PITNE Page #. INTROUTION capacitor can store energy in the form of potential energy in an electric field. In this chapter well discuss the capacity of conductors to hold charge and energy.. apacitance of an isolated conductor When a conductor is charged its potential increases. It is found that for an isolated conductor (conductor should e of finite dimension, so that potential of infinity can e assumed to e zero) potential of the conductor is proportional to charge given to it. = charge on conductor = potential of conductor = Isolated conductor Where is proportionally constant called capacitance of the conductor.. efinition of capacitance : apacitance of conductor is defined as charge reuired to increase the potential of conductor y one unit..3 Important point aout the capacitance of an isolated conductor : It is a scalar uantity. Unit of capacitance is farad in SI unis and its dimensional formula is M L I T 4 Farad : Farad is the capacitance of a conductor for which coulom charge increases potential y volt. Farad = oulom olt F = 6 F, nf = 9 F or pf = F apacitance of an isolated conductor depends on following factors : (a) Shape and size of the conductor : On increasing the size, capacitance increase. () On surrounding medium : (c) With increase in dielectric constant K, capacitance increases. Presence of other conductors: When a neutral conductor is placed near a charged conductor capacitance of conductors increases. apacitance of a conductor does not depend on (a) harge on the conductor () Potential of the conductor (c) Potential energy of the conductor..4 apacitance of an isolated Spheical onductor. Ex. Find out the capacitance of an isolated spherical conductor of radius R. Let there is charge Q on sphere. KQ Potential R Hence y formula : Q =
Page # PITNE KQ Q R = 4 R (i) (ii) If the medium around the conductor is vacuum or air.: vacuum = 4 R R = Radius of spherical conductor. (may e solid or hollow) If the medium around the conductor is a dielectric of constant K from surface of sphere to infinity then medium = 4 KR (iii) medium air / vaccum = K = dielectric constant.. PITOR : capacitor or condenser consists of two coductors separated y an insulator or dielectric. (i) When uncharged conductor is rought near to a charged conductor, the charge on conductors remains same ut its potential dcreases resulting in the increase of capacitance. (ii) In capacitor two conductors have eual ut opposite charges. (iii) The conductors are called the plates of the capacitor. The name of the capacitor depends on the shape of the capacitor. (iv) Formulae related with capacitors: (a) Q = Q Q Q Q = harge of positive plate of capacitor. = Potential difference etween positive and negative plates of capacitor = apacitance of capacitor. (v) The capacitor is represented as following :, (vi) ased on shape and arrangement of capacitor plates there are various types of capacitors: (a) Parallel plate capacitor () Spherical capacitor. (c) ylindrical capacitor (v) apacitance of a capacitor depends on (a) rea of plates. () istance etween the plates. (c) ielectric medium etween the plates.. Parallel Plate apacitor Two metallic parallel plates of any shape ut of same size and separated y small distance constitute parallel plate capacitor. Suppose the area of each plate is and the separation etween the two plates is d. lso assume that the space etween the plates contains vacuum.
PITNE Page # 3 We put a charge on one plate and a charge on the other. This can e done either y connecting one plate with the positive terminal and the other with negative plate of a attery (as shown in figure a ) or y connecting one plate to the earth and y giving a charge to the other plate only. This charge will induce a charge on the earthed plate. The charges will appear on the facing surfaces. The charges density on each of these surfaces has a magnitude = /. or (a) () If the plates are large as compard to the separation etween them, then the electric field etween the plates (at point ) is uniform and perpendicular to the plates except for a small region near the edge. The magnitude of this uniform field E may e calculated y using the fact that oth positive and negative plates produce the electric field in the same direction (from positive plate towards negative plate) of magnitude / and therefore, the net electric field etween the plates will e, E Outside the plates (at point and ) the field due to positive sheet of charge and negative sheet of charge are in opposite directions. Therefore, net field at these points is zero. The potential difference etween the plates is, E.d d d The capacitance of the parallel plate capacitor is, d or d. ylindrical apacitor ylindrical capacitor consists of two co-axial cylinders of radii a and and length l. If a charge is given to the inner cylinder, induced change will reach the inner surface of the outer cylinder. y symmetry, the electric field in region etween the cylinders is radially outwards. y Gauss s theorem, the electric field at a distance r from the axis of the cylinder is given y E l The potential difference etween the cylinders is given y a E dr l or, l a In a r dr r l a In l a
Page # 4 PITNE.3 Spherical apacitor spherical capacitor consists of two concentric spheres of radii a and as shown. The inner sphere is positively charged to potential and outer sphere is at zero potential. The inner surface of the outer sphere has an eual negative charge. The potential difference etween the spheres is Q Q 4 a 4 Hence, capacitance Q 4a a ( a) Ex. Find capacitance of the given system. Q Q =rea d rranging charges E Now, = Ed = Q Qd 3Q Q Q 3Q Q / d 3. ENERGY STORE IN HRGE PITOR Q Q d initially Finally middle state Work has to e done in charging a conductor against the force of repulsion y the already existing charges on it. The work is stored as a potential energy in the electric field of the conductor. Suppose a conductor of capacity is charged to a potential and let e the charge on the conductor at this instant. The potential of the conductor when (during charging) the charge on it was (< ) is, Now, work done in ringing a small charge d at this potential is, dw d d total work done in charging it from to is,
PITNE Page # 5 W dw d This work is stored as the potential energy, U Further y using = we can write this expression also as, U In general if a conductor of capacity is charged to a potential y giving it a charge, then U 3. Energy ensity of a harged apacitor This energy is localized on the charges or the plates ut is distriuted in the field. Since in case of a parallel plate capacitor, the electric field is only etween the plates, i.e., in a volume ( d), the energy density U E U volume d d d or U E d E v d E 3. alculation of apacitance The method for the calculation of capacitance involves integration of the electric field etween two conductors or the plates which are just euipotential surfaces to otain the potential difference a. Thus, a a a E. dr 3.3 Heat Generated : () Work done y attery W = Q Q = charge flow in the attery = EMF of attery () W = e (When attery discharging) W = e (When attery charging) (3) Q = ( = euivalent capacitance) so W = = a Now energy on the capacitor Energy dissipated in form of heat (due to resistance) H = Work done y attery {final energy of capacitor - initial energy of capacitor} E.dr
Page # 6 PITNE Ex.3 t any time S switch is opened and S is closed then find out heat generated in circuit. S S initially finally harge flow through attery = Q f Q i = = H = ( ) () H Ex.4 (a) Find the final charge on each capacitor if they are connected as shown in the figure. F 5F F 5 Initially c c 3 Finally let charge flows clockwise then Now applying KL ( ) ( ) 3 ( ) ( ) ( ) ( ) so finally 5 5 5 5 = 75 5 = 6 75/ 6 55 35 6 3 6
PITNE Page # 7 () Find heat loss in the aove circuit. H = Energy [initially finally] on capacitor 5 () (5) 55 6 75 5 6 35 6 6 J 4. ISTRIUTION OF HRGES ON ONNETING TWO HRGE PITORS : When two capacitors and are connected as shown in figure Q Q Q Q Initially Q Q Q Q Finally efore connecting the capacitors Parameter I st apacitor II nd apacitor apacitance harge Q Q Potential fter connecting the capacitors Parameter I st apacitor II nd apacitor apacitance harge Q Q Potential (a) ommon potential : y charge conservation on plates and efore and after connection. Q Q = Q Q = () Q (Q Q ) Q (Q Q ) (c) Heat loss during redistriution : H U i Uf ( ) Total charge Total capaci tance The loss of energy is in the form of Joule heating in the wire.
Page # 8 PITNE When plates of similar charges are connected with each other (with and with ) then put all values (Q, Q,, ) with positive sign. When plates of opposite polarity are connected with each other ( with ) then take charge and potential of one of the plate to e negative. erivation of aove formulae : Let potential of and is zero and common potential on capacitors is, then at and it will e. = H ( ) = ( ) ( ) H = ( ) ( ) when oppositely charged terminals are connected then = H ( ) Ex.5 Find out the following if is connected with and is connected with. (i) How much charge flows in the circuit. (ii) How much heat is produced in the circuit. F 3F
PITNE Page # 9 Let potential of and is zero and common potential on capacitors is, then at and it will e. y charge conservation, 3 = 4 3 5 = 7 = 4 volt harge flow = 4 8 = Q = Q =3 Now final charges on each plate is shown in the figure. 8 8 4 4 (ii) Heat produced = () 3 () 5 (4) = 4 5 49 = 55 49 = 6 J When capacitor plates are joined then the charge remains conserved. We can also use direct formula of redistriution as given aove. Ex.6 Repeat aove uestion if is connected with and is connected with. Q = Q =3 Initial 36 4 4 36 6 6 Final Let potential of and is zero and common potential on capacitors is, then at and it will e 3 = = volt Now charge on each plate is shown in the figure. Heat produced = 4 5 5 4 = 55 = 54 J Here heat produced is more. Think why? Ex.7 Three capacitors as shown of capacitance F, F and F are charged upto potential difference 3, and 5 respectively. If terminal is connected with, is connected with E and F is connected with. Then find out charge flow in the circuit and find the final charges on capacitors.
Page # PITNE 3 F F E 5 F F Let charge flow is. Now applying Kirchhoffs voltage low ( ) (3 ) 3 = 5 =.5 Final charges on plates 7.5 7.5 4.5 4.5 3 3 F 3 3 F F E F.5 7.5 7.5 5. PITOR IRUITS Ex.8 Find charge on each capacitor. F F 4F 5 harge on = = ( 5) = 3 harge on = = ( ()) = 6 harge on 3 = 3 3 = 4 ( ) = 4 5 O O O O 3 F F 4F Ex.9 Find charge on each capacitor. harge on = (x ) harge on = (x ) harge on 3 = (x ) 3 Now from charge conservation at node x (x ) (x ) (x ) 3 = x x 4x 8 = x = 5/ so Q 5 5 x O O 3 x O
PITNE Page # Q Q 3 5 5 (5 / 4) 4 3 5 5 5 5 3 3 Ex. In the given circuit find out the charge on each capacitor. (Initially they are uncharged) F F 5 F 3 F E G 5 5 3 3 Let potential at is, so at it is 3, at F it is and at point G potential is 5. Now apply Kirchhoffs I st law at point E. (total charge of all the plates connected to E must e same as efore i.e. ) (x ) (x 3) (x 5) = Final charges : 5x = x = 4 Q F = (3 4) = 5 Q F = ( 4) = 6 Q F = (4 (5)) = 58 4 3 6 6 5 5 58 58 5 5 F Ex. F 4 F F 4 F Find voltage across capacitor. x 3 y 4 4 5 4 O O O O Now from charge conservation at node x and y for x (x 4) (x ) (x y) 3 = (x 4) (x ) (x y) = 6x y =...() For y (y x) 3 [y (4)] 4 (y ) 5 = (y x) (y 4) y =