaller plate will be (A) Less than other (B) More than other TG: @Chalnaayaaar   Digital Pvt. Ltd. [11] Capacitance (C) Equal to other (D) Will depend upon the medium between them Solution: Equal to other, because the charges are produced due to induction and moreover the net charge of the capacitor should be zero. Capacitance of Parallel Plate Capacitor with dielectric Electric field intensity in vacuum 0E Electric field intensity in medium mE m 0 r 0 r Q E A   = = =      0 m r E E =  Potential difference between the plates m 0 r Qd V E d A =  =   Capacitance of a capacitor in presence of medium TG: @Chalnaayaaar   Digital Pvt. Ltd. [12] Capacitance m Q C V = ( )0 r m A Q C Qd   = 0 r m m r 0 A C C C d    =  =  Capacitance of Parallel Plate Capacitor (PPC) depends on (i) Overlapping area C A  (ii) Distance between plates 1 C d   (iii) Medium between plates rC   * Capacitance of a parallel plate capacitor does not depend on thickness and nature of metal of plates. Specific Inductive Capacitance It is the ratio of capacitance in medium to the capacitance in air. m air C K C = Fringing of Electric Field For the plates of finite area the electric field between the two plates will not be uniform and the field lines bend out ward at the edges. This is called "fringing of electric field" TG: @Chalnaayaaar   Digital Pvt. Ltd. [15] Capacitance Here, 0 F Q 2   =     2 2 0 0 Q Q A F A 2A 2   = = =   Again 2 0 (CV) F 2A =  0 0 A C A Cd d  =   = 2 2 2 2C V CV CV F F 2 Cd 2d 2d = =  = Pressure on each plate of a capacitor: As 2 0 A F 2  =  or 2 0 F P A 2  = =  This is known as electrostatic pressure. * Electrostatic pressure always act perpendicular to surface and outwards. Sharing of charges When two charged conductors are connected by a conducting wire then charge flows from a conductor at higher potential to that at lower potential. This flow of charge stops when the potential of both conductors become equal. Let the amounts of charges after the conductors are connected be Q1' and Q2' respectively and their common potential be V then Common potential According to the law of conservation of charge Q before connection = Q after connection C1V1 + C2V2 = C1V + C2V Common potential after connection 1 1 2 2 1 2 C V C V V C C + = + TG: @Chalnaayaaar   Digital Pvt. Ltd. [16] Capacitance Charges after connection 1 ' 1 2 1 1 1 1 2 1 2 Q Q C Q C V C Q C C C C    + = = =    + +    (Q = Total charge on the system) 2 ' 1 2 2 2 2 1 2 1 2 Q Q C Q C V C Q C C C C    + = = =    + +    Ratio of the charges after redistribution 1 1 1 2 2 2 Q ' C R (in case of spherical conductors) Q ' C R = = Energy stored in capacitor 2 2Q CV QV Energy = 2 C 2 2 = = Energy density In PPC, potential energy stored in the form of electric field i.e. in the space between two plates and volume of this space is (A × d). Energy density = 21 CV Energy 2 volume Ad = Here, 0A C d  = , V = Ed Energy density = 2 0A1 (Ed) 2 d Ad   2 0 1 Energy density E 2 = As, 0 E  =  So, 2 0 1 Energy density 2   = Charging of PPC by battery when a capacitor is charged by a battery then battery charges it till than its potential difference becomes equal to EMF of battery. Whenever any capacitor (Initially charged or uncharged) connected to battery then its final voltage is always equal to emf of battery. Final potential energy of capacitor = 21 CV 2 TG: @Chalnaayaaar   Digital Pvt. Ltd. [17] Capacitance Work done by battery W = Charge supplied by battery  EMF W = qV W = CV2 * Energy supplied by battery = CV2 From conservation of energy Wby battery = U Heat loss Change in P.E.  +  2 21 CV CV 0 heat loss 2   = − +    21 Heat loss CV 2  = Note : (1) When uncharged capacitor is connected to battery then 50% of energy supply by battery is stored in capacitor and remaining 50% will be lost. (2) Energy loss does not depends on resistance of circuit. When initially capacitor is charged then heat loss is not equal to 21 CV 2 , find heat loss by use of following concept Wby battery =  U + Heat loss Illustration 1: An uncharged capacitor having capacitance C is connected across a battery of emf V. Now the capacitor is disconnected and then reconnected across the same battery but with reversed polarity. Find Heat loss. Solution: Total charge flow Qf – Qi = (CV – (–CV)) = 2CV W.D. by battery = (2CV)(V) = 2CV2 Heat loss = W.D. by battery – Energy stored in capacitor = 2CV2 – 1 2 CV2 = 3 2 CV2 For Parallel Plate Capacitor * If like plate are connected or nothing is given TG: @Chalnaayaaar   Digital Pvt. Ltd. [20] Capacitance Ceq = Q V (c) Here same electrical effect refers to the charge flown through the battery or charge transferred between the terminals plates of capacitors connected to battery. It could also be understood in form of energy stored being same in two cases i.e. when original capacitor network is connected and the one when original network is replaced by equivalent capacitance. There Are Two Standard Combinations: (1) Series Combination (2) Parallel Combination (1) Series Combination When initially uncharged capacitors are connected as shown so that charges do not have any alternative path(s) to flow then the combination is called series combination. Derivation: To find equivalent capacitance of this combination lets connect a battery across its terminals. ≡ Ceq = Q V Lets assume that initially, the capacitors were uncharged and after connecting to battery, Q charge flows through the battery as shown in above figure. Applying Kirchhoff's voltage law A B C1 C2 C3 Ceq V BA TG: @Chalnaayaaar   Digital Pvt. Ltd. [21] Capacitance −( Q C1 ) − ( Q C2 ) − ( Q C3 ) + V = 0. ; V Q = 1 C1 + 1 C2 + 1 C3  1 Ceq = 1 C1 + 1 C2 + 1 C3 or in general 1 Ceq = n n 1 n 1 C=  Salient features of series combination (a) All capacitors will have same charge but can have different potential difference across them. (b) We can say that potential difference across capacitor is inversely proportional to its capacitance in series combination  1 C V1 = Q C1 , V2 = Q C2 , V3 = Q C3 … (c) V1 : V2 : V3 = 1 C1 : 1 C2 : 1 C3 In series combination the smallest capacitor gets maximum potential. (d) V1 = 1 C1 1 C1 + 1 C2 + 1 C3 +...... V V2 = 1 C2 1 C1 + 1 C2 + 1 C3 +...... V V3 = 1 C3 1 C1 + 1 C2 + 1 C3 +. . . . . . V Where V = V1 + V2 + V3 (e) In series : 1 Ceq = 1 C1 + 1 C2 + 1 C3 + . . . . . .. In series combination equivalent capacitance is always less than the smallest capacitor of combination. (f) Energy stored in the combination Ucombination = Q2 2C1 + Q2 2C2 + Q2 2C3 Ucombination= Q2 2Ceq Energy supplied by the battery in charging the combination Ubattery = Q × V = Q . Q Ceq = Q2 Ceq TG: @Chalnaayaaar   Digital Pvt. Ltd. [22] Capacitance Ucombination Ubattery = 1 2 Half of the energy supplied by the battery is stored in form of electrostatic energy and half of the energy is converted into heat through resistance. (if capacitors are initially uncharged) Illustration 1: Three initially uncharged capacitors are connected in series as shown in circuit with a battery of emf 30 V. Find out following: (i) charge flow through the battery, (ii) potential energy in 3F capacitor. (iii) Utotal in capacitors (iv) heat produced in the circuit Solution: 1 Ceq = 1 2 + 1 3 + 1 6 = 3 + 2 + 1 6 = 1 Ceq = 1F. (i) Q = Ceq V = 30 C. (ii) Charge on 3F capacitor = 30 C Energy = Q2 2C = 30×30 2×3 = 150J (iii) Utotal = 30×30 2 J = 450 J (iv) Heat produced = (30 C) (30) – 450 J = 450 J. Parallel Combination: When one plate of each capacitors (more than one) is connected together and the other plate of each capacitor is connected together, such combination is called parallel combination. Derivation Q = Q1 + Q2 + Q3 Q = C1V + C2V + C3V = V(C1 + C2 + C3) Q V = C1 + C2 + C3 Ceq = C1 + C2 + C3 In general Ceq = ∑ Cn n n=1 Salient Features of Parallel Combination (a) All capacitors have same potential difference but can have different charges. TG: @Chalnaayaaar   Digital Pvt. Ltd. [25] Capacitance Combination of Parallel Plates: Combinations of parallel plates can be grouped and equivalent capacitance can be found easily by redrawing the circuit by connecting plates to common terminals. This has been illustrated by following example. Four parallel plates each of area A and plate separation d. These four plates are forming three parallel plate capacitors, one between plates A and B, one between plates B and C and another between plates C and D. Capacitance of each of these capacitors is given as C = ∈0A d . If plates A and C are connected as terminal X and plates B and D are connected as terminal Y. Then calculation of equivalent capacitance across terminal X and Y can be done by following method. Let us assume that we have connected a battery between terminal X and Y. This battery supplies equal and opposite charges to terminal X and Y. For the detailed analysis let us say 1 and 2 are charge density on faces b and e. Similarly 3 and 4 are charge density on faces f and a. Now equal opposite charges will be there on opposite faces as shown in figure (as per Gauss Law). Also electric field at any point inside conductor must be zero therefore 4 must be zero. Vbc = Ved = Vfg So all the capacitors are in parallel combination. Ceq = C1 + C2 + C3 = 3A∈0 d Let us try to find out equivalent capacitance between A and B. (take each plate Area = A) in following case by using the same above concept. A bc d e fg h B C D a x y 4 1 2 3 –1 –2 –3 –4 X Y – + C1 C2 C3 b X de f g Y c TG: @Chalnaayaaar   Digital Pvt. Ltd. [26] Capacitance Assuming a battery between terminals A and B the positive and negative charges on the plates will be as shown in the figure fellow. Here charges on plates d and e is zero because there potential must be same so no electric field should be there in between these faces, so charge on faces d and e must be zero. These are only two capacitors Ceq = C1 + C2 = 2A∈0 d Illustration 1: Find out equivalent capacitance between A and B. (take each plate Area = A) Solution: The modified circuit is Ceq = 2C 3 = 2A∈0 3d A B d d d a b e f c d g h A B – + + – C1 C2 c A f g B b A B d d d A+ – B d d d 0 – + + – + – 0 1 2 3 4 A B C C C C TG: @Chalnaayaaar   Digital Pvt. Ltd. [27] Capacitance Other method: Let charge density as shown Ceq = Q V = 2xA V V = V2 – V4 = (V2 – V3) + (V3 – V4) = xd ∈0 + 2xd ∈0 = 3xd ∈0  Ceq = 2Ax ∈0 3xd = 2A ∈0 3d = 2C 3 . Illustration 2: Find out equivalent capacitance between A and B. Solution: Let C = A∈0 d Equivalent circuit : 1 Ceq = 1 C + 2 3C = 5 3C Ceq = 3C 5 = 3A ∈0 5d Illustration 3: What is the effective capacitance between the points X and Y? A B 1 2 3 4 –x +x +x –x +2x –2x A – + + – – – + + C – B + d d d d Area A TG: @Chalnaayaaar   Digital Pvt. Ltd. [30] Capacitance Polar dielectrics • In absence of external field the centres of positive and negative charge do not coincide-due to asymmetric shape of molecules. • Each molecule has permanent dipole moment. • The dipole are randomly oriented so average dipole moment per unit volume of polar dielectric in absence of external field is nearly zero. In presence of external field dipoles tends to align in direction of field. Example: Water, Alcohol CO2, EC, NH3 Note: Dipole moment of polar molecules depends on temperature. Non-polar dielectrics • In absence of external field the centre of positive and negative charge coincides in these atoms or molecules because they are symmetric. • The dipole moment is zero in normal state. • In presence of external field they acquire induced dipole moment. Example: Nitrogen, Oxygen, Benzene, Methane ( ) Cl – H+ H+ H+ O – – HCl molecule H2O molecule TG: @Chalnaayaaar   Digital Pvt. Ltd. [31] Capacitance Note: Induced electric dipole moment of non-polar molecules is independent of temperature. Polarisation: The alignment of dipole moments of permanent or induced dipoles in the direction applied electric field is called polarisation. • Polarisation vector ?⃗? This is a vector quantity which describes the extent to which molecules of dielectric become polarized by an electric field or oriented in direction of field. P⃗ = the dipole moment per unit volume of dielectric = np⃗ where n is number of atoms per unit volume of dielectric and P is dipole moment of an atom or molecule. P⃗ = np⃗ = qid Ad = ( qi A ) = i = induced surface charge density. Capacitors with Dielectric: (i) In absence of dielectric E = σ ∈0 (ii) When a dielectric fills the space between the plates then molecules having dipole moment align themselves in the direction of electric field. b = i induced (bound) charge density (called bound charge because it is not due to free electrons). The induced charge also produce electric field. Let E0 , V0 , C0 be electric field, potential difference and capacitance in absence of dielectric and E, V, C are electric field, potential difference and capacitance respectively in presence of dielectric . O– – C++++ O– – O– – C++++ O– – C++++ O– – O– – TG: @Chalnaayaaar   Digital Pvt. Ltd. [32] Capacitance Electric field in absence of dielectric E0 = V0 d = σ ε0 = Q ε0A Electric field in presence of dielectric E = E0 − Ei = σ−σb ε0 = Q−Qb ε0 = V d Capacitance in absence of dielectric C0 = Q V0 Capacitance in presence of dielectric C = Q−Qb V The dielectric constant or relative permittivity K or r = E0 E = V0 V = C C0 = Q Q−Qb = σ σ−σb = ε ε0 Form K = Q Q−Qb ⇒ Qb = Q (1 − 1 K ) and K = σ σ−σb ⇒ σb = σ (1 − 1 K ) (iii) Capacitance in the presence of dielectric C = σA V = σA σ K ∈0 . d = AK ∈0 d = AK ∈0 d Here capacitance is increased by a factor K. + + + + + + + + + + + – – – – – – – – – – – + – + + + + – – +Q0 – – Ed Electric field inside dielectric E0 –Q0 TG: @Chalnaayaaar   Digital Pvt. Ltd. [35] Capacitance 10 r 1 1 A C d   = 20 r 2 2 A C d   = These two are in parallel so, C = C1 + C2 1 20 r 1 0 r 2A A C d d     = + Special case : If A1 = A2 = A 2 Then, 1 2r r0A C d 2  +   =     Illustration 1: Find out capacitance between A and B if three dielectric slabs of dielectric constant K1 of area A and thickness d1, K2 of area A and thickness d2 and K3 of area A and thickness d3 are inserted between the plates of series plate capacitor of plate area A as shown in figure. (Given distance between the two plates d). Solution: AB 1 2 3 1 1 1 1 C C C C = + + 0 1 0 2 0 3 1 2 3 1 2 3 AK AK AK C ; C ;C d d d     = = =    Illustration 2: Find out capacitance between A and B if three dielectric slabs of dielectric constant K1 of area A1 and thickness d, K2 of area A2 and thickness d1 and K3 of area A2 and thickness d2 are inserted between the plates of parallel plate capacitor of plate area A as shown in figure. (Given distance between the two plates d). TG: @Chalnaayaaar   Digital Pvt. Ltd. [36] Capacitance Solution: 1 0 2 0 3 0 1 2 3 A A A K K K 2 2 2C C C d d d 2 2 2    = = = It is equivalent to 3 1 2 1 2 C C C C 2 C C = + + Illustration 3: Find out capacitance between A and B if three dielectric slabs of dielectric constant K1 of area A1 and thickness d, K2 of area A2 and thickness d1 and K3 of area A2 and thickness d2 are inserted between the plates of parallel plate capacitor of plate area A as shown in figure. (Given distance between the two plates d = d1+d2). TG: @Chalnaayaaar   Digital Pvt. Ltd. [37] Capacitance Solution: It is equivalent to C = C1 + C2C3 C2+C3 C = A1K1 ∈0 d1 + d2 + A2K2 ∈0 d1 . A2K3 ∈0 d2 A2K2 ∈0 d1 + A2K3 ∈0 d2 = A1K1 ∈0 d1 + d2 + A2 2K2K30 2 A2K2 ∈0 d2 + A2K3 ∈0 d1 = A1K1 ∈0 d1 + d2 + A2K2K3 ∈0 K2d2 + K3d1 Illustration 4: Find out capacitance between A and B if two dielectric slabs of dielectric constant K1 and K2 of thickness d1 and d2 and each of area A are inserted between the plates of parallel plate capacitor of plate area A as shown in figure. Solution: : C = σA V ; V = E1 d1 + E2 d2 = σd1 K1∈0 + σd2 K2∈0 = σ ∈0 ( d1 k1 + d2 k2 )  C = A∈0 d1 K1 + d2 K2  1 C = d1 AK1∈0 + d2 AK2∈0 This formula suggests that the system between A and B can be considered as series combination of two capacitors. TG: @Chalnaayaaar