Chapter 18: Electrostatics and Capacitance PDF

72 Electrostatics Capacitance. (1) Definition : We know that charge given to a conductor increases it’s potential i.e., Q  V  Q  CV capacitance is the ability of conductor to hold the charge. (2) Unit and dimensional formula : S.I. unit is 60 Where C is a proportionality constant, called capacity...

72 Electrostatics Capacitance. (1) Definition : We know that charge given to a conductor increases it’s potential i.e., Q  V  Q  CV capacitance is the ability of conductor to hold the charge. (2) Unit and dimensional formula : S.I. unit is 60 Where C is a proportionality constant, called capacity or capacitance of conductor. Hence Coulomb  Farad (F) Volt 1 pF  1 F  10 12 F ) C.G.S. unit is Stat Farad 1 F  9  10 11 Stat Farad. E3 Smaller S.I. units are mF, F, nF and pF ( 1mF  10 3 F , 1 F  10 6 F , 1nF  10 9 F , Dimension : [C ]  [M 1 L2 T 4 A 2 ]. ID (3) Capacity of an isolated spherical conductor : When charge Q is given to a spherical conductor of radius R, then potential at the surface of sphere is Hence it’s capacity C  Note  1  k   4  0   Q 1  4  0 R  C  4 πε 0 R .R V 9  10 9 CR + + + + + + + R O + + + + + Q + + + + + D YG in C.G.S. Q R U V  k. + : If earth is assumed to be spherical having radius R  6400 km. It’s theortical 1  6400  10 3  711 F. But for all practical purpose capacitance 9 9  10 of earth is taken infinity. capacitance C  (4) Energy of a charged conductor : When a conductor is charged it’s potential increases U from 0 to V as shown in the graph; and work is done against repulsion, between charge stored in the conductor and charge coming from the source (battery). This work is stored as “electrostatic potential energy” ST From graph : Work done = Area of graph  Hence potential energy U  U 1 QV 2 V 1 QV ; By using Q  CV , we can write 2 1 1 Q2 QV  CV 2  2 2 2C Q (5) Combination of drops : Suppose we have n identical drops each having – Radius – r, Capacitance – c, Charge – q, Potential – v and Energy – u. If these drops are combined to form a big drop of – Radius – R, Capacitance – C, Charge – Q, Potential – V and Energy – U then – Electrostatics 73 Q  nq (i) Charge on big drop : (ii) Radius of big drop : Volume of big drop = n  volume of a single drop i.e., 4 4 R 3  n  r 3 , 3 3 60 R  n1/ 3 r (iii) Capacitance of big drop : C  n 1 / 3 c V (iv) Potential of big drop : V  n 2 /3 v 1 1 CV 2  (n 1 / 3 c) (n 2 / 3 v ) 2 2 2 U  n 5 /3 u E3 (v) Energy of big drop : U  Q nq  C n1 / 3 c (6) Sharing of charge : When two conductors joined together through a conducting wire, charge begins to flow from one conductor to another till both have the same potential, due to ID flow of charge, loss of energy also takes place in the form of heat. Suppose there are two spherical conductors of radii r1 and r2 , having charge Q1 and Q 2 , potential V1 and V2 , energies U 1 and U 2 and capacitance C 1 and C 2 respectively, as shown in U figure. If these two spheres are connected through a conducting wire, then alteration of charge, potential and energy takes place. Q1 Q2 r2 r1 D YG C1 Q1 Q2 r2 C2 C1 V1 V2 V V U1 U2 U 1 U2 Q2=C2V2 Q1=C1 V1 r1 C2 Q2=C2V2 Q1=C1 V1 (B) (A) (i) New charge : According to the conservation of charge Q1  Q 2  Q1'  Q 2'  Q (say), also Q 2'  C1 V 4  0 r1 Q1' r , '  1  C 2 V 4  0 r2 Q 2 r2 U Q 1'  r  Q 2'  Q  2   r1  r2  ST   1 Q 1' Q 2' 1 r1  r2 Q 1'  Q 2'  r1  r2 r2  r  and similarly Q1'  Q  1   r1  r2  (ii) Common potential : Common potential V  Q 2' (V )  Total charge Total capacity  Q1  Q 2 Q '  Q 2'  1 C1  C 2 C1  C 2  C1 V1  C 2 V 2 C1  C 2 (iii) Energy loss : As electrical energy stored in the system before and after connecting the spheres is  C V  C 2 V2 1 1 1 1 U i  C 1 V12  C 2 V22 and U f  (C 1  C 2 ). V 2  (C 1  C 2 )  1 1 2 2 2 2  C1  C 2    2 74 Electrostatics so energy loss ΔU  U i  U f  C1 C 2 (V1  V2 )2 2(C 1  C 2 ) Concept  60 Capacity of a conductor is a constant term, it does not depend upon the charge Q, and potential (V) and nature of the material of the conductor. + + + + + + + + + + Examples based on sharing of charge, drops and general concept of capacity E3 + + + + + Example: 95 Eight drops of mercury of same radius and having same charge coalesce to form a big drop. Capacitance of big drop relative to that of small drop will be (a) 16 times (b) 8 times Solution: (d) By using relation C  n 1 / 3. c  (c) 4 times (d) 2 times C  (8 )1 / 3. c  2 c (b) 16  C from A to B (c) 32 C from B to A (d) 32  C from A to B U (a) 20  C from A to B ID Example: 96 Two spheres A and B of radius 4 cm and 6 cm are given charges of 80 C and 40 C respectively. If they are connected by a fine wire, the amount of charge flowing from one to the other is [MP PET 1991]  r1 r  1  r2 Solution: (d) Total charge Q  80  40  120  C. By using the formula Q 1 '  Q   . New charge on sphere A  D YG  r  4  is Q 'A  Q  A   120    48  C. Initially it was 80  C, i.e., 32  C charge flows from A to 4  6   rA  rB  B. Example: 97 Two insulated metallic spheres of 3 F and 5 F capacitances are charged to 300 V and 500 V respectively. The energy loss, when they are connected by a wire, is (a) 0.012 J (c) 0. 0375 J (d) 3.75 J C1 C 2 (V1  V 2 ) 2 ; ΔU  0.375 J 2(C 1  C 2 ) U Solution: (c) By using ΔU  (b) 0.0218 J Example: 98 64 small drops of mercury, each of radius r and charge q coalesce to form a big drop. The ratio of the surface density of charge of each small drop with that of the big drop is ST (a) 1 : 64 Solution: (d) (b) 64 : 1 (c) 4 : 1 (d) 1 : 4 2  Small q / 4 r 2  q  R        ; since R = n1/3r and Q = nq 2  Big Q / 4R  Q  r  So  Small 1  1/3  Big n   Small 1   Big 4 Tricky example: 14 Two hollow spheres are charged positively. The smaller one is at 50 V and the bigger one is at 100 V. How should they be arranged so that the charge flows from the Electrostatics 75 smaller to the bigger sphere when they are connected by a wire (a) By placing them close to each other (b) By placing them at very large distance from each other 60 (c) By placing the smaller sphere inside the bigger one (d) Information is insufficient E3 Solution: (c) By placing the smaller sphere inside the bigger one. The potential of the smaller one will now be 150 V. So on connecting it with the bigger one charge will flow from the smaller one to the bigger one. Capacitor. (1) Definition : A capacitor is a device that stores electric energy. It is also named ID condenser. or A capacitor is a pair of two conductors of any shape, which are close to each other and have equal and opposite charge. –Q U +Q – – + – + – D YG + + + + – – (2) Symbol : The symbol of capacitor are shown below or variable capacitor U (3) Capacitance : The capacitance of a capacitor is defined as the magnitude of the charge Q on the positive plate divided by the magnitude of the potential difference V between the Q V ST plates i.e., C  (4) Charging : A capacitor get’s charged when a battery is connected across the plates. The plate attached to the positive terminal of the battery get’s positively charged and the one joined to the negative terminal get’s negatively charged. Once capacitor get’s fully charged, flow of charge carriers stops in the circuit and in this condition potential difference across the plates of capacitor is same as the potential difference across the terminals of battery (say V). –Q +Q C + –  + V – + V – 76 Electrostatics (5) Charge on capacitor : Net charge on a capacitor is always zero, but when we speaks of the charge Q on a capacitor, we are referring to the magnitude of the charge on each plate. Charge distribution in making of parallel plate capacitor can easily be understand by reading 60 carefully the following sequence of figures – +Q Q 2  Q 2  Q 2 Q 2  Q 2    Q 2  X –Q +Q X Y X Y ID X  E3  (6) Energy stored : When a capacitor is charged by a voltage source (say battery) it stores the electric energy. If C = Capacitance of capacitor; Q = Charge on capacitor and V = 1 1 Q2 CV 2  QV  2 2 2C : In charging capacitor by battery half the energy supplied is stored in the capacitor D YG Note U Potential difference across capacitor then energy stored in capacitor U  and remaining half energy (1/2 QV) is lost in the form of heat. (7) Types of capacitors : Capacitors are of mainly three types as described in given table Parallel Plate Capacitor Spherical Capacitor Cylindrical Capacitor It consists of two parallel metallic plates (may be circular, rectangular, square) separated by a small distance It consists of two concentric conducting spheres of radii a and b (a < b). Inner sphere is given charge +Q, while outer sphere is earthed It consists of two concentric cylinders of radii a and b (a < b), inner cylinder is given charge +Q while outer cylinder is earthed. Common length of theb cylinders is l q –q a then +Q + + U – –Q – + – air – + – + + – ST A – – – +Q – + + – + – + a + – –Q – + + – – l – – + – b – – – d A = Effective overlaping area of each plate d = Separation between the plates Q = Magnitude of charge on the inner side of each plate Capacitance C  4 πε 0. ab b a ab. In the ba presence of dielectric medium (dielectric constant K) between in C.G.S. C  Capacitance C  2 πε 0 l b log e   a In the presence of dielectric Q – –– – a – – – – – – – b +Q Electrostatics 77 E = Electric field between the     plates    0  in C.G.S. : ε0 A d Special Case : If outer sphere is given a charge +Q while inner sphere is earthed medium (dielectric constant K) capacitance increases by K times and C'  2 0 Kl b  log e   a Induced charge on the inner sphere A C 4 πd a b2 Q '  . Q , C'  4 πε 0. b b a This arrangement is not a capacitor. But it’s capacitance is equivalent to the sum of capacitance of spherical capacitor and spherical conductor i.e. U If a dielectric medium of dielectric constant K is filled completely between the plates then capacitance increases by K times C'  KC ID Capacitance : C  ab b a 60 V = Potential difference across the plates the spheres C '  4  0 K E3  = Surface density of charge  Q of each plate     A b2 ab  4 0  4  0 b ba b a D YG 4  0. Concepts  It is a very common misconception that a capacitor stores charge but actually a capacitor stores electric energy in the electrostatic field between the plates.  Two plates of unequal area can also form a capacitor because effective overlapping area is considered. U A ST d  If two plates are placed side by side then three capacitors are formed. One between distant earthed bodies and the first face of the first plate, the second between the two plates and the third between the second face of the second plate and distant earthed objects. However the capacitances of the first and third capacitors are negligibly small in comparision to that between the plates which is the main capacitance.  Capacitance of a parallel plate capacitor depends upon the effective overlapping area of plates (C  A) ,   separation between the plates  C  1  and dielectric medium filled between the plates. While it is d independent of charge given, potential raised or nature of metals and thickness of plates.  The distance between the plates is kept small to avoid fringing or edge effect (non-uniformity of the field) at the bounderies of the plates. + + – – + – + – + + – –  Spherical conductor is equivalent to a spherical capacitor with it’s outer sphere of infinite radius. A spherical capacitor behaves as a parallel plate capacitor if it’s spherical surfaces have large radii and are close to each other. E3   60 78 Electrostatics The intensity of electric field between the plates of a parallel plate capacitor (E = /0) does not depends upon the distance between them.  The plates of a parallel plate capacitor are being moved away with some velocity. If the plate separation at   1 d2. ID any instant of time is ‘d’ then the rate of change of capacitance with time is proportional to Radial and non-uniform electric field exists between the spherical surfaces of spherical capacitor. Two large conducting plates X and Y kept close to each other. The plate X is given a charge Q 1 while plate Y is given a charge Q 2 (Q 1  Q 2 ) , the distribution of charge on the four faces a, b, c, d will be as shown in the D YG U following figure. + – + – + – Example based on simple concepts of capacitor Example: 99 The capacity of pure capacitor is 1 farad. In D.C. circuit, its effective resistance will be U (a) Zero (b) Infinite (c) 1 ohm (d) 1 ohm 2 ST Solution: (b) Capacitor does not work in D.C. for D.C. it’s effective resistance is infinite i.e. it blocks the current to flow in the circuit. Example: 100 A light bulb, a capacitor and a battery are connected together as shown here, with switch S initially open. When the switch S is closed, which one of the following is true s + – (a) The bulb will light up for an instant when the capacitor starts charging Electrostatics 79 (b) The bulb will light up when the capacitor is fully charged (c) The bulb will not light up at all (d) The bulb will light up and go off at regular intervals 60 Solution: (a) Current through the circuit can flow only for the small time of charging, once capacitor get’s charged it blocks the current through the circuit and bulb will go off. Example: 101 Capacity of a parallel plate condenser is 10 F when the distance between its plates is 8 cm. If the distance between the plates is reduced to 4cm, its capacity will be [CBSE 2001; Similar to CPMT 1997; AFMC 2000] Solution: (c) C  (b) 15F 0 A 1  d d  (c) 20F (d) 40F E3 (a) 10F C1 d  2 C2 d1 or C 2  d1 8  C 1   10  20  F d2 4 (b) 4.529  10 9 m 2 Solution: (a) By using the relation C  0 A d [BHU 2002; AIIMS 1998] (c) 9.281  10 m 9 (d) 12.981  10 9 m 2 2 3  5  10 3 Cd   1. 694  10 9 m 2. 0 8. 85  10 12 U (a) 1.694  10 9 m 2 ID Example: 102 What is the area of the plates of a 3 F parallel plate capacitor, if the separation between the plates is 5 mm  A D YG Example: 103 If potential difference of a condenser (6  F) is changed from 10 V to 20 V then increase in energy is (a) 2  10 4 J Solution: (d) Initial energy U i  (b) 4  10 4 J 1 CV12 ; 2  Increase in energy U  U f  U i  [CPMT 1997, 87] (c) 3  10 4 (d) 9  10 J Final energy U f  4 J 1 CV 22 2 1 1 C (V22  V12 )   6  10 6 (20 2  10 2 )  9  10 2 2 4 J. U Example: 104 A spherical capacitor consists of two concentric spherical conductors. The inner one of radius R1 maintained at potential V1 and the outer conductor of radius R 2 at potential V2. The potential at a point P at a distance x from the centre (where R 2  x  R1 ) is V1  V2 (x  R1 ) R 2  R1 ST (a) (c) V1  V2 x (R 2  R1 ) (b) V1 R 1 (R 2  x )  V2 R 2 (x  R 1 ) (R 2  R1 ) x (d) (V1  V 2 ) x (R 1  R 2 ) Solution: (b) Let Q1 and Q 2 be the charges on the inner and the outer sphere respectively. Now V1 is the total potential on the sphere of radius R1, So, V1  Q1 Q 2  R1 R 2 …….. (i) and V2 is the total potential on the surface of sphere of radius R 2 , 80 Electrostatics So, V2  Q 2 Q1  R2 R2 …….. (ii) If V be the potential at point P which lies at a distance x from the common centre then Substracting (ii) from (i) Q1 Q 2 (V  V2 )R1 R 2   (V1  V2 )R1 R 2  R 2 Q1  R1 Q 1  Q1  1 R1 R 2 R 2  R1 Now substituting it in equation (iii), we have V  V  V1 R1 (R 2  x )  V2 R 2 (x  R1 ) x (R 2  R 1 ) The diameter of each plate of an air capacitor is 4 cm. To make the capacity of this plate capacitor equal to that of 20 cm diameter sphere, the distance between the plates will be ID Example: 105 (R1  x )(V1  V2 )R1 R 2  V1 xR 1 (R 2  R1 ) E3 V1  V2  ……..(iii) 60 1 Q (R  x ) Q Q1 Q 2 Q 1 1    V1  1 1    V1  1  Q1    V1 R1 x R2 x xR 1 x R 1   V (a) 4  10 3 m (b) 1  10 3 m 0 A  4  0 R  d d (d) 1  10 3 cm A  (2  10 2 ) 2   1  10  3 m. 4R 4   10  10  2 U Solution: (b) According to the question (c) 1 cm D YG Example: 106 A spherical condenser has inner and outer spheres of radii a and b respectively. The space between the two is filled with air. The difference between the capacities of two condensers formed when outer sphere is earthed and when inner sphere is earthed will be (a) Zero (b) 4  0 a (c) 4  0 b Solution: (c) Capacitance when outer sphere is earthed C1  4 0. sphere is earthed C 2  4  0.  b  (d) 4  0 a   b a ab and capacitance when inner ba b2. Hence C 2  C1  4  0. b ba Example: 107 After charging a capacitor of capacitance 4  F upto a potential 400 V, its plates are connected with a resistance of 1 k . The heat produced in the resistance will be U (a) 0.16 J (b) 1.28 J (c) 0.64 J (d) 0.32 J ST Solution: (d) This is the discharging condition of capacitor and in this condition energy released 1 1 U  CV 2   4  10 6  (400 ) 2  0. 32 J  0.32 J. 2 2 Example: 108 The amount of work done in increasing the voltage across the plates of a capacitor from 5 V to 10 V is W. The work done in increasing it from 10 V to 15 V will be (a) 0.6 W (b) W (c) 1.25 W (d) 1.67 W 1 2 2 Solution: (d) As we know that work done  U final  U initial  C (V2  V1 ) 2 When potential difference increases from 5 V to 10 V then 1 ……..(i) W  C (10 2  5 2 ) 2 When potential difference increases from 10 V to 15 V then 1 ……..(ii) W '  C (15 2  10 2 ) 2 Electrostatics 81 On solving equation (i) and (ii) we get W '  1.67 W. Tricky example: 15 Q1 Q3 Q1  Q 2  Q 3  Q 4 2C (b) Q4 E3 Q2 (a) 60 In an isolated parallel plate capacitor of capacitance C, the four surface have charges Q1 , Q 2 , Q 3 and Q 4 as shown. The potential difference between the plates is Q2  Q3 2C (c) Q2  Q3 2C (d) Q1  Q 4 2C ID Solution: (c) Plane conducting surfaces facing each other must have equal and opposite charge densities. Here as the plate areas are equal, Q 2  Q 3. The charge on a capacitor means the charge on the inner surface of the positive plate (here U it is Q 2 ) D YG Potential difference between the plates   Dielectric. Q 2Q 2 charge  2  capacitanc e C 2C Q 2  (Q 2 ) Q 2  Q 3 . 2C 2C Dielectrics are insulating (non-conducting) materials which transmits electric effect without conducting we know that in every atom, there is a positively charged nucleus and a negatively charged electron cloud surrounding it. The two oppositely charged regions have their U own centres of charge. The centre of positive charge is the centre of mass of positively charged ST protons in the nucleus. The centre of negative charge is the centre of mass of negatively charged electrons in the atoms/molecules. (1) Type of Dielectrics : Dielectrics are of two types – (i) Polar dielectrics : Like water, Alcohol, CO 2 , (ii) Non polar dielectric : Like NH 3 , Benzene, Methane etc. are made of non-polar atoms/molecules. In non-polar molecules, when HCl etc. are made of polar atoms/molecules. In polar molecules when no electric field is O– – applied centre of positive charges does not coincide with the centre of negative charges. 105o H+  P H+ N2 , O2 , no electric field is applied the centre of positive charge coincides with the centre of negative charge in the molecule. Each molecule has zero dipole moment in its normal state. – + +  – P  0 82 Electrostatics When electric field is applied, positive also. But a polar dielectric has net dipole moment is zero in the absence of electric field because polar molecules are randomly oriented as shown in figure. charge experiences a force in the direction of electric field and negative charge experiences a force in the direction opposite to the field i.e., molecules becomes induced electric dipole. In the presence of electric field polar molecules tends to line up in the direction of electric field, and the substance has finite dipole moment. Note E3 60 A polar molecule has permanent electric  dipole moment (p) in the absence of electric field  In general, any non-conducting, ID : D YG U material can be called as a dielectric but broadly non conducting material having non polar molecules referred to as dielectric because induced dipole moment is created in the non polar molecule. (2) Polarization of a dielectric slab : It is the process of inducing equal and opposite charges on the two faces of the dielectric on the application of electric field. Suppose a dielectric slab is inserted between the plates of a capacitor. As shown in the figure. E + + + + + + Induced electric field inside the dielectric is E i , hence this + induced electric field decreases the main field E to E  E i i.e., + + + + – + – + + – + Ei – + – + – + – + – + – + – – – – – – – – – – U New electric field between the plates will be E '  E  E i. ST (3) Dielectric constant : After placing a dielectric slab in an electric field. The net field is decreased in that region hence If E  Original electric field and E '  Reduced electric field. Then E  K where K is called E' dielectric constant K is also known as relative permittivity ( r ) of the material or SIC (Specific Inductive Capacitance) The value of K is always greater than one. For vacuum there is no polarization and hence E  E ' and K  1 Electrostatics 83 (4) Dielectric breakdown and dielectric strength : If a very high electric field is created in a dielectric, the outer electrons may get detached from their parent atoms. The dielectric then behaves like a conductor. This phenomenon is known as dielectric breakdown. The maximum value of electric field (or potential gradient) that a dielectric material can S.I. unit of dielectric strength of a material is 60 tolerate without it’s electric breakdown is called it’s dielectric strength. V kV but practical unit is. m mm E3 Variation of Different Variables (Q, C, V, E and U) of Parallel Plate Capacitor. Suppose we have an air filled charged parallel plate capacitor having variables are as follows : +Q Surface charge density –   Q , Capacitance – C   0 A A d ID Charge – Q, Potential difference across the plates – V  E. d + A + Electric field between the plates – E    Q Energy stored – U  1 CV 2 2  – – – Ai r – + – + – d A 0 U 0 –Q + + V Q2 1  QV 2C 2 C '  KC D YG (1) When dielectric is completely filled between plates : If a dielectric slab is fills K 0 A completely the gap between the plates, capacitance increases by K times i.e., C '   d The effect of dielectric on other variables such as charge. Potential difference field and energy associated with a capacitor depends on the fact that whether the charged capacitor is disconnected from the battery or battery is still connected. ST U Quantity Battery is Removed A Battery Remains connected K A d K d V Capacity C = KC C = KC Charge Q = Q (Charge is conserved) Q = KQ Potential V = V/K V = V (Since Battery maintains the potential difference) Intensity E = E/K E = E 84 Electrostatics Energy Note U = U/K U = U/K :  If nothing is said it is to be assumed that battery is disconnected. 60 (2) When dielectric is partially filled between the plates : If a dielectric slab of thickness t (t  d ) is inserted between the plates as shown below, then E  Main electric field between the plates, E i  Induced electric field in dielectric. E '  (E  E i )  The reduced value of electric field in V '  E (d  t)  E ' t  E (d  t )  E.t K + +  V '  E  d  t  t     d  t  t   Q  d  t  t  K 0  A 0  K Now capacitance of the capacitor Q V'  0 A C'  d t t2 t2 D YG t1 t K K1 A K2 K3 + – + – + – + – + – + – + – d A t1 t2 t3 t4 K1 K2 K3 K4 d d C'  + – + – + – U C'  K E – – Ei A ID  E3 the dielectric. Potential difference between the two plates of capacitor is given by 0 A  t  t t d  (t1  t 2  t3 ........)   1  2  3 ........   K1 K 2 K 3  C'  0 A  t1 t2 t3 t4    K  K  K  K  2 3 4   1 U (3) When a metallic slab is inserted between the plates : ST t A A K= d d Capacitance C '  circuited) K= 0 A (d  t) or A d C '   (In this case capacitor is said to be short Electrostatics 85 (4) When separation between the plates is changing : If separation between the plates 1 changes then it’s capacitance also changes according to C . The effect on other variables d depends on the fact that whether the charged capacitor is disconnected from the battery or 60 battery is still connected. Quantity E3 (i) Separation is increasing Battery remains connected Battery is removed A ID A d Remains constant because a battery is not present D YG Charge 1 i.e., C '  C d Decreases because C  Q'  Q i.e., C'  C Decreases i.e., the battery. 1 i.e., Increases because V  Q  V  Electric Remains constant because E    Q C C U i.e., ST A 0 E'  E Decrease because E  i.e., Increases because U  Energy i.e., V   V (Since Battery maintains the potential difference) V'  V 0 Decreases because battery is present i.e., Q '  Q Remaining charge (Q  Q ' ) goes back to Potential differenc e field V U Capacity d Q2 2C  U 1 C U'  U Q A 0 E'  E Decreases because U  i.e.,  EQ 1 CV 2 2  UC U'  U (ii) Separation is decreasing Quantity Battery is removed Capacity Increases because C  Charge Remains constant because battery is not present Battery remains connected 1 i.e., C '  C d Increases i.e., C'  C Increases because battery is present i.e., Q'  Q 86 Electrostatics Remaining charge (Q '  Q) supplied from Q'  Q i.e., the battery. V  1 Q V  C C V  V i.e., (Since Remains constant because E    Q i.e., Increases because E  Q A 0 Energy  U 1 C U'  U i.e., E'  E i.e., Q2 2C the  EQ A 0 E'  E Decreases because U  maintains potential difference) V' V 0 Battery 60 Electric field Decreases because E3 Potential differenc e Increases because U  1 CV 2 2  UC U'  U i.e., ID Force Between the Plates of a Parallel Plate Capacitor. Field due to charge on one plate on the other is E  | F|   A Q  2 0 2 0 A 2 2 D YG   2    A 2 0  + + U   F  A    2 0  , hence the force F  QE 2 0 + A + – – Ai r – – – + + – d Energy Density Between the Plates of a Parallel Plate Capacitor. The energy stored in a capacitor is not localised on the charges or the plates but is distributed in the field. And as in case of a parallel plate capacitor field is only between the plates i.e. in a volume (A× d), the so called energy density. 2  1 V  1 0    0 E2. 2 d 2 Concepts ST U 1 CV 2 1  A  V 2 Energy 2 Hence Energy density     0  Ad Volume 2  d  Ad  t   In the expression of capacitance of parallel plate capacitor filled partially with dielectric term  d  t   is K   known as effective air separation between the plates.  When dielectric is partially filled between the plates of a parallel plate capacitor then it’s capacitance increases but potential difference decreases. To maintain the capacitance and potential difference of capacitor  A  as before (i.e., c  0 , V  d ) separation between the plates has to be increased. Suppose separation is d 0 increased by d ' so in this case 0 A  A  0 which gives us t   d  d  d 't   K  K  t t  d' Example based on capacitor with dielectric Electrostatics 87 Example: 109The mean electric energy density between the plates of a charged capacitor is (here Q = Charge on the capacitor and A = Area of the capacitor plate) Q2 (b) 2 0 A 2 Q (c) 2 0 A 2 Q2 2 0 A (d) None of these 60 (a) 2 1 1  Q  Q2  . Solution: (a) Energy density   0 E 2   0  2 2  A 0  2 0 A 2 Solution: (b) Given C  (b) 20  F 0 A  15  F d Then by using C '  0 A dt (c) 30  F ……..(i) t K 0 A   (d) 25  F 2   0 A  10 3 ; From equation (i) C '  20  F. 3 ID (a) 15  F E3 Example: 110 Plate separation of a 15  F capacitor is 2 mm. A dielectric slab (K = 2) of thickness 1 mm is inserted between the plates. Then new capacitance is given by 2  10  3  10  3  3 10 2 Example: 111 There is an air filled 1 pF parallel plate capacitor. When the plate separation is doubled and U the space is filled with wax, the capacitance increases to 2 pF. The dielectric constant of wax is [MNR 1998] (b) 4 D YG (a) 2 (c) 6 (d) 8 Solution: (b) Given that capacitance C  1 pF After doubling the separation between the plates C '  C 2 and when dielectric medium of dielectric constant k filled between the plates then C '  According to the question, C '  KC 2 2  KC 2 K  4. U Example: 112 If a slab of insulating material 4  10 5 m thick is introduced between the plate of a parallel plate capacitor, the distance between the plates has to be increased by 3.5  10 5 m to ST restore the capacity to original value. Then the dielectric constant of the material of slab is (a) 10 Solution: (d) By using K  (b) 12 (c) 6 (d) 8 t ; here t  4  10 5 m ; d '  3.5  10 5 m  t  d' K  4  10 5 4  10 5  3. 5  10 5 =8 Example: 113 The force between the plates of a parallel plate capacitor of capacitance C and distance of separation of the plates d with a potential difference V between the plates, is (a) CV 2 2d (b) C 2V 2 2d 2 (c) C 2V 2 d 2 (d) V 2d C 88 Electrostatics Solution: (a) Since F  Q2 C 2 V 2 CV 2  F . 2 0 A 2 0 A 2d Example: 114 A capacitor when filled with a dielectric K  3 has charge Q 0 , voltage V0 and field E 0. If the (a) 3 Q0 , 3 V0 , 3 E0 (b) Q0 , 3 V0 , 3 E 0 (c) Q 0 , 60 dielectric is replaced with another one having K = 9, the new values of charge, voltage and field will be respectively V0 , 3 E0 3 (d) Q 0 , V0 E 0 , 3 3 E3 Solution: (d) Suppose, charge, potential difference and electric field for capacitor without dielectric medium are Q, V and E respectively With dielectric medium of K  3 Charge Q 0  Q With dielectric medium of K  9 Charge Q '  Q  Q 0 Electric field E 0  V 3 Potential difference V '  V V  0 9 3 ID Potential difference V0  E 3 Electric field E'  E E0.  9 3 D YG U Example: 115 A slab of copper of thickness b is inserted in between the plates of parallel plate capacitor d then the ratio of as shown in the figure. The separation between the plates is d. If b  2 capacities of the capacitor after and before inserting the slab will be d 2 :1 Solution: (b) Capacitance before inserting the slab C  (c) 1 : 1 0 A d (d) 1 : 2 and capacitance after inserting the slab 0 A. dt ST C'  b Cu (b) 2 : 1 U (a) A= Area Where t  b  2 A d C' 2 so C '  0 hence, . 2 d C 1 Example: 116 The capacity of a parallel plate condenser is C 0. If a dielectric of relative permitivity  r and thickness equal to one fourth the plate separation is placed between the plates, then its C capacity becomes C. The value of will be C0 (a) 5 r 4 r  1 (b) 4 r 3 r  1 (c) 3 r 2 r  1 (d) 2 r r  1 Electrostatics 89 ……..(i) By dividing equation (ii) by equation (i) Finally capacitance C  4 r C  C0 3 r  1 0 A ……..(ii) d d /4 d  r 4 60 0 A d Solution: (b) Initially capacitance C 0  Tricky example: 16 An air capacitor of capacity C  10  F is connected to a constant voltage battery of 12 (a) 120  C E3 V. Now the space between the plates is filled with a liquid of dielectric constant 5. The charge that flows now from battery to the capacitor is (b) 600  C (c) 480  C (d) 24  C Solution: (c) Initially charge on the capacitor Q i  10  12  120  C ID When dielectric medium is filled, so capacitance becomes K times, i.e. new capacitance C '  5  10  50  C Final charge on the capacitor Q f  50  12  600  C U Hence additional charge supplied by the battery  Q f  Q i  480  C. D YG Grouping of Capacitors. Series grouping (1) Charge on each capacitor remains same and equals to the main charge supplied by the C2 C3 C1 battery +Q +Q +Q + + + Q + – – – – –Q + + + + V1 – – – – –Q + + + + V2 V = V1 + V2 + V3 + – – – – –Q – U If n identical  C2 C  1  C2 and V 2    capacitors  . V  each Q1 Q2 Q = Q1 + Q2 + Q3 Q Q3 +Q1 + + + + – – – – +Q2 + + + + – – – – +Q3 + + + + – – – – –Q1 –Q2 –Q3 V ST (5) Potential difference across each capacitor remains same and equal to the applied potential difference V (3) In series combination potential difference and energy distribution in the reverse ratio of 1 1 capacitance i.e., V  and U . C C (4) If two capacitors having capacitances C1 and C2 are connected in series then C1 C 2 Multiplication C eq   C1  C 2 Addition  . V  (1) V3 (2) Equivalent capacitance 1 1 1 1 or C eq  (C 1 1  C 2 1  C 31 )  1    Ceq C1 C2 C3  C1 V 1   C  1  C2 Parallel grouping having (2) Ceq = C1 + C2 + C3 (3) In parallel combination charge and energy distributes in the ratio of capacitance i.e. Q  C and U  C (4) If two capacitors having capacitance C1 and C2 respectively are connected in parallel then C eq  C 1  C 2  C1 Q 1    C1  C 2   C2 . Q and Q 2     C1  C 2  . Q  (5) If n identical capacitors are connected in 90 Electrostatics parallel Equivalent capacitance C eq  nC and Charge on each capacitor Q '  Redistribution of Charge Between Two Capacitors. Q n 60 capacitances C are connected in series with supply voltage V then Equivalent capacitance C and Potential difference across each C eq  n V capacitor V ' . n    C1 C 2 Q1  Q 2 C 1 V1  C 2 V2  and loss of energy U  (V1  V2 ) 2 C1  C 2 C1  C 2 2 (C 1  C 2 ) D YG The common potential V  U  C2 Q 2'  Q   C1  C 2 ID E3 When a charged capacitor is connected across an uncharged capacitor, then redistribution of charge occur to equalize the potential difference across each capacitor. Some energy is also wasted in the form of heat. Suppose we have two charged capacitors C1 and C 2 after – – – – – – C1 C2 disconnecting these two from their respective batteries. These two ++ + ++ + Q1 Q2 capacitors are connected to each other as shown below (positive V1 V2 plate of one capacitor is connected to positive plate of other while negative plate of one is connected to negative plate of other)  C1   , Charge on capacitors redistributed and new charge on them will be Q1'  Q   C1  C 2  Note :  Two ST U capacitors of capacitances C1 and C2 are charged to potential of V1 and V2 respectively. After disconnecting from batteries they are again connected to each other with reverse polarity i.e., positive plate of a capacitor connected to negative plate of Q  Q 2 C 1 V1  C 2 V2 other. So common potential V  1 . C1  C 2 C1  C 2 Electrostatics 89 Circuit With Resistors and Capacitors. (1) A resistor may be connected either in series or in parallel with the capacitor as shown below C Parallel RC Circuit 60 Series RC Circuit R R V0 S E3 C S V0 Resistor has no effect on the charging of to charge. capacitor. The charging current is maximum in the beginning; it decreases with time and becomes zero after a long time. Resistor provides an alternative path for the electric current. U (2) Three states of RC circuits ID In this combination capacitor takes longer time (i) Initial state : i.e., just after closing the switch or just after opening the switch. D YG (ii) Transient state : or instantaneous state i.e., any time after closing or opening the switch. (iii) Steady state : i.e., a long time after closing or opening the switch. In the steady state condition, the capacitor is charged or discharged. U (3) Charging and discharging of capacitor in series RC circuit : As shown in the following figure (i) when switch S is closed, capacitor start charging. In this transient state potential difference appears across capacitor as well as resistor. When capacitor gets fully charged the entire potential difference appeared across the capacitor and nothing is left for the resistor. [shown in figure (ii)] ST (i) + C – R V V Transient state i S + – V0 (ii ) R C + Steady state – V0 V0  S + Q0 (Max charge) C – V0 (i) Charging : In transient state of charging charge on the capacitor at any t t    Q  Q 0  1  e RC  and potential difference across the capacitor at any instant V  V0  1  e RC       Q0 Q V0 Q = Q0(1 – e– t/RC ) VC VC = V0(1 – e– t/RC ) instant     90 Electrostatics 60 (ii) Discharging : After the completion of charging, if battery is removed capacitor starts discharging. In transient state charge on the capacitor at any instant Q  Q 0 e  t / RC and potential E3 difference cross the capacitor at any instant V  V0 e  t / CR. Q0 V0 Q VC O t VC = V0 e–t/RC O t ID Q= Q0 e–t/RC (iii) Time constant () : The quantity RC is called the time constant i.e.,   RC. U In charging : It is defined as the time during which charge on the capacitor rises to 0.63 times (63%) the maximum value. That is when t    RC , Q  Q 0 (1  e 1 )  0.639 Q 0 or D YG In discharging : It is defined as the time during which charge on a capacitor falls to 0.37 times (37%) of the initial charge on the capacitor that is when t    RC , Q  Q 0 (e 1 )  0.37 Q 0 (iv) Mixed RC circuit : In a mixed RC circuit as shown below, when switch S is closed current flows through the branch containing resistor as well as through the branch contains capacitor and resistor (because capacitor is in the process of charging) U R1 ST i C S R2 R V0 Transient state R1 i C i S R2 R No current V0 Steady state When capacitor gets fully charged (steady state), no current flows through the line in which V0 capacitor is connected. Therefore the current through resistor R 1 is , hence potential R1  r  Electrostatics 91 V0 R. The same potential difference will R1  r  1 difference across resistance will be equal to CV0 R1 R1  r  E3 60 appear across the capacitor, hence charge on capacitor in steady state Q  Concepts In series combination equivalent capacitance is always lesser than that of either of the individual capacitors e.g. If two capacitors having capacitances 3F and 6F respectively are connected in series then equivalent capacitance 36 C eq   2 F Which is lesser than the smallest capacitance (3F) of the 36 network.  In parallel combination, equivalent capacitance is always greater than the maximum capacitance of either capacitor in network.  If n identical plates arranged as+ shown constitutes + – below, + – + – + –(n – –1) capacitors in series. If each capacitors + – 0 A  0 A– + – + – + having capacitance then C eq  + – + – + – + – d (n  1)d – + – + – + D YG + U ID  – In this situation except two extreme plates each plate is common to adjacent capacitors.  If n identical plates are arranged such that even no. of plates are connected together and odd number of plates are connected together, then (n – 1) capacitors will be formed and they will be in parallel grouping. ST U 2 1 Equivalent capacitance C'  (n  1) C  4 3 5 6 7 Where C = capacitance of a capacitor  0 A d If n identical capacitors are connected in parallel which are charged to a potential V. If these are separated and connected in series then potential difference of combination will be nV. Network Solving. To solve capacitive network for equivalent capacitance following guidelines should be followed. 92 Electrostatics Guideline 1. calculated. Guideline 2. Identify the two points across which the equivalent capacitance is to be Connect (Imagine) a battery between these points. 60 Guideline 3. Solve the network from the point (reference point) which is farthest from the points between which we have to calculate the equivalent capacitance. (The point is likely to be not a node) (1) Simple circuits : Suppose equivalent capacitance is to be determined in the following networks between points A and B (i) E3 Suppose equivalent capacitance is to be determined in the following networks between points A and B 36 6 F 3 F   A 3 F (ii) 3 F 6 F A (iii)  6 F B 2 F 9 F ST 6 F B 9 F 9 F 6 F 6  3 F 2 A 6 F A 9 F  B 9 F 9 F 6 F + – 3 + 3 = 6 F  9 F 6  3 F  3 F C 2 AB  3  2  5 μF 6 F 6 F 6 F B 2 F 9 F B 3 F 6 F 3 F U 9 F A 6 F D YG 6 F  2  3  5 F B U 3 F AB 3 F A A B C 2 F ID 6 F 3 F  2 F 36 2 F A 9 F 6 F 9 F 9 F Series 9 3 9 F  3 F A 2 F 9 F 9 F B A  9 F 9 F  9 F B 9 F 9 F 9 F 3 F Parallel 6 + 3 = 9 F 9 F 6 F + – B 6 F 6 F + – 9 F By similar process CAB = 3F A B Electrostatics 93 (2) Circuits with extra wire : If there is no capacitor in any branch of a network then every point of this branch will be at same potential. Suppose equivalent capacitance is to be determine in following cases A C C C  A B C C B A A C A C  B A B B C 60 (i) A B C B (ii) A A C  C A B (iii) C B B A U  B  C A CAB = 2C D YG C A C B + – C C B A U B A Parallel C + C = 2C C + –  2C/3 Parallel 2C 5C C  3 3 Series  C 2C  C  2C 3 C B – 2C  C 2C C A + C C C B C ST  C A 5C between A and B is 3 C B A Hence equivalent capacitance B CAB = 3C C B C A B A  C C A B C A A C C C B A C (iv)  C A No p.d. across vertical branch so it is removed C ID C CAB = 3C B C C A E3 + – B A + – (v) Since there is no capacitor in the path APB, the points A, P and B are electrically same i.e., C C  C B A P the input and output points are directly connected (short circuited). Thus, entire charge will prefer to flow along path APB. It means that the capacitors connected in the circuit will not receive any charge for storing. Thus 94 Electrostatics equivalent capacitance of this circuit is zero. (i) (ii) D C2 E C2 A B C4 (iii) C5 C2 D C3 E3 C3 A C5 C5 C5 A C1 D C1 C1 60 (3) Wheatstone bride based circuit : If in a network five capacitors are arranged as shown in following figure, the network is called wheatstone bridge type circuit. If it is balanced then C1 C 3 hence C5 is removed and equivalent capacitance between A and B  C2 C4 B B C4 C3 E C AB  ID E C4 C1 C 2 C3C4  C1  C 2 C 3  C 4 (4) Extended wheatstone bridge : The given figure consists of two wheatstone bridge U connected together. One bridge is connected between points AEGHFA and the other is connected between points EGBHFE. D YG This problem is known as extended wheatstone bridge problem, it has two branches EF and GH to the left and right of which symmetry in the ratio of capacities can be seen. It can be seen that ratio of capacitances in branches AE and EG is same as that between the capacitances of the branches AF and FH. Thus, in the bridge AEGHFA; the branch EF can be removed. Similarly in the bridge EGBHFE branch GH can be removed C C E A F C E C C G B H A B C F C C H C AB  C ST C C  C U C C G 2C 3 (5) Infinite chain of capacitors : In the following figure equivalent capacitance between A and B (i) A C1 C1 C1  C2 C2 C2 B Suppose the effective capacitance between A and B is CR. Since the network is infinite, even if we remove one pair of capacitors from the chain, remaining network would still have infinite C1 A C2 of C1 X A CR pair CR Parallel (C2 + CR ) capacitors, i.e., Series C1 (C 2  C R )  CR (C2 + CR ) C1  C 2  C R effective Electrostatics 95 Hence equivalent capacitance between A and B C1 (C 2  C R )  CR C1  C 2  C R  C AB   C2   C   1  4 1   1  2   C2    E3 C AB  60 capacitance between X and Y would also be CR A C1 C C2 Suppose there are n sections between A and B and the network is terminated by C0 with equivalent capacitance CR. Now if C0  we add one more sections to the network between D and C (as shown in the following figure), the equivalent capacitance of the network CR will be independent of number of sections if the D D YG C C1 C2 C2 C2 B C1 C1 C1 U A ID (ii) For what value of C0 in the circuit shown below will the net effective capacitance between A and B be independent of the number of sections in the chain Parallel C2 + C 0 CR C0 B C0 C2 capacitance between D and C still remains C0 i.e., D C0  Hence C1  (C 2  C 0 ) C1  C 2  C 0  C 0  C 2 C 0  C1 C 2  0 U 2 ST On simplification C0   C2   C   1  4 1   1  2   C2    (6) Network with more than one cell : (i) E1 C1 E1 – E2 C2  C1 Potential difference across C1 is  C2    (E1  E 2 ) and potential difference  C1  C 2   C1   (E1  E 2 ) across C2 is  C  C 1 2   C2 E2 (ii) Potential difference between the ends of this arrangement is given V. C1 C1 A E A C2 B + E –  + V (E – V) C2 B  – C1 C2 96 Electrostatics K1 K2 d/ 2 d/ 2 C1  1 K1 K 2 2 d (iii) K K 2 d/ 2 C 22 ST K1 A/ 2 C1 d/ 2  2 K1 K 2 1 1 1  C eq     C eq C1 C 2  K1  K 2  0 A .  d K1  K 2 2 In this case C1 and C2 are in series while this combination is in parallel with C3 A A A K 3 0 K1 0 K1 0 K  A K  A 2  K 3 0 A 2  1 0 , C  2  2 0 and C  C1  3 2 d d d d d 2d 2 2 2 k 1 0 A k 2  0 A  C1 C 2 k  A d d Hence, C eq   C3   3 0  k 1 0 A k 2  0 A 2d C1  C 2  d d  k1k 2 k3  0 A . C eq    2  d  k1  k 2 U 3 2 K1 K 2 K1  K 2 K 2 0 A and d 2 In this case these two capacitors are in parallel and K  A K A C1  1 0 , C 2  2 0 2d 2d  K  K2  0 A Hence, C eq  C1  C 2  C eq   1 . 2   d Also K eq  C3 A/ 2 C2  D YG A/ 2 K1 0 A , d 2 Also K eq  (ii) A/ 2 The system can be assumed to be made up of two capacitors C1 and C2 which may be said to connected in series E3 A 2 U 1 ID (i) 60 (7) Advance case of compound dielectrics : If several dielectric medium filled between the plates of a parallel plate capacitor in different ways as shown. k k k Also k eq   3  1 2  2 k1  k 2    96 Electrostatics Example based on series and parallel grouping of capacitors Example: 117 Three capacitors of 2 f, 3 f and 6 f are joined in series and the combination is charged by means of a 24 volt battery. The potential difference between the plates of the 6  f capacitor is (a) 4 volts (b) 6 volts (c) 8 volts 1 1 1 1    C eq 2 3 6 Solution: (a) Equivalent capacitance of the network is (d) 10 volts 2 F 3 F 6 F E3 C eq  1F 60 [MP PMT 2002 Similar to MP PMT 1996] Q Charge supplied by battery Q = Ceq.V  1 × 24 = 24 C + – 24V 24  4 volt. 6 ID Hence potential difference across 6 F capacitor  Example: 118 Two capacitors each of 1  f capacitance are connected in parallel and are then charged by 200 V D.C. supply. The total energy of their charges in joules is (a) 0.01 (b) 0.02 1 C eq V 2 2 D YG Here C eq  2 F  (d) 0.06 1 F U Solution: (c) By using formula U  (c) 0.04 1 F 1 U   2  10 6  (200 ) 2 2 + – 200V  0.04 J Example: 119 Five capacitors are connected as shown in the figure. The equivalent capacitance between the point A and B is 2 f [MP PMT 2002; SCRA 1996; Pantnagar 1987] A ST U 1 f B (a) 1  f 2 f (b) 2  f Solution: (b) 2 f A + – 1 f 2 f 1 f (c) 3  f Series 2  1 F 2 (d) 4  f Parallel 1 + 1 = 2 F A  1 f 2 f 2 f  Parallel 1 + 1 = 2 F A 1 f 1 f 1 f B B B 2 f 1 f 1 f A  2 f 1 f B 2 f Series 2  1 F 2 Electrostatics 97 Hence equivalent capacitance between A and B is 2F. Example: 120In the following network potential difference across capacitance of 4.5 F is [RPET 2001; MP PET 1992 3 F 60 4.5 F 6 F – E3 + 12 V (b) 4 V Solution: (a) Equivalent capacitance C eq  (c) 2 V 9  4.5  3 F 9  4.5 (d) 6 V ID (a) 8 V Charge supplied by battery Q = Ceq  V = 3 × 12 = 36 C 4.5 F 36 = 8V. 4.5 D YG U Hence potential difference across 4.5 F  3 F Parallel 3 + 6 = 9 F 6 F + – 12 V Example: 121 A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants K1, K2 and K3 as shown in fig. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant K is given by [IIT Screening 2000] A/2 A/2 K1 K2 ST U d (a) K3 A A = Area plates of 1 1 1 1    K K1 K 2 2 K 3 (c) K  d/ 2 (b) K1 K 2  2K 3 K1  K 2 Solution: (b) The effective capacitance is given by 1 1 1   K K1  K 2 2 K 3 (d) K  K1  K 2  2K 3 1 d  C eq 0 A  1  1    2 K ( K  K ) 1 2   3 The capacitance of capacitor with single dielectric of dielectric constant K is C  K 0 A d 98 Electrostatics 0 A According to question C eq  C i.e.,  1  1 d    2 K 3 K1  K 2   K 0 A d 1 1 1  . K 2 K 3 K1  K 2  C2 C3 + – ID 2V E3 C1  32  (c)    10 6 J  3  (b) 11  10 6 J Solution: (b) Equivalent capacitance  C eq   16  (d)    10 6 J  3  C1 C 2 26  C3   4  5.5 F C1  C 2 8 U (a) 22  10 6 J 60 Example: 122 Two capacitors C1 = 2 F and C2 = 6F in series, are connected in parallel to a third capacitor C 3 = 4F. This arrangement is then connected to a battery of e.m.f. = 2 V, as shown in the fig. How much energy is lost by the battery in charging the capacitors ? 1 1 C eq.V 2   5.5  (2) 2  11  10 6 J 2 2 D YG U Example: 123 In the circuit shown in the figure, each capacitor has a capacity of 3 F. The equivalent capacity between A and B is B A (a) 3 F 4 (b) 3 F U A Solution: (d) A ST A (c) 6 F B 3 F  B B B Hence equivalent capacitance C eq 3 F A (d) 5 F 3 F  B 3 F 3 F 36   3  5 F. 36 A Parallel 3 + 3 = 6 F 3 F 6 F B Example: 124 Given a number of capacitors labelled as 8F, 250 V. Find the minimum number of capacitors needed to get an arrangement equivalent to 16 F, 1000 V (a) 4 (b) 16 (c) 32 (d) 64 Solution: (c) Let C = 8 F, C = 16 F and V = 250 volt, V = 1000 V Suppose m rows of given capacitors are connected in parallel which each row contains n capacitor then Electrostatics 99 Potential difference across each capacitors V  V' and equivalent capacitance of network n mC. n On putting the values, we get n = 4 and m = 8. Hence total capacitors = m × n = 8 × 4 = 32. C'  2 2 n C'  V '    . Here C V 60 Short Trick : For such type of problem number of capacitors n  16  1000     32 8  250  (a) V E3 Example: 125 Ten capacitors are joined in parallel and charged with a battery up to a potential V. They are then disconnected from battery and joined again in series then the potential of this combination will be [RPET 2000] (b) 10 V (c) 5 V ID Solution: (b) By using the formula V '  nV  V '  10 V. (d) 2 V Example: 126 For the circuit shown, which of the following statements is true V1 = V2 = – – +30V +20V S3 C1 = 2pF C2 = 3pF S2 D YG U S1 (a) With S1 closed, V1 = 15 V, V2 = 20 V (b) With S3 closed, V1 = V2 = 25 V (c) With S1 and S2 closed V1 = V2 = 0 V (d) With S1 and S3 closed V1 = 30 V, V2 = 20 Solution: (d) When S3 is closed, due to attraction with opposite charge, no flow of charge takes place through S3. Therefore, potential difference across capacitor plates remains unchanged or V1 = 30 V and V2 = 20 V. Alternate Solution U Charges on the capacitors are – q 1  (30 ) (2)  60 pC , q 2  (20 ) (3)  60 pC or + ST – + 2PF The situation is similar as the two capacitors in series are first charged with a battery of emf 50 V and then disconnected. When S3 is closed, q q q 1  q 2  q (say) – 3PF  q = 60pC q = 60pC + – + – V1 = 30V V2 = 20V + V1 = 30 V and V2 = 20 V. – 50V Example: 127 A finite ladder is constructed by connecting several sections of 2  F,4  F capacitor combinations as shown in the figure. It is terminated by a capacitor of capacitance C. What value should be chosen for C, such that the equivalent F points capacitance of the ladder between 4the 4 F A and B becomes 4 F independent of the number A of sections in between [MP PMT 1999] 2 F B 2 F 2 F C 100 Electrostatics (c) 18  F (b) 2  F Solution: (a) By using formula C C C2   1  4 1 C 2   2   C1  4  F    1 ; C 2  2 F    (d) 6  F We get 60 (a) 4  F C  4  F. E3 Example: 128Figure shows two capacitors connected in series and joined to a battery. The graph shows the variation in potential as one moves from left to right on the branch containing the – PMT 1999] capacitors. +[MP V (a) C1  C 2 (b) C1  C 2 C1 X ID C2 (c) C1  C 2 (d) The information is insufficient to decide the relation between C 1 and C 2 U Solution: (c) According to graph we can say that potential difference across the capacitor C 1 is more than that across C 2. Since charge Q is same i.e., Q  C1 V1  C 2 V2  D YG (V1  V2 ). C1 V  2 C2 V1  C1  C 2 Example: 129 Two condensers of capacity C and 2C are connected in parallel and these are charged upto V volt. If the battery is removed and dielectric medium of constant K is put between the plates of first condenser, then the potential at each condenser is (a) V K2 (b) 2  Solution: (d) Initially (c) 2V K2 (d) 3V K2 C C eq  3 C Equivalent capacitance of the system 2C Q ST U K 3V Finally Hence common potential + – V KC Total charge Q  (3C)V Equivalent capacitance of the system C eq  KC  2C 2C V Q 3CV 3V  . (KC  2C) (K  2)C K  2 Example: 130Condenser A has a capacity of 15 F when it is filled with a medium of dielectric constant 15. Another condenser B has a capacity 1 F with air between the plates. Both are charged separately by a battery of 100V. after charging, both are connected in parallel without the battery and the dielectric material being removed. The common potential now is Electrostatics 101 (a) 400V (b) 800V (c) 1200V Solution: (b) Charge on capacitor A is given by Q1  15  10 6 (d) 1600V  100  15  10 4 C Charge on capacitor B is given by Q 2  1  10 6  100  10 4 C 15  10 6  1F 15 60 Capacity of capacitor A after removing dielectric  Now when both capacitors are connected in parallel their equivalent capacitance will be Ceq  1  1  2 F (15  10 4 )  (1  10 4 ) 2  10 6  800 V. E3 So common potential  Example: 131 A capacitor of 20 F is charged upto 500V is connected in parallel with another capacitor of 10 F which is charged upto 200V. The common potential is (a) 500V V  (c) 300V (d) 200V C1 V1  C 2 V2 ; C1 = 20 F, V1 = 500 V, C2 = 10 F and V2 = 200 V C 1 V2 ID Solution: (b) By using V  (b) 400V 20  500  10  200  400 V. 20  10 [DCE 1995] U Example: 132 In the circuit shown D YG 12 V + – C1 = 4 F C2 = 8 F – + 6V (a) The charge on C2 is greater than that of C1 C1 (b) The charge on C2 is smaller than that of U (c) The potential drop across C1 is smaller than C2 (d) across C1 is greater than C2 ST 4 8 8  F 12 3 V1 8 So Q   6  16 C 3 drop V2 C2 = 8 F C1 = 4 F Hence potential difference V1  potential 12 – 6 = 6 V + – Solution: (d) Given circuit can be redrawn as follows C eq  The 16 16  4 volt and V2   2 volt i.e. V1 > V2 4 8 Example: 133 As shown in the figure two identical capacitors are connected to a battery of V volts in parallel. When capacitors are fully charged, their stored energy is U 1. If the key K is opened and a material of dielectric constant K  3 is inserted in each capacitor, their stored energy is now U 2. U1 will be U2 [IIT 1983] K + C V – A C B 102 Electrostatics (b) 5 (c) 3 3 5 (d) 1 3 60 (a) 3 Solution: (a) Initially potential difference across both the capacitor is same hence energy of the system is 1 1 CV 2  CV 2  CV 2 2 2 ……..(i) E3 U1  ID In the second case when key K is opened and dielectric medium is filled between the plates, capacitance of both the capacitors becomes 3C, while potential difference across A is V and V potential difference across B is hence energy of the system now is 3 2 U2  1 1 10 V  (3 C)V 2  (3 C)    CV 2 6 2 2 3 So, U 1  3 U2 U 5 …….(ii) Example: 134 In the following figure the resultant capacitance between A and B is 1F. The capacitance C is D YG [IIT 1977] 1 F C A 8 F U 32 F 11 (b) 12 F 2 F 2 F (a) 4 F 6 F B 11 F 32 (c) 23 F 32 ST Solution: (d) Given network can be simplified as follows A A 4 F 6 F 8 F Parallel 2 + 2 = 4 F 4 F 4 F 8 F  32 F 23 1 F C 1 F C (d) Series 12 F 2 F 2 F 6  12 6  12  4 F Series 84 B 84 C 8 8 32 F   3 9 9 8 3 F 8 F 9 B  B 8 F 3  1 F C Parallel A Parallel 4  4  8 F 4 F Series 18 8  F 18 9 A  8 F 8 F 3 B Electrostatics 103  C 60 B A 32 F 9 Given that equivalent capacitance between A and B i.e., CAB  1F But C AB 32 9  1  C  32 F. 32 23 9 E3 32 C 9 hence  32 C C 9 C Example: 135 A 1F capacitor and a 2 F capacitor are connected in parallel across a 1200 volts line. The ID charged capacitors are then disconnected from the line and from each other. These two capacitors are now connected to each other in parallel with terminals of unlike signs together. The charges on the capacitors will now be (c) 800 C and 400 C (d) 800 C and 800 C U (b) 400 C and 800 C (a) 1800 C each Solution: (b) Initially charge on capacitors can be calculated as follows Q1 + – 1F Finally when battery is disconnected and unlike plates are connected together then common potential D YG C1 Q1 = 1 × 1200 = 1200 C and Q2 = 2 × 1200 = 2400 C Q2 + – 2 F C2 V'  Q 2  Q1 2400  1200  C1  C 2 12 1 F + – + – + – C1  400 V + – 1200 V – + 2 F – + – + C2 Hence, New charge on C 1 is 1  400  400 C And New charge on C 2 is 2  400  800 C. V U Example: 136 The two condensers of capacitances 2 F and 3 F are in series. The outer plate of the first ST condenser is at 1000 volts and the outer plate of the second condenser is earthed. The potential of the inner plate of each condenser is (a) 300 volts (b) 500 volts (c) 600 volts (d) 400 volts Solution: (d) Here, potential difference across the combination is V A  VB  1000 V Equivalent capacitance C eq  23 6  F 23 5 Hence, charge on each capacitor will be Q  C eq  (V A  VB )  So potential difference between A and C, 3 F 2 F + 1000 V V A  VC  A C B 0 V 6  1000  1200 C 5 1200  600 V 2  1000  VC  600  Vc  400 V Example: 137 Four identical capacitors are connected in series with a 10V battery as shown in the figure. The point N is earthed. The potentials of points A and B are 104 Electrostatics – + 10V N CA C B C (c) 5 V  5 V (b) 7.5 V  2.5 V (a) 10 V,0 V B C 60 A (d) 7.5 V,2.5 V 10  2.5 V 4 Hence potential difference between A & N i.e., V A  VN  2.5  2.5  2.5  7.5 V  V N  VB  2. 5 V A  0  V A  7.5 V While E3 Solution: (b) Potential difference across each capacitor will be  0  VB  2. 5  VB  2.5 V Example: 138In the figure below, what is the potential difference between the points A and B and between B and C respectively in steady state 1 F ID 3 F B 3 F 1 F U 1 F A D YG 100 V (a) 100 volts both (c) V AB  25 volts, VBC  75 volts 2 0  1 0 C  (b) V AB  75 volts, VBC  25 volts (d) V AB  50 volts VBC  50 volts Solution: (c) In steady state No current flows in the given circuit hence resistances can be eliminated Parallel 3+3= 6 F ST U 3 F A A B B B 3 F 1 F Parallel 1+1= 2 F A C 1 F A C  2 F B C Line (2) 1 F A 100 V Line (1) V2=VBC V1=VAB C 1 F 6 F + – 100 V C By using the formula to find potential difference in series combination of two capacitor   C2  V1   C C  2  1    C1 .V and V2  V C 2  C 2    2  V1  V AB     100  25 V ; 26  6  V2  V BC     100  75 V. 26 Example: 139 A capacitor of capacitance 5 F is connected as shown in the figure. The internal resistance of the cell is 0.5. The amount of charge on the capacitor plate is Electrostatics 105 1 1 5 F 2 (a) 0 C 2.5 V – (c) 10 C (b) 5 C the battery 1 1 Line (1) 5 F 2 E3 Solution: (c) In steady state current drawn from 2.5 i  1A (1  1  0.5) (d) 25 C 60 + In steady state capacitor is fully charged hence No current will flow through line (2) i 0.5  ID Hence potential difference across line (1) is V  1  2  2volt , the same potential difference appears across the capacitor, so charge on capacitor Q  5  2  10 C Line (2) 2. 5 Example: 140When the key K is pressed at time t  0. Which of the following statements about the current i in the resistor AB of the adjoining circuit is true V C=1F D YG K B 1000  A 1000  U 2V (a) i  2mA at all t (b) i oscillates between 1mA and 2mA (c) i = 1mA at all t to 1mA (d) At t = 0, i = 2mA and with time it goes ST U Solution: (d) At t  0 whole current passes through capacitance; so effective resistance of circuit is 2 1000  and current i   2  10 3 A  2mA. After sufficient time, steady state is reached; 1000 then there is no current in capacitor branch; so effective resistance of circuit is 2 1000  1000  2000  and current i   1  10  3 A  1mA i.e., current is 2mA at t  0 and 2000 with time it goes to 1mA. Example: 141 The plates of a capacitor are charged to a potential difference of 320 volts and are then connected across a resistor. The potential difference across the capacitor decays exponentially with time. After 1 second the potential difference between the plates of the capacitor is 240 volts, then after 2 and 3 seconds the potential difference between the plates will be [MP PET 1998] (a) 200 and 180 volts (b) 180 and 135 volts (c) 160 and 80 volts (d) 140 and 20 volts Solution: (b) During discharging potential difference across the capacitor falls exponentially as V  V0 e  t ( = 1/RC) Where V = Instantaneous P.D. and V0  max. P.D. across capacitor 106 Electrostatics After 1 second V1 = 320 (e–)  240 = 320 (e–)  e   3 4 2 3 After 2 seconds V2 = 320 (e–)2  320     180 volt 4 3 60 3 After 3 seconds V3 = 320 (e–)3 = 320     135 volt 4 1 2 3 4 E3 Example: 142 Five similar condenser plates, each of area A. are placed at equal distance d apart and are connected to a source of e.m.f E as shown in the following diagram. The charge on the plates 1 and 4 will be – 5 V (a)  0 A 2 0 A d , (b) d ID +  0 AV 2 0 AV , d d (c)  0 AV 3 0 AV d , (d) d  0 AV 4  0 AV d , d U Solution: (b) Here five plates are given, even number of plates are connected together while odd number of plates are connected together so, four capacitors are formed and they are in parallel combination, hence redrawing the figure as shown below. D YG Capacitance of each Capacitor is C  1 0 A d U Charge on plate (1) is  0 A d + – + – 3 + – 4 + – + – V + – + –  0 AV 5 + – 4 d d + 2 AV 2   0. d ST While charge on plate 4 is   0 AV 2 3 + – 2 Potential difference across each capacitor is V So charge on each capacitor Q  + – + – + – – V Example: 143 Four plates are arranged as shown in the diagram. If area of each plate is A and the distance between two neighbouring parallel plates is d, then the capacitance of this system between A and B will be d A B d (a) 4 0 A d (b) 3 0 A d (c) 2 0 A d (d) 0 A d Electrostatics 107 Solution: (c) To solve such type of problem following guidelines should be follows 1 2 A Guideline 1. Mark the number (1,2,3……..) on the plates B 3 A 60 4 1 2 B 3 2 E3 Guideline 2. Rearrange the diagram as shown below 4 3 ID Guideline 3. Since middle capacitor having plates 2, 3 is short circuited so it should be eliminated from the circuit 1 2 A Tricky example: 17 U Hence equivalent capacitance between A and B C AB  2 0 A B 4 3 d D YG A capacitor of capacitance C1 = 1F can withstand maximum voltage V1 = 6 KV (kilovolt) and another capacitor of capacitance C2 = 3F can withstand maximum voltage V2 = 4KV. When the two capacitors are connected in series, the combined system can withstand a maximum voltage of (a) 4 KV (b) 6 KV [MP PET 2001] (c) 8 KV (d) 10 KV Solution: (c) We know Q = CV U Hence (Q1)max = 6 mC while (Q2)max = 12 mC ST However in series charge is same so maximum charge on C2 will also be 6 mC (and not 12 6 mC  2 KV and as in series V = mC) and hence potential difference across C2 will be V2  3 F V1 + V2 So Vmax  6 KV  2 KV  8 KV