2.5 A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10–12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

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.

1.20 A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 103 N/C and points radially inward, what is the net charge on the sphere? 43 Reprint 2025-26 Physics 1.21 A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 mC/m2. (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere? 1.22 An infinite line charge produces a field of 9 × 104 N/C at a distance of 2 cm. Calculate the linear charge density. 1.23 Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10–22 C/m2. What is E: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates? Reprint 2025-26 Chapter Two ELECTROSTATIC POTENTIAL AND CAPACITANCE 2.12.12.12.12.1 IIINTRODUCTIONIINTRODUCTIONNTRODUCTIONNTRODUCTIONNTRODUCTION In Chapters 5 and 7 (Class XI), the notion of potential energy was introduced. When an external force does work in taking a body from a point to another against a force like spring force or gravitational force, that work gets stored as potential energy of the body. When the external force is removed, the body moves, gaining kinetic energy and losing an equal amount of potential energy. The sum of kinetic and potential energies is thus conserved. Forces of this kind are called conservative forces. Spring force and gravitational force are examples of conservative forces. Coulomb force between two (stationary) charges is also a conservative force. This is not surprising, since both have inverse-square dependence on distance and differ mainly in the proportionality constants – the masses in the gravitational law are replaced by charges in Coulomb’s law. Thus, like the potential energy of a mass in a gravitational field, we can define electrostatic potential energy of a charge in an electrostatic field. Consider an electrostatic field EEEEE due to some charge configuration. First, for simplicity, consider the field E due to a charge Q placed at the origin. Now, imagine that we bring a test charge q from a point R to a point P against the repulsive force on it due to the charge Q. With reference Reprint 2025-26 Physics to Fig. 2.1, this will happen if Q and q are both positive or both negative. For definiteness, let us take Q, q > 0. Two remarks may be made here. First, we assume that the test charge q is so small that it does not disturb the original configuration, namely the charge Q at the origin (or else, we keep Q fixed at the origin by some unspecified force). Second, in bringing the charge q fromFIGURE 2.1 A test charge q (> 0) is moved from the point R to the R to P, we apply an external force Fext just enough to point P against the repulsive counter the repulsive electric force FE (i.e, Fext= –FE). force on it by the charge Q (> 0) This means there is no net force on or acceleration of placed at the origin. the charge q when it is brought from R to P, i.e., it is brought with infinitesimally slow constant speed. In this situation, work done by the external force is the negative of the work done by the electric force, and gets fully stored in the form of potential energy of the charge q. If the external force is removed on reaching P, the electric force will take the charge away from Q – the stored energy (potential energy) at P is used to provide kinetic energy to the charge q in such a way that the sum of the kinetic and potential energies is conserved. Thus, work done by external forces in moving a charge q from R to P is WRP = – = (2.1) This work done is against electrostatic repulsive force and gets stored as potential energy. At every point in electric field, a particle with charge q possesses a certain electrostatic potential energy, this work done increases its potential energy by an amount equal to potential energy difference between points R and P. Thus, potential energy difference ∆U = U P − U R = W RP (2.2) (Note here that this displacement is in an opposite sense to the electric force and hence work done by electric field is negative, i.e., –WRP .) Therefore, we can define electric potential energy difference between two points as the work required to be done by an external force in moving (without accelerating) charge q from one point to another for electric field of any arbitrary charge configuration. Two important comments may be made at this stage: (i) The right side of Eq. (2.2) depends only on the initial and final positions of the charge. It means that the work done by an electrostatic field in moving a charge from one point to another depends only on the initial and the final points and is independent of the path taken to go from one point to the other. This is the fundamental characteristic of a conservative force. The concept of the potential energy would not be meaningful if the work depended on the path. The path-independence of work done by an electrostatic field can be proved using the 46 Coulomb’s law. We omit this proof here. Reprint 2025-26 Electrostatic Potential and Capacitance (ii) Equation (2.2) defines potential energy difference in terms of the physically meaningful quantity work. Clearly, potential energy so defined is undetermined to within an additive constant.What this means is that the actual value of potential energy is not physically significant; it is only the difference of potential energy that is significant. We can always add an arbitrary constant a to potential energy at every point, since this will not change the potential energy difference: (U P + α) − (U R + α) = U P − U R Put it differently, there is a freedom in choosing the point where potential energy is zero. A convenient choice is to have electrostatic potential energy zero at infinity. With this choice, if we take the point R at infinity, we get from Eq. (2.2) Count Alessandro Volta (1745 – 1827) Italian W ∞ P = U P − U ∞ = U P (2.3) physicist, professor at Since the point P is arbitrary, Eq. (2.3) provides us with a Pavia. Volta established that the animal electri- COUNTdefinition of potential energy of a charge q at any point. city observed by LuigiPotential energy of charge q at a point (in the presence of field Galvani, 1737–1798, indue to any charge configuration) is the work done by the experiments with frog external force (equal and opposite to the electric force) in muscle tissue placed in bringing the charge q from infinity to that point. contact with dissimilar metals, was not due to 2.2 ELECTROSTATIC POTENTIAL any exceptional property of animal tissues but ALESSANDROConsider any general static charge configuration. We define was also generated potential energy of a test charge q in terms of the work done whenever any wet body on the charge q. This work is obviously proportional to q, since was sandwiched between the force at any point is qE, where E is the electric field at that dissimilar metals. This VOLTA point due to the given charge configuration. It is, therefore, led him to develop the convenient to divide the work by the amount of charge q, so first voltaic pile, orthat the resulting quantity is independent of q. In other words, battery, consisting of a (1745 work done per unit test charge is characteristic of the electric large stack of moist disks of cardboard (electro-field associated with the charge configuration. This leads to lyte) sandwiched the idea of electrostatic potential V due to a given charge between disks of metal –1827) configuration. From Eq. (2.1), we get: (electrodes). Work done by external force in bringing a unit positive charge from point R to P  U P − U R  = VP – VR = (2.4)  q  where VP and VR are the electrostatic potentials at P and R, respectively. Note, as before, that it is not the actual value of potential but the potential difference that is physically significant. If, as before, we choose the potential to be zero at infinity, Eq. (2.4) implies: Work done by an external force in bringing a unit positive charge from infinity to a point = electrostatic potential (V ) at that point. 47 Reprint 2025-26 Physics In other words, the electrostatic potential (V ) at any point in a region with electrostatic field is the work done in bringing a unit positive charge (without acceleration) from infinity to that point. The qualifying remarks made earlier regarding potential energy also apply to the definition of potential. To obtain the work done per unit test charge, we should take an infinitesimal test charge FIGURE 2.2 Work done on a test charge q dq, obtain the work done dW in bringing it from by the electrostatic field due to any given infinity to the point and determine the ratio charge configuration is independent dW/dq. Also, the external force at every point of the of the path, and depends only on path is to be equal and opposite to the electrostatic its initial and final positions. force on the test charge at that point. 2.3 POTENTIAL DUE TO A POINT CHARGE Consider a point charge Q at the origin (Fig. 2.3). For definiteness, take Q to be positive. We wish to determine the potential at any point P with position vector r from the origin. For that we must calculate the work done in bringing a unit positive test charge from infinity to the point P. For Q > 0, the work done against the repulsive force on the test charge is positive. Since work done is independent of the path, we choose a convenient path – along the radial direction from infinity to the point P. At some intermediate point P¢ on the path, the electrostatic force on a unit positive charge is FIGURE 2.3 Work done in bringing a unit positive test charge from infinity to the Q × 1 rˆ ′ (2.5) point P, against the repulsive force of 2 4 πε0r ' charge Q (Q > 0), is the potential at P due to the charge Q. where ˆ′r is the unit vector along OP¢. Work done against this force from r¢ to r¢ + Dr¢ is Q ∆W = − 2 ∆′r (2.6) 4 πε0r ' The negative sign appears because for Dr¢ < 0, DW is positive. Total work done (W) by the external force is obtained by integrating Eq. (2.6) from r¢ = ¥ to r¢ = r, r Q Q r Q = dr ′ = ε 0r ′ 2 4 πε0r ′ ∞ 4 πε0r (2.7) W = − ∫4∞ π This, by definition is the potential at P due to the charge Q Q V (r ) = (2.8) 48 4 πε0r Reprint 2025-26 Electrostatic Potential and Capacitance Equation (2.8) is true for any sign of the charge Q, though we considered Q > 0 in its derivation. For Q < 0, V < 0, i.e., work done (by the external force) per unit positive test charge in bringing it from infinity to the point is negative. This is equivalent to saying that work done by the electrostatic force in bringing the unit positive charge form infinity to the point P is positive. [This is as it should be, since for Q < 0, the force on a unit positive test charge is attractive, so that the electrostatic force and the displacement (from infinity to P) are FIGURE 2.4 Variation of potential V with r [in units of in the same direction.] Finally, we (Q/4pe0) m-1] (blue curve) and field with r [in units of (Q/4pe0) m-2] (black curve) for a point charge Q.note that Eq. (2.8) is consistent with the choice that potential at infinity be zero. Figure (2.4) shows how the electrostatic potential ( 1/r) and the electrostatic field (1/r 2 ) varies with r. Example 2.1 (a) Calculate the potential at a point P due to a charge of 4 × 10–7C located 9 cm away. (b) Hence obtain the work done in bringing a charge of 2 × 10–9 C from infinity to the point P. Does the answer depend on the path along which the charge is brought? Solution (a) = 4 × 104 V (b) W = qV = 2 × 10–9C × 4 × 104V = 8 × 10–5 J No, work done will be path independent. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements: One along EXAMPLE r and another perpendicular to r. The work done corresponding to the later will be zero. 2.1