2.14 COMBINATION OF CAPACITORS We can combine several capacitors of capacitance C1, C2,…, Cn to obtain a system with some effective capacitance C. The effective capacitance depends on the way the individual capacitors are combined. Two simple possibilities are discussed below. 2.14.1 Capacitors in series Figure 2.26 shows capacitors C1 and C2 FIGURE 2.26 Combination of two combined in series. capacitors in series. The left plate of C1 and the right plate of C2 are connected to two terminals of a battery and have charges Q and –Q , respectively. It then follows that the right plate of C1 has charge –Q and the left plate of C2 has charge Q. If this was not so, the net charge on each capacitor would not be zero. This would result in an electric field in the conductor connecting C1and C2. Charge would flow until the net charge on both C1 and C2 is zero and there is no electric field in the conductor connecting C1 and C2. Thus, in the series combination, charges on the two plates FIGURE 2.27 Combination of n (±Q) are the same on each capacitor. The total capacitors in series. 71 Reprint 2025-26 Physics potential drop V across the combination is the sum of the potential drops V1 and V2 across C1 and C2, respectively. Q Q + (2.55) V = V1 + V2 = C1 C 2 V 1 1 i.e., = + , (2.56) Q C1 C 2 Now we can regard the combination as an effective capacitor with charge Q and potential difference V. The effective capacitance of the combination is Q C = (2.57) V We compare Eq. (2.57) with Eq. (2.56), and obtain 1 1 1 = + (2.58) C C1 C 2 The proof clearly goes through for any number of capacitors arranged in a similar way. Equation (2.55), for n capacitors arranged in series, generalises to Q Q Q V = V1 + V 2 + ... + V n = + + ... + (2.59) C1 C 2 C n Following the same steps as for the case of two capacitors, we get the general formula for effective capacitance of a series combination of n capacitors: 1 1 1 1 1 = + + + ... + (2.60) C C1 C 2 C 3 C n 2.14.2 Capacitors in parallel Figure 2.28 (a) shows two capacitors arranged in parallel. In this case, the same potential difference is applied across both the capacitors. But the plate charges (±Q1) on capacitor 1 and the plate charges (±Q2) on the capacitor 2 are not necessarily the same: Q1 = C1V, Q2 = C2V (2.61) The equivalent capacitor is one with charge Q = Q1 + Q2 (2.62) and potential difference V. Q = CV = C1V + C2V (2.63) The effective capacitance C is, from Eq. (2.63), C = C1 + C2 (2.64) The general formula for effective capacitance C for parallel combination of n capacitors [Fig. 2.28 (b)] follows similarly, Q = Q1 + Q2 + ... + Qn (2.65) FIGURE 2.28 Parallel combination of i.e., CV = C1V + C2V + ... CnV(2.66) (a) two capacitors, (b) n capacitors. which gives C = C1 + C2 + ... Cn (2.67) 72 Reprint 2025-26 Electrostatic Potential and Capacitance Example 2.9 A network of four 10 mF capacitors is connected to a 500 V supply, as shown in Fig. 2.29. Determine (a) the equivalent capacitance of the network and (b) the charge on each capacitor. (Note, the charge on a capacitor is the charge on the plate with higher potential, equal and opposite to the charge on the plate with lower potential.) FIGURE 2.29 Solution (a) In the given network, C1, C2 and C3 are connected in series. The effective capacitance C¢ of these three capacitors is given by 1 1 1 1 = + + C ′ C1 C 2 C 3 For C1 = C2 = C3 = 10 mF, C¢ = (10/3) mF. The network has C¢ and C4 connected in parallel. Thus, the equivalent capacitance C of the network is  10  C = C¢ + C4 =  3 + 10 mF =13.3mF (b) Clearly, from the figure, the charge on each of the capacitors, C1, C2 and C3 is the same, say Q. Let the charge on C4 be Q¢. Now, since the potential difference across AB is Q/C1, across BC is Q/C2, across CD is Q/C3 , we have Q Q Q + + = 500 V . C1 C 2 C 3 Also, Q¢/C4 = 500 V. This gives for the given value of the capacitances, 10 −3 Q = 500 V × µ F = 1.7 × 10 C and EXAMPLE 3 Q ′ = 500 V × 10 µ F = 5.0 × 10 −3 C 2.9

2.8 In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10–3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor? 79 Reprint 2025-26 Physics 2.9 Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, (a) while the voltage supply remained connected. (b) after the supply was disconnected. 2.10 A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor? 2.11 A 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process? Reprint 2025-26 Chapter Three CURRENT ELECTRICITY 3.1 INTRODUCTION In Chapter 1, all charges whether free or bound, were considered to be at rest. Charges in motion constitute an electric current. Such currents occur naturally in many situations. Lightning is one such phenomenon in which charges flow from the clouds to the earth through the atmosphere, sometimes with disastrous results. The flow of charges in lightning is not steady, but in our everyday life we see many devices where charges flow in a steady manner, like water flowing smoothly in a river. A torch and a cell-driven clock are examples of such devices. In the present chapter, we shall study some of the basic laws concerning steady electric currents. 3.2 ELECTRIC CURRENT Imagine a small area held normal to the direction of flow of charges. Both the positive and the negative charges may flow forward and backward across the area. In a given time interval t, let q+ be the net amount (i.e., forward minus backward) of positive charge that flows in the forward direction across the area. Similarly, let q– be the net amount of negative charge flowing across the area in the forward direction. The net amount of charge flowing across the area in the forward direction in the time interval t, then, is q = q+– q–. This is proportional to t for steady current Reprint 2025-26 Physics and the quotient q I = (3.1) t is defined to be the current across the area in the forward direction. (If it turn out to be a negative number, it implies a current in the backward direction.) Currents are not always steady and hence more generally, we define the current as follows. Let DQ be the net charge flowing across a cross- section of a conductor during the time interval Dt [i.e., between times t and (t + Dt)]. Then, the current at time t across the cross-section of the conductor is defined as the value of the ratio of DQ to Dt in the limit of Dt tending to zero, ∆Q lim (3.2) I (t ) ≡ t 0 ∆→ ∆t In SI units, the unit of current is ampere. An ampere is defined through magnetic effects of currents that we will study in the following chapter. An ampere is typically the order of magnitude of currents in domestic appliances. An average lightning carries currents of the order of tens of thousands of amperes and at the other extreme, currents in our nerves are in microamperes. 3.3 ELECTRIC CURRENTS IN CONDUCTORS An electric charge will experience a force if an electric field is applied. If it is free to move, it will thus move contributing to a current. In nature, free charged particles do exist like in upper strata of atmosphere called the ionosphere. However, in atoms and molecules, the negatively charged electrons and the positively charged nuclei are bound to each other and are thus not free to move. Bulk matter is made up of many molecules, a gram of water, for example, contains approximately 1022 molecules. These molecules are so closely packed that the electrons are no longer attached to individual nuclei. In some materials, the electrons will still be bound, i.e., they will not accelerate even if an electric field is applied. In other materials, notably metals, some of the electrons are practically free to move within the bulk material. These materials, generally called conductors, develop electric currents in them when an electric field is applied. If we consider solid conductors, then of course the atoms are tightly bound to each other so that the current is carried by the negatively charged electrons. There are, however, other types of conductors like electrolytic solutions where positive and negative charges both can move. In our discussions, we will focus only on solid conductors so that the current is carried by the negatively charged electrons in the background of fixed positive ions. Consider first the case when no electric field is present. The electrons will be moving due to thermal motion during which they collide with the fixed ions. An electron colliding with an ion emerges with the same speed as before the collision. However, the direction of its velocity after the collision is completely random. At a given time, there is no preferential 82 direction for the velocities of the electrons. Thus on the average, the Reprint 2025-26 Current Electricity number of electrons travelling in any direction will be equal to the number of electrons travelling in the opposite direction. So, there will be no net electric current. Let us now see what happens to such a piece of conductor if an electric field is applied. To focus our thoughts, imagine the conductor in the shape of a cylinder of radius R (Fig. 3.1). Suppose we now take two thin circular discs FIGURE 3.1 Charges +Q and –Q put at the ends of a dielectric of the same radius and put of a metallic cylinder. The electrons will drift positive charge +Q distributed over one disc because of the electric field created to and similarly –Q at the other disc. We attach neutralise the charges. The current thus the two discs on the two flat surfaces of the will stop after a while unless the charges +Q cylinder. An electric field will be created and and –Q are continuously replenished. is directed from the positive towards the negative charge. The electrons will be accelerated due to this field towards +Q. They will thus move to neutralise the charges. The electrons, as long as they are moving, will constitute an electric current. Hence in the situation considered, there will be a current for a very short while and no current thereafter. We can also imagine a mechanism where the ends of the cylinder are supplied with fresh charges to make up for any charges neutralised by electrons moving inside the conductor. In that case, there will be a steady electric field in the body of the conductor. This will result in a continuous current rather than a current for a short period of time. Mechanisms, which maintain a steady electric field are cells or batteries that we shall study later in this chapter. In the next sections, we shall study the steady current that results from a steady electric field in conductors.

2.15 ENERGY STORED IN A CAPACITOR A capacitor, as we have seen above, is a system of two conductors with charge Q and –Q. To determine the energy stored in this configuration, consider initially two uncharged conductors 1 and 2. Imagine next a process of transferring charge from conductor 2 to conductor 1 bit by 73 Reprint 2025-26 Physics bit, so that at the end, conductor 1 gets charge Q. By charge conservation, conductor 2 has charge –Q at the end (Fig 2.30 ). In transferring positive charge from conductor 2 to conductor 1, work will be done externally, since at any stage conductor 1 is at a higher potential than conductor 2. To calculate the total work done, we first calculate the work done in a small step involving transfer of an infinitesimal (i.e., vanishingly small) amount of charge. Consider the intermediate situation when the conductors 1 and 2 have charges Q¢ and FIGURE 2.30 (a) Work done in a small –Q¢ respectively. At this stage, the potential difference step of building charge on conductor 1 V¢ between conductors 1 to 2 is Q¢/C, where C is thefrom Q¢ to Q¢ + d Q¢. (b) Total work done in charging the capacitor may be capacitance of the system. Next imagine that a small viewed as stored in the energy of charge d Q¢ is transferred from conductor 2 to 1. Work electric field between the plates. done in this step (d W), resulting in charge Q¢ on conductor 1 increasing to Q¢+ d Q¢, is given by Q ′ δW = V ′δQ ′ = δQ ′ (2.68) C Integrating eq. (2.68) Q 2 Q 2 Q ′ 1 Q ′ Q δ Q ’ = = W = ∫ C 2 2C 0 C 0 We can write the final result, in different ways Q 2 1 2 1 W = = CV = QV (2.69) 2C 2 2 Since electrostatic force is conservative, this work is stored in the form of potential energy of the system. For the same reason, the final result for potential energy [Eq. (2.69)] is independent of the manner in which the charge configuration of the capacitor is built up. When the capacitor discharges, this stored-up energy is released. It is possible to view the potential energy of the capacitor as ‘stored’ in the electric field between the plates. To see this, consider for simplicity, a parallel plate capacitor [of area A (of each plate) and separation d between the plates]. Energy stored in the capacitor 1 Q 2 ( Aσ)2 d = = × (2.70) 2 C 2 ε0 A The surface charge density s is related to the electric field E between the plates, σ E = (2.71) ε0 From Eqs. (2.70) and (2.71) , we get Energy stored in the capacitor 2 × A d (2.72) U = (1/2 ) ε0 E 74 Reprint 2025-26 Electrostatic Potential and Capacitance Note that Ad is the volume of the region between the plates (where electric field alone exists). If we define energy density as energy stored per unit volume of space, Eq (2.72) shows that Energy density of electric field, u =(1/2)e0E 2 (2.73) Though we derived Eq. (2.73) for the case of a parallel plate capacitor, the result on energy density of an electric field is, in fact, very general and holds true for electric field due to any configuration of charges. Example 2.10 (a) A 900 pF capacitor is charged by 100 V battery [Fig. 2.31(a)]. How much electrostatic energy is stored by the capacitor? (b) The capacitor is disconnected from the battery and connected to another 900 pF capacitor [Fig. 2.31(b)]. What is the electrostatic energy stored by the system? FIGURE 2.31 Solution (a) The charge on the capacitor is Q = CV = 900 × 10–12 F × 100 V = 9 × 10–8 C The energy stored by the capacitor is = (1/2) CV 2 = (1/2) QV = (1/2) × 9 × 10–8C × 100 V = 4.5 × 10–6 J (b) In the steady situation, the two capacitors have their positive EXAMPLE plates at the same potential, and their negative plates at the same potential. Let the common potential difference be V¢. The 2.10 75 Reprint 2025-26 Physics charge on each capacitor is then Q¢ = CV¢. By charge conservation, Q¢ = Q/2. This implies V¢ = V/2. The total energy of the system is 1 1 − 6 = 2 × Q ' V ' = QV = 2.25 × 10 J 2 4 Thus in going from (a) to (b), though no charge is lost; the final energy is only half the initial energy. Where has the remaining energy gone? 2.10 There is a transient period before the system settles to the situation (b). During this period, a transient current flows from the first capacitor to the second. Energy is lost during this time in EXAMPLE the form of heat and electromagnetic radiation. SUMMARY 1. Electrostatic force is a conservative force. Work done by an external force (equal and opposite to the electrostatic force) in bringing a charge q from a point R to a point P is q(VP–VR), which is the difference in potential energy of charge q between the final and initial points. 2. Potential at a point is the work done per unit charge (by an external agency) in bringing a charge from infinity to that point. Potential at a point is arbitrary to within an additive constant, since it is the potential difference between two points which is physically significant. If potential at infinity is chosen to be zero; potential at a point with position vector r due to a point charge Q placed at the origin is given is given by 1 Q V ( r ) = 4 πεo r 3. The electrostatic potential at a point with position vector r due to a point dipole of dipole moment p placed at the origin is 1 p.rˆ V ( r ) = 2 4 πεo r The result is true also for a dipole (with charges –q and q separated by 2a) for r >> a. 4. For a charge configuration q1, q2, ..., qn with position vectors r1, r2, ... rn, the potential at a point P is given by the superposition principle 1 q1 q 2 qn V = ( + + ... + ) 4 πε0 r1P r2P rnP where r1P is the distance between q1 and P, as and so on. 5. An equipotential surface is a surface over which potential has a constant value. For a point charge, concentric spheres centred at a location of the charge are equipotential surfaces. The electric field E at a point is perpendicular to the equipotential surface through the point. E is in the direction of the steepest decrease of potential. Reprint 2025-26 Electrostatic Potential and Capacitance 6. Potential energy stored in a system of charges is the work done (by an external agency) in assembling the charges at their locations. Potential energy of two charges q1, q2 at r1, r2 is given by 1 q1 q 2 U = 4 πε0 r12 where r12 is distance between q1 and q2. 7. The potential energy of a charge q in an external potential V(r) is qV(r). The potential energy of a dipole moment p in a uniform electric field E is –p.E. 8. Electrostatics field E is zero in the interior of a conductor; just outside the surface of a charged conductor, E is normal to the surface given by σ E = nˆ where ˆn is the unit vector along the outward normal to the ε0 surface and s is the surface charge density. Charges in a conductor can reside only at its surface. Potential is constant within and on the surface of a conductor. In a cavity within a conductor (with no charges), the electric field is zero. 9. A capacitor is a system of two conductors separated by an insulator. Its capacitance is defined by C = Q/V, where Q and –Q are the charges on the two conductors and V is the potential difference between them. C is determined purely geometrically, by the shapes, sizes and relative positions of the two conductors. The unit of capacitance is farad:, 1 F = 1 C V –1. For a parallel plate capacitor (with vacuum between the plates), A C = ε0 d where A is the area of each plate and d the separation between them. 10. If the medium between the plates of a capacitor is filled with an insulating substance (dielectric), the electric field due to the charged plates induces a net dipole moment in the dielectric. This effect, called polarisation, gives rise to a field in the opposite direction. The net electric field inside the dielectric and hence the potential difference between the plates is thus reduced. Consequently, the capacitance C increases from its value C0 when there is no medium (vacuum), C = KC0 where K is the dielectric constant of the insulating substance. 11. For capacitors in the series combination, the total capacitance C is given by 1 1 1 1 = + + + ... C C1 C 2 C 3 In the parallel combination, the total capacitance C is: C = C1 + C2 + C3 + ... where C1, C2, C3... are individual capacitances. 77 Reprint 2025-26 Physics 12. The energy U stored in a capacitor of capacitance C, with charge Q and voltage V is 1 1 2 1 Q 2 U = QV = CV = 2 2 2 C The electric energy density (energy per unit volume) in a region with electric field is (1/2)e0E2. Physical quantity Symbol Dimensions Unit Remark Potential or V [M1 L2 T–3 A–1] V Potential difference is physically significant Capacitance C [M–1 L–2 T–4 A2] F Polarisation P [L–2 AT] C m-2 Dipole moment per unit volume Dielectric constant K [Dimensionless] POINTS TO PONDER 1. Electrostatics deals with forces between charges at rest. But if there is a force on a charge, how can it be at rest? Thus, when we are talking of electrostatic force between charges, it should be understood that each charge is being kept at rest by some unspecified force that opposes the net Coulomb force on the charge. 2. A capacitor is so configured that it confines the electric field lines within a small region of space. Thus, even though field may have considerable strength, the potential difference between the two conductors of a capacitor is small. 3. Electric field is discontinuous across the surface of a spherical charged ˆn outside. Electric potential is, however shell. It is zero inside and σε0 continuous across the surface, equal to q/4pe0R at the surface. 4. The torque p × E on a dipole causes it to oscillate about E. Only if there is a dissipative mechanism, the oscillations are damped and the dipole eventually aligns with E. 5. Potential due to a charge q at its own location is not defined – it is infinite. 6. In the expression qV (r) for potential energy of a charge q, V (r) is the potential due to external charges and not the potential due to q. As seen in point 5, this expression will be ill-defined if V (r) includes potential 78 due to a charge q itself. Reprint 2025-26 Electrostatic Potential and Capacitance 7. A cavity inside a conductor is shielded from outside electrical influences. It is worth noting that electrostatic shielding does not work the other way round; that is, if you put charges inside the cavity, the exterior of the conductor is not shielded from the fields by the inside charges. EXERCISES