C
APACITANCE
INTRODUCTION
A capacitor can store energy in the form of potential energy in an electric field. In this chapter we'll discuss the capacity of conductors to hold charge and energy.
CAPACITANCE OF AN ISOLATED CONDUCTOR
When a conductor is charged its potential increases. It is found that for an isolated conductor (conductor should be of finite dimension, so that potential of infinity can be assumed to be zero). Potential of the conductor is proportional to charge given to it.
q = charge on conductor V = potential of conductor q V
q = CV
Where C is proportionality constant called capacitance of the conductor.
Definition of capacitance :
Capacitance of conductor is defined as charge required to increase the potential of conductor by one unit.
Important points about the capacitance of an isolated conductor :
(i) It is a scalar quantity.
(ii) Unit of capacitance is farad in SI units and its dimensional formula is M–1 L–2 2 T4
(iii) 1 Farad : 1 Farad is the capacitance of a conductor for which 1 coulomb charge increases potential by 1 volt.
1 Farad = Volt 1 Coulomb 1 1 F = 10–6 F, 1nF = 10–9 F or 1 pF = 10–12 F (iv) Capacitance of an isolated conductor depends on following factors :
(a) Shape and size of the conductor : On increasing the size, capacitance increases. (b) On surrounding medium : With increase in dielectric constant K, capacitance increases.
(c) Presence of other conductors : When a neutral conductor is placed near a charged conductor capacitance of conductors increases.
(v) Capacitance of a conductor do not depend on (a) Charge on the conductor
(b) Potential of the conductor
POTENTIAL ENERGY OR SELF ENERGY OF AN ISOLATED CONDUCTOR
Work done in charging the conductor to the charge on it against its own electric field or total energy stored in electric field of conductor is called self energy or self potential energy of conductor.
Electric potential energy (Self Energy) :
U = C 2 q2 = 2 1 CV2 = 2 qV
q = Charge on the conductor V = Potential of the conductor C = Capacitance of the conductor.
Self energy is stored in the electric field of the conductor with energy density (Energy per unit volume)
dV dU
= 2 1
0 E2 [The energy density in a medium is 2 1
0 r E2 ] where E is the electric field at that point.
In case of charged conductor energy stored is only out side the conductor but in case of charged insulating material it is outside as well as inside the insulator.
SOLVED EXAMPLES
01. (i) When 10 coulomb charge is assigned to an isolated conductor its potential becomes 5 volt, find out capacitance of the conductor?
(ii) If now further 20 coulomb charge is supplied to it then what is the new potential on conductor? Sol. (i) C = V Q = 5 10 = 2 Farad. (ii) V = 2 30 C Q = 15 volt.
02. An isolated conductor of 10 F capacitance is given 10 C charge. Find out stored energy and its potential?
Sol. Stored energy U = 2 1 CV2 = C Q . 2 1 2 = F 10 ) C 1 ( . 2 1 2 = 20 C 1 = 0.05 J Potential V = C Q = F 10 C 1 = 10 1 Voltolt
CAPACITANCE OF AN ISOLATED SPHERICAL CONDUCTOR
(i) If the medium around the conductor is vacuum or air.CVacuum = 40R
R = Radius of spherical conductor. (may be solid or hollow.)
(ii) If the medium around the conductor is a dielectric of constant K from surface of sphere to infinity. Cmedium = 40KR (iii) vaccum / air medium C C = K = dielectric constant.
SOLVED EXAMPLES
01. 8 similar charged drops combine to from a bigger drop. The ratio of the capacity of bigger drop to that of smaller drop will
be-(A) 2 : 1 (B) 8 : 1 (C) 4 : 1 (D) 16 : 1
Sol. Cbigger drop = (Csmall drop)n1/3 .... (1)
n = 8 .... (2) drop small drop begger C C = 1 2
Hence the correct answer will be (A).
CAPACITOR
A capacitor or condenser consists of two conductors separated by an insulator or dielectric.
(i) When uncharged conductor is brought near to a charged conductor, the charge on conductors remains same but its potential decreases resulting in the increase of capacitance.
(ii) In capacitor two conductors have equal but 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 = CV C = A B B B A A V V Q V V Q V Q
Q = Charge of positive plate of capacitor.
V = Potential difference between positive and negative plates of capacitor C = Capacitance of capacitor.
(b) Energy stored in the capacitor
U = 2 1 CV2 = C 2 Q2 = 2 QV
This energy is stored inside the capacitor in its electric field with energy density
dV dU = 2 1 E2 or 2 1 r E2
(v) The capacitor is represented as following: ,
(vi) Based on shape and arrangement of capacitor plates there are various types of capacitors. (a) Parallel plate capacitor.
(b) Spherical capacitor. (c) Cylindrical capacitor.
(vii) Capacitance of a capacitor depends (a) Area of plates.
(b) Distance between the plates.
(c) Dielectric medium between the plates.
(viii) Capacitance of a parallel plate capacitor (air filled) is given by following formula C =
d A
0
where A = area of the plates. d = distance between plates.
(ix) Electric field intensity between the plates of capacitors (air filled ) E = /0 = V/d
(x) Force experienced by any plate of capacitor F = q2/2A
0
SOLVED EXAMPLES
01. Two parallel conducting plates of a capacitor of capacitance C containing charges Q and – 2Q at a distance d apart. Find out potential difference between the plates of capacitors.
Sol. Capacitance = C Electric field = 0 A 2 Q 3 V = 0 A 2 Qd 3 V = 2C Q 3 = C 2 Q 3
02. A capacitor of capacitance C which is initially uncharged is connected with a battery. Find out heat dissipated in the circuit during the process of charging.
Sol. Final status
Let potential at point A is 0, so at B also 0 and at C and D it is . finally, charge on the capacitor QC = C Ui = O Uf = 2 1 CV2 = 2 1 C2
Work done by battery =
pdtw =
idt =
idt = . Q = . C = C2(Now onwards remember that w.d. by battery = Q if Q has flown out of the cell from high potential and w.d. on battery is Q if Q has flown into the cell through high potential)
Heat produced = W – (Uf – Ui ) = 2C – 2 1 2C = 2 C2
03. A capacitor of capacitance C which is initially charged upto a potential difference is connected with a battery of emf such that the positive terminal of battery is connected with positive plate of capacitor. Find out heat loss in the circuit during the process of charging.
Sol.
Since the initial and final charge on the capacitor is same before and after connection. Here no charge will flow in the circuit so heat loss = 0
04. A capacitor of capacitance C which is initially charged upto a potential difference is connected with a battery of emf such that the positive terminal of battery is connected with positive plate of capacitor. After a long time (i) Find out total charge flow through the battery
(ii) Find out total work done by battery
(iii) Find out heat loss in the circuit during the process of charging. Sol. Let potential of A is 0 so at B it is
2
. So final charge on capacitor = C/2 Charge flow through the capacitor = (C/2 – C) = –C/2
So charge is entering into battery. finally,
Change in energy of capacitor = Ufinal – Uinitial
= 2 1 C 2 2 – 2 C 2 = 8 1 2C – 2 1 2C = – 8 C 32
Work done by battery = 2 × 2 C = – 4 C 2
Work done by battery = Change in energy of capacitor + Heat produced Heat produced = 8 C 32 – 4 C 2 = 8 C 2
Before connecting the capacitors
Parameter Capacitor Is t Capacitor IIn dCapac itance C1 C2
Charge Q1 Q2
Potential V1 V2
After connecting the capacitors
Parameter Is t Capacitor IIn d Capacitor
Capac itance C1 C2
Charge Q’1 Q’2
Potential V V
(a) Common potential :
By charge conservation of plates A and C before and after connection. Q1 + Q2 = C1V + C2V V = 2 1 2 1 C C Q Q = 2 1 2 2 1 1 C C V C V C
= TotalTotalcapacichargtanece
(b) Q1'= C1V = 2 1 1 C C C (Q1 + Q2) Q2' = C2 V = 1 2 2 C C C (Q1 +Q2)
(c) Heat loss during redistribution : H = Ui – Uf = 2 1 2 1 2 1 C C C C (V1 – V2) 2
The loss of energy is in the form of Joule heating in the wire.
Note :
(i) When plates of similar charges are connected with each other (+ with + and – with –) then put all values (Q1, Q2, V1, V2) with positive sign.
(ii) When plates of opposite polarity are connected with each other (+ with –) then take charge and potential of one of the plate to be negative.
SOLVED EXAMPLES
01. Find out the following if A is connected with C and B is connected with D.Sol.
Let potential of B and D is zero and common potential on capacitors is V, then at A and C it will be V 3V + 2V = 40 + 30
5V = 70 V = 14 volt Charge flow = 40 – 28 = 12 C Now final charges on each plate
(ii) Heat produced = 2 1 × 2 × (20)2 + 2 1 × 3 ×(10)2 – 2 1 × 5 × (14)2 = 400 + 150 – 490 = 550 – 490 = 60 J Note 1. When capacitor plates are joined then the charge remains conserved.
Note 2. We can also use direct formula of redistribution as given above. 02. Repeat above question if A is connected with D and B is connected with C.
Sol. Let potential of B and C is zero and common potential on capacitors is V, then at A and D it will be V 2V + 3V = 10 V = 2
Now charge on each plate Heat produced = 400 + 150 –
2 1
× 5 × 42 = 550 – 10 = 540 J Note : Here heat produced is more. Think why?
03. Three capacitors as shown of capacitance 1F, 2F and 2F are charged upto potential difference 20 V, 10 V and 15 V respectively. If terminal A is connected with D, C is connected with E and F is connected with B. Then find out charge flow in the circuit and find the final charges on capacitors.
Sol. Let charge flow is q.
Now applying kirchhoff's voltage low – 2 ) 20 q ( – 2 ) q 30 ( – 1 q 30 = 0 –2q = – 25 q = 12.5 C
Final charges on plates
04. In the given circuit find out the charge on each capacitor. (Initially they are uncharged)
Sol. Let potential at A is 0, so at D it is 30 V, at F it is 10 V and at point G potential is –25V. Now apply kirchhoff’s Ist law at point E. ( total charge of all the plates connected to 'E' must be same as before i.e. 0)
(x – 10) + (x – 30)2 + (x + 25)2 = 0 5x = 20 x = 4 Final charges : Q2F = (30 – 4)2 = 52 C Q1F = (10 – 4) = 6C Q2F = (4 – (–25))2 = 58 C
COMBINATION OF CAPACITORS
Series Combination :(i) When initially uncharged capacitors are connected as shown in the combination, it is called series combination.
(ii) All capacitors will have same charge but different potential difference across them. (iii) We can say that
V1 =
1
C Q
V1 = potential across C1
Q = charge on positive plate of C1 C1 = capacitance of capacitor similarly V2 = 2 C Q , V3 = 3 C Q ... (iv) V1 : V2 : V3 = 1 C 1 : 2 C 1 : 3 C 1
We can say that potential difference across capacitor is inversely proportional to its capacitance in series combination.
V
C 1
Note : In series combination the smallest capacitor gets maximum potential.
(v) Equivalent Capacitance : Equivalent capacitance of any combination is that capacitance which when connected in place of the combination stores same charge and energy that of the combination.
In series : eq C 1 = 1 C 1 + 2 C 1 + 3 C 1 ...
Note : In series combination equivalent is always less the smallest capacitor of combination. (vi) Energy stored in the combination
Ucombination = 1 2 C 2 Q + 2 2 C 2 Q + 3 2 C 2 Q Ucombination = eq 2 C 2 Q
Energy supplied by the battery in charging the combination
Ubattery = Q × V = Q . eq C Q = eq 2 C Q battery n combinatio U U = 2 1
Note : 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.
SOLVED EXAMPLES
01. Three initially uncharged capacitors are connected in series as shown in circuit with a battery of emf 30V. 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
Sol. eq C 1 = 2 1 + 3 1 + 6 1 = 6 1 2 3 = 1 Ceq = 1F. (i) Q = Ceq V = 30C.
(ii) charge on 3F capacitor = 30C
energy = C 2 Q2 = 3 2 30 30 = 150J (iii) Utotal = 2 30 30 J = 450 J
(iv) Heat produced = (30 C) (30) – 450 J = 450 J
02. Two capacitors of capacitance 1 F and 2F are charged to potential difference 20V and 15V as shown in figure. If now terminal B and C are connected together terminal A with positive of battery and D with negative terminal of battery then find out final charges on both the capacitor
Sol. Now applying kirchoff voltage law
1 ) q 20 ( – 2 q 30 + 30 = 0 – 40 – 2q – 30 – q = – 60 3q = –10 Charge flow = –10/3 C. Parallel Combination :
(i) 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.
(iii) We can say that : Q1 = C1V
Q1 = Charge on capacitor C1 C1 = Capacitance of capacitor C1 V = Potential of across capacitor C1 (iv) Q1 : Q2 : Q3 = C1 : C2 : C3
The charge on the capacitor is proportional to its capacitance Q C (v) Q1 = 3 2 1 1 C C C C Q Q2 = 1 2 3 2 C C C C Q Q3 = 1 2 3 3 C C C C Q Where Q = Q1 + Q2 + Q3 ...
Note : Maximum charge will flow through the capacitor of largest value. (vi) Equivalent capacitance of parallel combination Ceq = C1 + C2 + C3
In general
n 1 n n eq C CNote : Equivalent capacitance is always greater than the largest capacitor of combination. (vii) Energy stored in the combination :
Vcombination = 2 1 C1V2 + 2 1 C2V2 + .... = 2 1 (C1 + C2 + C3 ...) V2 = 2 1 CeqV2 Ubattery = QV = CV2 = battery n combinatio U U = 2 1
Note : 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.
SOLVED EXAMPLES
01. Three initially uncharged capacitors are connected to a battery of 10 V is parallel combination find out following (i) charge flow from the battery
(ii) total energy stored in the capacitors (iii) heat produced in the circuit
(iv) potential energy in the 3F capacitor. Sol. (i) Q = (30 + 20 + 10)C = 60 C
(ii) Utotal =
2 1
× 6 × 10 × 10 = 300 J (iii) heat produced = 60 × 10 – 300 = 300 J (iv) U3F =
2 1
02. In the given circuit find out charge on 6F and 1F capacitor. Sol. It can be simplified as
Ceq =
9 18
= 2F
charge flow through the cell = 30 × 2 C Q = 60 C
Now charge on 3F = Charge on 6F= 60 C Potential difference across 3F
= 60/ 3= 20 V
Charge on 1F = 20 C.
Mixed Combination :
The combination which contains mixing of series parallel combinations or other complex combinations fall in mixed category.
SOLVED EXAMPLES
01. Two condensers of same capacity are first connected in parallel and then in series. The ratio of resultant capacities in two cases will
be-(A) 1 : 4 (B) 4 : 1 (C) 1 : 2 (D) 2 : 1 Sol. CP = C + C = 2C CS = C C CC = 2 C S P C C = 2 / C C 2 = 1 4
Hence the correct answer will be (B).
02. The equivalent capacity in the adjoining figure between the point X and Y will be
(A) 4.5 F (B) 9F (C) 1 F (D) 6F
Sol. Equivalent circuit (All capacitors are connected between same p.d. x and y)
All the three condensers are connected in parallel, hence the resultant capacity = 3 + 3 + 3 = 9F
03. In the adjoining diagram if the capacity of each condenser is 1F then the resultant capacity between the points P and Q will be
(A) (B) zero (C) 2F (D) 8F
Sol. The different branches of the condensers have number of condensers in G.P., hence the equivalent capacity C´ = 1 + 2 1 + 4 1 + 8 1 + ... C´ = 2 1 1 1 = 2F..
Hence the correct answer will be (C).
Wheatstone bridge : If 4 1 C C = 5 3 C C or C1C5 = C3C4 then
(i) Such combination is called balanced Wheatstone bridge. (ii) In this case VB = VD .
(iii) Change on C2 = 0
(iv) The equivalent can be converted in to given circuits:
or (v) Ceq = 5 3 2 1 5 3 4 1 C C C C ) C C )( C C ( = 3 1 3 1 C C C C + 4 5 5 4 C C C C (vi) If C1 = C4 = C3 = C5 = C then Ceq = C
SOLVED EXAMPLES
01. The equivalent capacity between the points X and Y in the following circuit (Figure) will be
Sol. Because the bridge is balanced, hence the central capacitance between Z and T is ineffective. C1 and C2 are connected in series, hence their resultant C´ =
2 C
= 3F similarly C3 and C4 are connected in series, hence their
resultant C´´ =
2 C
= 3F. Now the two branches are connected in parallel Ceq. = 3 + 3 = 6F
Hence the correct answer will be (A).
Other important circuit solving techniques :
(Applicable in both capacitive and resistive networks).
(a) Equipotential Technique : All the junctions which are at equal potential (such as junctions connected by a connecting wire) can be replaced by a single junction. So redraw the circuit to get it simplified
(b) Infinite Circuits : Assume equivalent capacitance/resistance to be Ceq/Req of whole network, then add one more branch (repetitive) in infinite network. Calculate equivalent of this new circuit. It should again be equal to Ceq/Req.
Note : If all the resistance/capacitances of a circuit are made K times then equivalent will also become K-times.
SOLVED EXAMPLES
01. In the adjoining figure, the effective capacity of the group of condensers will be
(A) nC (B) (C) zero (D) 0.62 C
Sol. As the combination is spreading upto infinity, hence the capacity will remain same at last but one step also. Let the capacity of the combination is C´.
´ C 1 = C 1 + ´ C C 1 C´ 1 = ] C ´ C [ C C 2 ´ C or C´2 + 2CC´ – CC´– C´ = 0 or C´2 + CC´ – C2 = 0
This is a quadratic equation in C´ C´ = 2
5 C C
Negative capacity is impossible C´ = 0.62 C Hence the correct answer will be (D).
(c) Kirchhoff's Laws :
(i) Junction Rule : Sum of charges present on the plates of capacitors connected at a junction is equal to zero (If initially all the capacitors are uncharged) (while adding charges of different plates battery can be neglected as net charge on battery is zero).
(ii) Loop Rule : In any closed loop the algebraic sum of potential drops across different elements is equal to zero.
CAPACITORS WITH DIELECTRIC
(i) In absence of dielectric
E =
(ii) When a dielectric fills the space between the plates then molecules having dipole moment align themselves in the direction of electric field.
rb = induced charge density (called bound charge because it is not due to free electrons). * For polar molecules dipole moment 0
* For non-polar molecules dipole moment 0
For non-polar molecules the molecule of substance arranged as given below : (iii) Capacitance is presence of dielectric
C = V A = d . K A 0 = d AK0 = d AK0
Here capacitance is increased by a factor K. C =
d AK0
(iv) Polarisation of material : When nonpolar substance is placed in electric field then dipole moment is induced in the molecule. This induction of dipole moment is called polarisation of material. The induced charge also produce electric field.
b induced (bound) charge density.
Ein = E – Eind = 0 b 0
It is seen the ratio of electric field between the plates in absence of dielectric and in presence of dielectric is constant for a material of dielectric. This ratio is called 'Dielectric constant' of that material. It is repre-sented by r or k. = 0 K b = K 1 1
(v) If the medium does not filled between the plates completely then. Electric field will be as shown in figure Case : (1)
The total electric field produced by bound induced charge on the dielectric outside the slab is zero because they cancel each other.
(vi) Comparison of E (electric field), (surface charges density), Q (charge), C (capacitance) and F (force between the plates) before and after inserting a dielectric slab between the plates of a parallel plate capacitor.
C = d A 0 C' = d K A0 Q = CV Q' = C'V E = 0 = 0 A Cv E' = K0 = 0 A Cv = d V = d V also
Here potential difference between the plates, Here potential difference between the plates
Ed = V E’d E = d V E’ = d V d V = 0 d V = 0 K ' Equating both 0 0 K ' ' = K
In the presence of dielectric, i.e. in case IInd capacitance of capacitor is more. (vii)Energy density of a dielectric = 0 rE2
2 1
SOLVED EXAMPLES
01. If a dielectric slab of thickness t and area A is inserted in between the plates of a parallel plate capacitor of plate area A and distance between the plates d (d > t) then find out capacitance of system. What do you predict about the dependence of capacitance on location of slab ?
Sol. C = v Q = v A V = 0 1 t + 0 K t + 0 2 t ( t1 + t2 = d – t)
= 0 k t t t1 2 V = 0 k t t d = C Q = C A C = K / t t d A 0
Note: (i) Capacitance does not depend upon the position of dielectric (it can be shifted up or down still capacitance does not change).
(ii) If the slab is of metal slab then C – t d A 0
02. 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.
Sol. C = V A ; V = E1 d1 + E2 d2 = 0 1 1 K d + 0 2 2 K d = 2 2 1 1 0 k d k d C = 2 2 1 1 0 K d K d A 0 2 2 0 1 1 AK d AK d C 1
This formula suggests that the system between A and B can be considered as series combination of two capacitors.
03. Find out capacitance between A and B if two dielectric slabs of dielectric constant K1 and K2 of area A1 and A2 and each of thickness d are inserted between the plates of parallel plate capacitor of plate area A as shown in figure. Sol. C1 = d K A1 10 , C2 = d K A2 20 E1 = d V = 0 1 1 K , E2 = d V = 0 2 2 K 1 = d V K10 2 = d V K20
1 1 K = 2 2 K 1b = 1(1 – 1 k 1 ) 2b = 2(1 – 2 k 1 ) The combination is equivalent to :
C1 + C2
04. 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) Sol. It is equivalent to C = C1 + 3 2 3 2 C C C C C = 2 1 0 2 1 d d K A + 2 0 3 2 1 0 2 2 1 0 3 2 1 0 2 2 d K A d K A d K A . d K A = 2 1 0 1 1 d d K A + 1 0 3 2 2 0 2 2 0 3 2 2 2 d K A d K A K K A = 2 1 0 2 1 d d K A + 1 3 2 2 0 3 2 2 2 d K d K K K A
05. A dielectric of constant K is slipped between the plates of parallel plate condenser in half of the space as shown in the figure. If the capacity of air condenser is C, then new capacitance between A and B will be
(A) 2 C (B) K 2 C (C) 2 C [1 + K] (D) C ] K 1 [ 2
Sol. This system is equivalent to two capacitors in parallel with area of each plate 2 A . C´ = C1 + C2 = 2 K ) 2 / A ( d 2 / A 0 0 = d A 0 [1 + K] = 2 C [1 + K] Hence the correct answer will be (C).
06. In the adjoining figure two dielectrics of constants K1 and K2 are filled in a parallel plate condenser. The capacity of the condenser will be :
(A) C = d A 0 (B) C = d(K K ) A K K 2 2 1 0 2 1 (C) C = A K K 2 ) K K ( 2 0 2 1 2 1 (D) 2 1 2 1 K K K K 2
Sol. Two capacitors, each of area A and plate separation
2 d
, are connected in series.
C = 2 1 2 1 C C C C C1 = d A K 2 10 C2 = d A K 2 20 C = d A K K 2 1 20 C = ) K K ( d A K K 2 2 1 0 2 1
07. The capacity of a parallel plate capacitor in air is 50F and on immersing it into oil it becomes 110 F. The dielectric constant of oil is
(A) 0.45 (B) 0.55 (C) 1.10 (D) 2.20 Sol. K = 0 C C K = 50 110 = 5 11 = 2.20
08. Two parallel plate condensers with capacities C and 2C are connected in parallel and are charged to potential difference V. Now the battery is removed and a dielectric of constant K is inserted between the plates of condenser C. Now the potential difference across each condenser will be
(A) 2 K V (B) 2 K V 2 (C) 2 K V 3 (D) 3V K 2 Sol. C´ = C + 2C = 3C q = C´V = 3CV
When dielectric is inserted C” = KC + 2C = (K + 2) C
the potential difference across the capacitors
Vf = e capacitanc total charge total = " C q = ) K 2 ( C CV 3 = 2 K V 3
09. A parallel plate condenser is charged to a certain potential and then disconnected. The separation of the plates is now increased by 2.4 mm and a plate of thickness 3 mm is inserted into it keeping its potential constant. The dielectric constant of the medium will be
Sol. As charge and potential of the condenser both are constant in two cases, hence its capacity must also remain constant C0 = C or d A 0 = K 1 1 t ' d A 0 or d = d´ – t K 1 1 or (d´ – d) = t K 1 1 or 2.4 × 10–3 = 3 × 10–3 K 1 1 or 1 – K 1 = 0.8 or K 1 = 0.2 K = 5
Force on a dielectric due to charged capacitor
If dielectric is completely inside the capacitor then force is equal to zero.
Case I – Voltage source remains connected V = constant. U = 2 1 CV2 F = dx dU = 2 V2 dx dC where C = d K xb0 + d b ) x ( 0 F = d b 0 [Kx + – x] = d b 0 (K – 1) F = d 2 V ) 1 K ( b 2 0
Case II : When charge on capacitor is constant C = d K xb0 + d b ) x ( 0 , U = C 2 Q2 F = dx dU = 2 2 C 2 Q . dx dC dx dC = d b 0 (K – ) = dx dC . C 2 Q 2 2
10. Find V and E at : (Q is a point charge kept at the centre of the nonconducting neutral thick sphere of inner radius 'a' and outer radius 'b')
(i) 0 < r < a (ii) a r < b (iii) r b
Sol. –q and +q charge will induce on inner and outer surface respectively E(0 < r < a) = 2 r KQ E (r b) = 2 r KQ E (a r < b) = 2 r KQ – 2 r Kq = 2 rr KQ q = Q r 1 1 . V (r b) = r KQ (a r b) VA = VP +
r b 2 r ) dr ( r KQ = b 1 r 1 kQ b kQ r V (r a) VB = VC +
r a 2 ( dr) r KQ = b kQ + r kQ b 1 a 1 + kQ a 1 r 111. What is potential at a distance r (<R) in a dielectric sphere of uniform charge density , radius R and dielectric constant r.. Sol. VA = VB + q WBA V = R 4 Q 0 +
r R 0 r ) dr ( 3 r = R 4 Q 0 + 0 r 2 2 3 ) r R ( Voutiside = r KQCOMBINATION OF PARALLEL PLATES
01. Find out equivalent capacitance between A and B.Sol. Put numbers on the plates The charges will be as shown in the figure. V12 = V32 = V34
so all the capacitors are in parallel combination. Ceq = C1 + C2 + C3
02. Find out equivalent capacitance between A and B.
Sol. These are only two capacitors. Ceq = C1 + C2
Sol.
The modified circuit is
Ceq = 3 C 2 Other method : Ceq = V Q = V xA 2 V = V2 – V4 = (V2 – V3) + (V3 – V2) = 0 xd + 0 xd 2 = 0 xd 3 Ceq = xd 3 xAx 3 0 = d 3 A 2 0 = 3 C 2 .
04. 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
(A) d A 0 , d A 20 (B) d A 0 , d AV 20 (C) d AV 0 , d AV 30 (D) d AV 0 , d AV 40
Sol. Equivalent circuit diagram Charge on first plate Q = CV Q = d AV 0
Charge on fourth plate Q´ = C(–V) Q´ = d AV 0
As plate 4 is repeated twice, hence charge on 4 will be Q´´ = 2Q´ Q´ = –
d AV 20
OTHER TYPES OF CAPACITORS
Spherical capacitor :
This arrangement is known as spherical capacitor.
V1 – V2 = b KQ a KQ – b KQ b KQ = a KQ – b KQ C = 2 1 V V Q = b KQ a KQ Q = ) a b ( K ac = b a ab 4 0 C = a b ab 4 0 If b >> a C = 40a C = a b ab 4 0 r2 Cylindrical capacitor :
There are two co-axial conducting cylindrical surfaces >> a
>> b
where a and b is radius of cylinders.
C = V = b a n K 2 = a b n 2 4 0 = a b n 2 0
Capacitance per unit length =
a b n 2 0 F/m
CIRCUIT SOLUTION FOR R–C CIRCUIT AT t = 0 (INITIAL STATE) AND AT t = (FINAL STATE)
Note :
(i) Charge on the capacitor does not change instantaneously or suddenly if there if a resistance in the path (series) of the capacitor.
(ii) When an uncharged capacitor is connected with battery then its charge is zero initially hence potential difference across it is zero initially. At this time the capacitor can be treated as a conducting wire
(iii) The current will become zero finally (that means in steady state) in the branch which contains capacitor.
SOLVED EXAMPLES
01. Find out current in the circuit and charge on capacitor which isinitially uncharged in the following situations. (a) Just after the switch is closed.
(b) After a long time when switch was closed. Sol. For just after closing the switch:
potential difference across capacitor = 0 QC = 0
i = 2 10
= 5A After a long time
at steady state current i = 0
and potential difference across capacitor = 10 V QC = 3 × 10 = 30 C
02. Find out current i1, i2, i3, charge on capacitor and
dt dq
of capacitor in the circuit which is initially uncharged in the following situations.
(a) Just after the switch is closed
(b) After a long time when switch is closed.
Sol. Initially the capacitor is uncharged so its behaviour is like a conductor Let potential at A is zero so at B and C also zero and at F it is . Let potential at E is x so at D also x.
Apply Kirchhoff’s st law at point E :
R x + R 0 x + R 0 x = 0 R x 3 = R x = 3 Qc = 0 1 = R 3 / = R 3 2 , 2 = R 3 , 3 = R 3
Alternatively i1 = eq R = 2 R R = R 3 2 , i2 = i3 = 2 i1 = R 3 at t (finally)
capacitor completely charged so their will be no current through it. 2 = 0, 1 = 3 = R 2 Ve – VB = VD – VC = (/2R)R = /2 QC = 2 C , dt dQ = 2 = 0 1 2 3 Q dQ /dt Time t = 0 R 3 2 R 3 R 3 0 R 3 Finally t = 2R 0 2R 2 C 0
03. At t = 0 switch S1 is closed and remains closed for a long time and S2 remains open. Now S1 is opened and S2 is closed assuming that capacitor is initially uncharged find out
(i) The current through the capacitor immediately after that moment (ii) Charge on the capacitor long after that moment.
(iii) Total charge flown through the cell of emf 2 after S2 is closed. Sol. (i) Let Potential at point A is zero. Then at point B and C it will be
(because current through the circuit is zero). VB – VA = C( – 0)
Charge on capacitor = C( – 0) = C
(ii) Now S2 is closed and S1 is open. (p.d. across capacitor and charge on it will not change suddenly)
Potential at A is zero so at D it is – 2. current through the capacitor =
R ) 2 ( = R 3 (B to D) (iii) after a long time i = 0
VB – VA = VD – VA = – 2 Q = C (–2) = –2C
The charge on the lower plate (which is connected to the battery) changes from –C to 2C.
this charge will come form the battery, charge flown from that cell is 3C downward.
CHARGING AND DISCHARGING OF A CAPACITOR
Charging of a condenser :
(i) In the following circuit. If key 1 is closed then the condenser gets charged. Finite time is taken in the charging process. The quantity of charge at any instant of time t is given by q = q0[1 – e–(t/RC)]
Where q0 = maximum final value of charge at t = . According to this equations the quantity of charge on the condenser increases exponentially with increase of time. (ii) If t = RC = 1 then q = q0 [1 – e–(RC/RC)] = q0 e 1 1 or q = q0 (1 – 0.37) = 0.63 q0 = 63% of q0 (iii) Time t = RC is known as time constant.
i.e. the time constant is that time during which the charge rises on the condenser plates to 63% of its maximum value.
(iv) The potential difference across the condenser plates at any instant of time is given by V = V0[1 – e–(t/RC)] volt
(v) The potential curve is also similar to that of charge. During charging process an electric current flows in the circuit for a small interval of time which is known as the transient current. The value of this current at any instant of time is given by = 0[e–(t/RC)] ampere
According to this equation the current falls in the circuit exponentially (Fig.).
(vi) If t = RC = = Time constant = 0e(–RC/RC) =
e
0
= 0.37 0 = 37% of 0
i.e. time constant is that time during which current in the circuit falls to 37% of its maximum value.
Derivation of formulae for charging of capacitor it is given that initially capacitor is uncharged. let at any time
Applying kirchoff voltage law – iR – C q = 0 iR = C q C i = CR q C dt dq = CR q C q C CR dq = dt
q O C q dq =
t ORC dt – ln (C – q) + ln C = RC t ln q C C = et/RC C – q = C . e–t/RC q = C(1 – e–t/RC)RC = time constant of the RC series circuit. After one time constant
q = C e 1 1 C (1 – 0.37) = 0.63 C. Current at any time t
i = dt dq = C RC 1 e t/RC = t/RC e R
Voltage across capacitor after one time constant V = 0.63 Q = CV
VC = e(1 – e–t/RC)
Voltage across the resistor VR = iR = e–t/RC
Heat (by energy conservation = work done by battery – Ucapacitor ) = c() – ( 2 1 (2 – 0) = 2 1 c2 Alternatively : Heat = H =
0 2 Rdt i =
0 2 2 R RC t 2 e R dt = R 2
0 RC / t 2 e dt = 2 C 2 SOLVED EXAMPLES
01. Without using the formula of equivalent, find out charge on capacitorand current in all the branches as a function of time. Sol. Applying KVL in ABDEA
– iR = C 2 q i = CR 2 q R = CR 2 q C 2 q C 2 dq = 2CR dt q C 2 dq = 2CR dt –
q 0 (2C – q) dq = 2CR t C 2 q C 2 = e–t/2RC q = 2C (1 – e–t/2RC) q1 2 q C (1 – e–t/2RC) i 1 = 2R e–t/2RC q2 = 2 q C (1 – e–t/2RC) i 2 = 2R e–t/RC Alternate solution by equivalentTime constant of circuit = 2C × R = 2RC maximum charge on capacitor = 2C × = 2C
Hence equations of charge and current are as given below q = 2C (1 – e–t/2RC) q1 2 q C (1 – e–t/2RC) i 1 = 2R e–t/2RC q2 = 2 q C (1 – e–t/2RC) i 2 = 2R e–t/RC
02. A capacitor is connected to a 12 V battery through a resistance of 10. It is found that the potential difference across the capacitor rises to 4.0 V in 1s. Find the capacitances of the capacitor.
Sol. The charge on the capacitor during charging is given by Q = Q0(1 – e–t/RC). Hence, the potential difference across the capacitor is V = Q/C = Q0/C (1 – e–t/RC).
Here, at t = 1 s, the potential difference is 4V whereas the steady potential difference is Q0/C = 12V. So, 4V = 12V(1 – e–t/RC) or, 1 – e–t/RC = 3 1 or, e–t/RC = 3 2 or, 2 3 n RC t = 0.405 or, RC = 405 . 0 t = 45 . 0 s 1 = 2.469 s or, C = 10 s 469 . 2 = 0.25 F..
03. Initially the capacitor is uncharged find the charge on capacitor as a function of time, if switch is closed at t = 0.
Sol. Applying KVL in loop ABCDA – iR – (i – i1) R = 0 – 2iR + i1R = 0
Applying KVL in loop ABCEFDA – iR – i1R – C q = 0 2 R i 2 R i 2 1 1 = C q C – 3i1RC = 2q C – 2q = 3 dt dq . RC
q 0 C 2q dq =
t 03RC dt – 2 1 ln C q 2 C = RC 3 t q = 2 C
1 e2t/3RC
Method for objective :
In any circuit when there is only one capacitor then q = Qst
t/
e
1 ; Qst = steady state charge on capacitor (has been found in article 6 in this sheet) = Reff. C
Reffective is the resistance between the capacitor when battery is replaced by its internal resistance.
Discharging of a condenser :
(i) In the above circuit (in article 8.1) if key 1 is opened and key 2 is closed then the condenser gets discharged.
(ii) The quantity of charge on the condenser at any instant of time t is given by q = q0 e–(t/RC)
i.e. the charge falls exponentially. (iii) If t = RC = = time constant, then q =
e q0
= 0.37q0 = 37% of q0
i.e. the time constant is that time during which the charge on condenser plates discharge process falls to 37% (iv) The dimensions of RC are those of time i.e. MºLºT1 and the dimensions of
RC 1
are those of frequency i.e. M0L0T–1.
(v) The potential difference across the condenser plates at any instant of time t is given by V = V0e–(t/RC) Volt. (vi) The transient current at any instant of time is given by = –0e–(t/RC) ampere.
i.e. the current in the circuit decreases exponentially but its direction is opposite to that of charging current. Derivation of equation of discharging circuit :
Applying K.V.L. 0 iR C q i = CR q
q Q q dq =
t 0CR dt – ln Q q = + RC t q = Q . et/RC i = e t/RC RC Q dt dq SOLVED EXAMPLES
01. At t = 0 Sw is closed, if initially C1 is uncharged and C2 is charged to apotential difference 2 then find out following (Given C1 = C2 = C) (a) Charge on C1 and C2 as a function of time.
(b) Find out current in the circuit as a function of time. (c) Also plot the graphs for the relations derived in part (b). Sol. Let q charge flow in time 't' from the battery as shown.
The charge on various plates of the capacitor is as shown in the figure. Now applying KVL – C q –iR – C C 2 q = 0 3 = C q 2 + iR 3 – iRC = 2q
dt dq RC = 3C – 2q
q 03 C 2q dq =
t 0 RC dt C 3 q 2 C 3 ln = – RC t 2 q = C 2 3 (1 – e–2t/RC) And, i = dt dq = R 3 e–2t/RC On the plate B 2 2 q 2 C C C t t q' q q’ = 2C – q = 2C – 2 3 C + 2 3 C e–2t/RC = 2 C + 2 3 Ce–2t/RC = 2 C
2t/RC
e 3 1 02. The electric field between the plates of a parallel–plate capacitance 2.0 F drops to one third of its initial value in 4.4 s when the plates are connected by a thin wire. Find the resistance of the wire.
Sol. The electric field between the plates is
E = 0 A Q = 0 0 A Q e–t/RC or, E = E0e –t/RC
In the given problem, E =
3 1 E0 at t = 4.4 s. Thus, RC s 4 . 4 e 3 1 or, RC s 4 . 4 = In 3 = 1.1 or, R = F 0 . 2 1 . 1 s 4 . 4 = 2.0
EXERCISE # I
01. When 30C charge is given to an isolated conductor of capacitance 5F. Find out following (i) Potential of the conductor
(ii) Energy stored in the electric field of conductor
(iii) If this conductor is now connected to other isolated conductor of total charge 50 C and capacity 10 F then find out common potential of both the conductors.
(iv) Find out heat dissipated during the process of charge distribution. (v) Find out ratio of final charges on conductors.
(vi) Find out final charges on each conductor.
02. A and B are two spherical isolated conductors placed at large distance having radius ratio
B A r r = 2 1 and charge ratio B A Q Q = 1 2
. Find out following if they are joined by a conducting wire.
(i) Ratio of final charges
B A ' Q ' Q
. (ii) Ratio of final charge densities
B A
.
03. (i) Calculate the capacitance of a parallel plate capacitor having 20 cm × 20 cm square metal plates which are separated by a distance 8.85 mm.
(ii) If the plates contain equal but opposite charge of magnitude 20C. then find out energy stored between the plates of capacitor.
04. An air-filled parallel-plate capacitor is to be constructed which can store 12 mC of charge when operated at 1200 V. What can be the minimum plate area of the capacitor? The dielectric strength of air is 3 × 106 V/m.
05. Plate A of a parallel air filled capacitor is connected to a spring having force constant k and plate B is fixed. If a charge + q is placed on plate A and charge – q on plate B then find out extension in the spring in equilibrium. Assume area of plate is ‘A’.
06. A parallel-plate capacitor with the plate area 100 cm2 and the separation between the plates 1.0 cm is connected across a battery of emf 24 volts. Find the force of attraction between the plates.
07. (i) Find the charge supplied by the battery in the arrangement shown in figure. (ii) Find out charge on each capacitor.
08. Find the potential difference Va – Vb between the points a and b shown in each part of the figure.
09. Three uncharged capacitors of capacitance C1 = 1F, C2 = 2F and C3 = 3F are connected as shown in figure to one another and to points A, B and D potentials
A = 10 V,
B = 25 V and
D = 20 V. Determine the potential (
0) at point O.10. A capacitor with stored energy 4.0 J is connected with and identical capacitor with no electric field in between. Find the total energy stored in the two capacitors finally.
11. A 5.0 F capacitor is charged to 12 V. The positive plate of this capacitor is now connected to the negative terminal of a 12 V battery and vice versa. Calculate the heat developed in the connecting wires.
12. Five capacitors are connected as shown in figure below. Initially S is opened and all capacitors are uncharged. When S is closed, steady state is obtained. Then find out potential difference between the points M and N.
13. Three capacitors having capacitances 20F, 30F and 40F are connected in series with a 12 V battery. (i) Find the charge on each of the capacitors.
(ii) How much work has been done by the battery in charging the capacitors ?
14. Consider the situation shown in the figure. The switch S is open for a long time and then closed. (a) Find the charge flown through the battery when the switch S is closed.
(b) Find the work done by the battery.
(c) Find the change in energy stored in the capacitors. (d) Find the heat developed in the system.
15. If you have several 2.0
F capacitors, each capable of withstanding 200 volts without breakdown, how would you assemble a combination having an equivalent capacitance of;(a) 0.40
F or of (b) 1.2
F, capable of withstanding 1000 volts.16. A capacitor of capacitance C1 = 1.0
F charged upto a voltage V = 110 V is connected in a parallel to the terminals of a circuit consisting of two uncharged capacitor connected in series and possessing the capacitance C2 = 2.0
F and C3 = 3.0
F. What charge will flow through the connecting wires?17. Take the potential of the point B as shown in figure to be zero. (a) Find the potentials at the point C and D.
(b) If an uncharged capacitor is connected between C and D, what charge will appear in this capacitor?
18. Find the potential difference Va – Vb between the points a and b shown in each part of the figure.
19. Convince yourself that parts (a), (b) and (c) of figure are identical. Find the capacitance between the point A and B of the assembly.
20. Find the equivalent capacitances of the combinations shown in the figure between the indicated points.
21. (i) Find the equivalent capacitance of the infinite ladder shown in the figure between the points A and B. (ii) If now each capacitor is replaced by a capacitor which
is double in capacitance then repeat the question.
22. (i) Find the charge appearing on each of the three capacitors shown in figure. (ii) Find out total work done by the battery during charging process.
23. The two parallel plates of a capacitor have equal and opposite charges Q. The dielectric has a dielectric constant K and resistivity . Show that the "leakage" current carried by the dielectric is given by the relationship i =
o Q
24. The parallel plates of a capacitor have an area 0.2 m2 and are 102 m apart. The original potential difference between them is 3000 V, and it decreases to 1000 V when a sheet of dielectric is inserted between the plates filling the full space. Compute: (
0 = 9 x 1012 S. . units)(i) Original capacitance C0. (ii) The charge Q on each plate. (iii) Capacitance C after insertion of the dielectric. (iv) Dielectric constant K.
(v) Permittivity
of the dielectric. (vi) The original field E0 between the plates. (vii) The electric field E after insertion of the dielectric.25. A parallel plate isolated condenser consists of two metal plates of area A and separation 'd'. A slab of thickness 't' and dielectric constant K is inserted between the plates with its faces parallel to the plates and having the same surface area as that of the plates. Find the capacitance of the system. If K = 2, for what value of t/d will the capacitance of the system be 3/2 times that of the condenser with air filling the full space? Calculate the ratio of the energy in the two cases and account for the energy change.
26. Hard rubber has a dielectric constant of 2.8 and a dielectric strength of 18 x 106 volts/meter. If it is used as the dielectric material filling the full space in a parallel plate capacitor. What minimum area may the plates of the capacitor have in order that the capacitance be 7.0 x 102
f and that the capacitor be able to withstand apotential difference of 4000 volts.
27. Two parallel plate air capacitors filling the full space C were connected in series to a battery with e.m.f.
. Then one of the capacitors was filled up with uniform dielectric with relative permittivity k. How many times did the electric field strength in that capacitor decrease? What amount of charge flows through the battery?28. A parallel-plate capacitor of capacitance 5 F is connected to a battery of emf 6 V. The separation between the plates is 2 mm.
(a) find the electric field between the plates. (b) Find the electric field between the plates.
(c) A dielectric slab of thickness 1 mm and dielectric constant 5 is inserted into the gap to occupy the lower half of it. Find the capacitance of the new combination.
(d) How much charge has flown through the battery after the slab is inserted ?
29. Find the capacitances of the capacitors shown in figure. The plates area is A and the separation between the plates is d. Different dielectric slabs in a particular part of the figure are of the same thickness and the entire gap between the plates is field with the dielectric slabs.
30. Consider the situation shown in figure. The width of each plate is b. The capacitor plates are rigidly clamped in the laboratory and connected to a battery of emf . All surface are frictionless. Calculate the value of M for which the dielectric slab will stay in equilibrium.
31. Consider the situation shown in figure. The plates of the capacitor have plate area A and are clamped in the laboratory. The dielectric slab is released from rest with length a inside the capacitor. Neglecting any effect of friction or gravity, show that the slab will execute periodic motion and find its time period.
K a
d
32. A spherical condenser has 10 cm and 12 cm as the radii of inner and outer spheres. The space between the two spherical is filled with a dielectric of dielectric constant 5. Find the capacity when;
(i) the outer sphere is earthed. (ii) the inner sphere is earthed. 33. In the following figure the effective area of each plate of a mica condenser is A
and distance between two consecutive plates is d. If the dielectric constant of mica is k, then find the capacity of the condenser.
34. Find the potential difference between the points A and B and between the points B and C of figure in steady state.
35. A capacitance C, a resistance R and an emf are connected in series at t = 0. What is the maximum value of (a) the potential difference across the resistor, (b) the current in the circuit,
(c) the potential difference across the capacitor, (d) the energy stored in the capacitors. (e) the power delivered by the battery and (f) the power converted into heat. 36. (i) Find the charge on the capacitor shown in figure.
(ii) Find out values of i1, i2 and i3 in steady state.
37. Find the final charges on the four capacitor of capacitance 1F, 2F, 3F and 4F shown in figure. (Assuming initially they are uncharged).
38. How many time constants will elapse before the current in a charging RC circuit drops to half of its initial value? Answer the same question for a discharging RC circuit.
39. A capacitor of capacitance 100 F is connected across a battery of emf 6.0 V through a resistance of 20 k for 4.0 s. The battery is then replaced by a thick wire. What will be the charge on the capacitor 4.0 s after the battery is disconnected ?
40. The electric field between the plates of a parallel–plate capacitance 2.0 F drops to one third of its initial value in 4.4 s when the plates are connected by a thin wire. Find the resistance of the wire.
41. A capacitor charged to 50 V is discharged by connecting the two plates at t = 0. If the potential difference across the plates drops to 1.0 V at t = 10 ms, what will be the potential difference at t = 20 ms ?
42. A 5.0 F capacitor having a charge of 20 C is discharge through a wire of resistance 5.0 . Find the that dissipated in the wire between 25 to 50 s after the connections are made.
43. A capacitor of capacity 1
F is connected in closed series circuit with a resistance of 107 ohms, an open key and a cell of 2 V with negligible internal resistance:(i) When the key is switched on at time t = 0, find; (a) The time constant for the circuit.
(b) The charge on the capacitor at steady state.
(c) Time taken to deposit charge equal to half that at steady state.
(ii) If after fully charging the capacitor, the cell is shorted by zero resistance at time t = 0, find the charge on the capacitor at t = 50 s.
EXERCISE # II
01. The capacitance of a spherical conductor is proportional to
(A) C R2 (B) C R–2 (C) C R (D) C R–1
02. Stored energy in a charged conductor is
(A) 1CV2 2 (B) 2 2 1 Q V 2 (C) 2 2 1 Q 2 C (D) 2 1 Q 2 C 03. If the two plates of the charged capacitor are connected by a wire, then
(A) potential will become infinite (B) charge will become infinite
(C) capacitor will get discharged (D) charge will become double that of earlier one 04. The capacity of a parallel plate capacitor is 0.5mF. When a mica sheet is placed between the plates the potential
difference reduces to 1 8
of the previous value. The dielectric constant of mica is
(A) 1.6 (B) 5 (C) 8 (D) 40
05. Unit of capacitance is
(A) coulomb (B) volt (C) henry (D) farad
06. The capacitance of a capacitor is
(A) directly proportional to the dielectric constant of the medium between the plates (B) inversely proportional to the dielectric constant of the medium between the plates (C) proportional to the square of the dielectric constant of the medium between the plates (D) independent of the dielectric constant of the medium between the plates
07. If the energy of a capacitor of capacitance 2 F is 0.16 joule, then its potential difference will be
(A) 800 V (B) 400 V (C) 4
16 10 V (D) 16 10 4V
08. The capacitance of a parallel plate capacitor is 12 F . If the distance between its plates is reduced to half and the area of plates is doubled, then the capacitance of the capacitor will become
(A) 24 F (B) 12 F (C) 16 F (D) 48 F
09. A capacitor of 6 F is charged to such an extent that the potential difference between the plates becomes 50 V.. The work done in this process will be
(A) 7.5 10 2J (B) 7.5 10 3J (C) 3 10 6J (D) 3 10 3J
10. The radius of the circular plates of a parallel plate capacitor is R. Air is dielectric medium between the plates. If the capacitance of the capacitor is equal to the capacitance of a sphere of radius R, then the distance between the plates is
(A) R
4 (B)
R
2 (C) R (D) 2R
11. A capacitor of capacitance 500 F is charged at the rate of 100 C/s . The time in which the potential difference will become 20V, is
12. The capacitances of spherical iron, copper and alluminium conductors of same radii are C ,C and 1 2 C , then3 (A) C2 C1C3 (B) C3 C2 C1 (C) C1C2 C3 (D) C1C2 C3
13. 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 (C) 4 times (D) 2 times
14. 64 drops each charged to a potential of 100 volt, coalesce to form a big drop. The potential of the big drop will be
(A) 6400 volt (B) 3200 volt (C) 1600 volt (D) 800 volt
15. Two spheres of capacitances 3 F and 5 F are charged to 300 V and 500 V respectively and are connected together. The common potential will be
(A) 400 V (B) 425 V (C) 350 V (D) 375 V
16. The equivalent capacitance between A and B of the combination, shown in the figure, will be
3 F 2 F 1 F A B (A) 1.5 F (B) 3.0 F (C) 6 F 11 (D) 6 F
17. The potential of earth is zero because it is
(A) unchanged (B) an object of zero capacitance
(C) an object of infinite capacitance (D) having infinite charge
18. If a thin metal foil of same area is placed between the two plates of a parallel plate capacitor of capacitance C, then new capacitance will be
(A) C (B) 2C (C) 3C (D) 4C
19. On charging a capacitor of 20 F upto 500 V and a capacitor of 10 F upto 200 V, they are connected in parallel, their common potential will be
(A) 500 V (B) 400 V (C) 350 V (D) 250 V
20. Two capacitors of capacitances 1 F and 2 F are connected in series and this combination is charged upto a potential difference of 120 V. The potential difference on the capacitor of 1 F will be
(A) 40 V (B) 60 V (C) 80 V (D) 120 V
21. To reduce the capacitance of a parallel plate capacitor, the space between the plates is
(A) filled with dielectric material (B) reduced and area of the plates is increased (C) increased and area of the plates is decreased (D) increased and area is increased relatively 22. In the given circuit, the equivalent capacitance between A and B is
(A) C (B) 2C (C) 3C (D) 4C
23. In the given figure the capacitance between A and B will be (A) C
(B) 2C (C) 3C (D) 4C
24. The vertical plates of a parallel plate capacitor are just in front of each other and the capacitance is C. If the plates are shifted relatively, then capacitance
(A) will remain C (B) will be more than C (C) will be less than C (D) nothing can be said 25. Three capacitors of same capacitance are connected according to the following figures. For which combination
the equivalent capacitance will be maximum?
(A) in C and A (B) in A and D (C) in B (D) in D
26. In the above question, in which combination the energy stored will be maximum?
(A) in B and C (B) in D (C) in C and D (D) in A and B
27. Three capacitors of capacitances 12 F each are available. The minimum and maximum capacitances, which may be obtained from these are
(A) 12 F, 36 F (B) 4 F, 12 F (C) 4 F, 36 F (D) 0 F, F
28. n Capacitors each having capacitance C and breakdown voltage V are joined in series. The capacitance and the breakdown voltage of the combination is
(A) C and V (B) nC and nV (C) nC and V
n (D)
C
n and nV
29. The SI unit of is0
(A) farad (B) farad/metre (C) henry/metre (D) none of the above
30. Capacitors are used for
(A) smoothing rectified current from power supplies (B) elimination of sparking in switches
(C) storing large quantities of charge for use in research such as nuclear fusion (D) all of the above
31. Five capacitors of 10 F capacity each are connected to a D.C. potential of 100 volts as shown in figure. The equivalent capacitance between the points A and B will be equal to
(A) 40 F (B) 20 F
32. A parallel plate air capacitor is connected to a battery. the quantities charge, voltage, electric field and energy associated with this capacitor are given by Q , V ,E and 0 0 0 U respecitvely. A dielectric slab is now introduced0 to fill the space between the plates with battery still in connection. The corresponding quantities now given by Q, V, E and U are related to the previous ones as
(A) QQ ,U0 U0 (B) VV ,Q0 Q0 (C) EE ,U0 U0 (D) VV ,E0 E0
33. For the section AB of a circuit shown in figure, C1 , 1 F C2 , 2 F E10V and the potential difference
A B
V V 10 volt. Then, the charge on capacitor C is1
E C2 C1 A + – B (A) 0 C (B) 20 C 3 (C) 40 C 3
(D) none of the above
34. Two spherical conductors A and B of radii a and b
ba
are placed concentrically in air. A is given a charge +Q while B is earthed. Then the equivalent capacitance of the system is(A) 0 ab 4 b a (B) 40
ba
(C) 40b (D) 2 0 4 b b a 35. Force acting upon a charged particle kept between the plates of a charged condenser is F. If one of the plates of the condenser is removed, force acting on the same particle becomes
(A) 0 (B) F
2 (C) F (D) 2F
EXERCISE # III
ONLY ONE CORRECT
01. A capacitor of capacitance C is charged to a potential difference V from a cell and then disconnected from it. A charge +Q is now given to its positive plate. The potential difference across the capacitor is now
(A) V (B) V + Q C (C) V + Q 2C (D) V – Q C, if V < CV
02. In the circuit shown, a potential difference of 60V is applied across AB. The potential difference between the point M and N is
(A) 10 V (B) 15 V
(C) 20 V (D) 30 V
03. In the circuit shown in figure, the ratio of charges on 5F and 4F capacitor is : (A) 4/5
(B) 3/5 (C) 3/8 (D) 1/2
04. The minimum number of capacitors each of 3 F required to make a circuit with an equivalent capacitance 2.25 F is
(A) 3 (B) 4 (C) 5 (D) 6
05. From a supply of identical capacitors rated 8
F, 250 V, the minimum number of capacitors required to form a composite 16
F, 1000 V is :(A) 2 (B) 4 (C) 16 (D) 32
06. In the circuit shown, the energy stored in 1
F capacitor is(A) 40
J (B) 64
J(C) 32
J (D) none07. If charge on left plane of the 5F capacitor in the circuit segment shown in the figure is –20C, the charge on the right plate of 3F capacitor is
(A) + 8.57
C (B) – 8.57
C (C) + 11.42
C (D) –11.42
C 08. What is the equivalent capacitance of the system of capacitorsbetween A & B
(A) 7
6 C (B) 1.6 C
09. Two capacitor having capacitances 8 F and 16 F have breaking voltages 20 V and 80 V. They are combined in series. The maximum charge they can store individually in the combination is
(A) 160
C (B) 200
C (C) 1280
C (D) none of these10. Three plates A, B and C each of area 0.1 m2 are separated by 0.885 mm from
each other as shown in the figure. A 10 V battery is used to charge the system. The energy stored in the system is
(A) 1
J (B) 10–1
J (C) 10–2
J (D) 10–3
J11. A capacitor of capacitance C is initially charged to a potential difference of V volt. Now it is connected to a battery of 2V Volt with opposite polarity. The ratio of heat generated to the final energy stored in the capacitor will be
(A) 1.75 (B) 2.25 (C) 2.5 (D) 1/2
12. Five conducting parallel plates having area A and separation between them d, are placed as shown in the figure. Plate number 2 and 4 are connected wire and between point A and B, a cell of emf E is connected. The charge flown through the cell is
(A) 3 0AE 4 d (B) 2 0AE 3 d (C) 4 0AE d (D) 0AE 2d
13. Three long concentric conducting cylindrical shells have radii R, 2R and 2 2R. Inner and outer shells are connected to each other. The capacitance across middle and inner shells per unit length is:
(A) 0 1 3 ln2 (B) 6 0 ln 2 (C) 2ln20 (D) None
14. The plates S and T of an uncharged parallel plate capacitor are connected across a battery. The battery is then disconnected and the charged plates are now connected in a system as shown in the figure. The system shown is in equilibrium. All the strings are insulating and massless. The magnitude of charge on one of the capacitor plates is: [Area of plates = A]
(A) 2mgA 0 (B) 4mgA 0
k (C) mgA 0 (D) 0 2mgA k
15. Four metallic plates arearranged as shown in the figure. If the distance between each plate then capacitance of the given system between points A and B is (Given d << A)
(A) 0A d (B) 2 0A d (C) 3 0A d (D) 4 0A d
