14.1 In an n-type silicon, which of the following statement is true: (a) Electrons are majority carriers and trivalent atoms are the dopants. (b) Electrons are minority carriers and pentavalent atoms are the dopants. (c) Holes are minority carriers and pentavalent atoms are the dopants. (d) Holes are majority carriers and trivalent atoms are the dopants.

1.6 FORCES BETWEEN MULTIPLE CHARGES The mutual electric force between two charges is given by Coulomb’s law. How to calculate the force on a charge where there are not one but several charges around? Consider a system of n stationary charges q1, q2, q3, ..., qn in vacuum. What is the force on q1 due to q2, q3, ..., qn? Coulomb’s law is not enough to answer this question. Recall that forces of mechanical origin add according to the parallelogram law of addition. Is the same true for forces of electrostatic origin? Experimentally, it is verified that force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. The individual forces are unaffected due to the presence of other charges. This is termed as the principle of superposition. To better understand the concept, consider a system of three charges q1, q2 and q3, as shown in Fig. 1.5(a). The force on one charge, say q1, due to two other charges q2, q3 can therefore be obtained by performing a vector addition of the forces due to each one of these charges. Thus, if the force on q1 due to q2 is denoted by F12, F12 is given by Eq. (1.3) even though other charges are present. FIGURE 1.5 A system of 1 q1q 2 (a) three charges Thus, F12 = 2 ˆr12 (b) multiple charges. 11 4 πε0 r12 Reprint 2025-26 Physics In the same way, the force on q1 due to q3, denoted by F13, is given by 1 q1q 3 F13 = 2 rˆ13 4 πε0 r13 which again is the Coulomb force on q1 due to q3, even though other charge q2 is present. Thus the total force F1 on q1 due to the two charges q2 and q3 is given as 1 q1q 2 1 q1q 3 F1 = F12 + F13 = rˆ12 + rˆ13 (1.4) 4 π ε0 r122 4 πε0 r132 The above calculation of force can be generalised to a system of charges more than three, as shown in Fig. 1.5(b). The principle of superposition says that in a system of charges q1, q2, ..., qn, the force on q1 due to q2 is the same as given by Coulomb’s law, i.e., it is unaffected by the presence of the other charges q3, q4, ..., qn. The total force F1 on the charge q1, due to all other charges, is then given by the vector sum of the forces F12, F13, ..., F1n: i.e., 1  q1q 2 q1q 3 q1q n  F1 = F12 + F13 + ...+ F1n =  2 rˆ12 + 2 rˆ13 + ... + 2 1rˆ n  4 πε0  r12 r13 r1n  q1 n q i = ˆr1i 2 (1.5) 4πε0 =∑i 2 r1 i The vector sum is obtained as usual by the parallelogram law of addition of vectors. All of electrostatics is basically a consequence of Coulomb’s law and the superposition principle. Example 1.5 Consider three charges q1, q2, q3 each equal to q at the vertices of an equilateral triangle of side l. What is the force on a charge Q (with the same sign as q) placed at the centroid of the triangle, as shown in Fig. 1.6? FIGURE 1.6 1.5 Solution In the given equilateral triangle ABC of sides of length l, if we draw a perpendicular AD to the side BC, AD = AC cos 30º = ( 3 /2 ) l and the distance AO of the centroid O EXAMPLE 12 from A is (2/3) AD = (1/ 3 ) l. By symmatry AO = BO = CO. Reprint 2025-26 Electric Charges and Fields Thus, 3 Qq Force F1 on Q due to charge q at A = 2 along AO 4 πε0 l 3 Qq 2 along BOForce F2 on Q due to charge q at B = 4 πε0 l 3 Qq 2 along COForce F3 on Q due to charge q at C = 4 πε0 l 3 Qq 2 along OA, by theThe resultant of forces F2 and F3 is 4 πε0 l 3 Qq parallelogram law. Therefore, the total force on Q = 2 ( rˆ − rˆ ) 4 πε0 l = 0, where ˆr is the unit vector along OA. It is clear also by symmetry that the three forces will sum to zero. Suppose that the resultant force was non-zero but in some direction. EXAMPLE Consider what would happen if the system was rotated through 60° about O. 1.5 Example 1.6 Consider the charges q, q, and –q placed at the vertices of an equilateral triangle, as shown in Fig. 1.7. What is the force on each charge? FIGURE 1.7 Solution The forces acting on charge q at A due to charges q at B and –q at C are F12 along BA and F13 along AC respectively, as shown in Fig. 1.7. By the parallelogram law, the total force F1 on the charge q at A is given by F1 = F 1ˆr where 1ˆr is a unit vector along BC. The force of attraction or repulsion for each pair of charges has the q 2 same magnitude F = 4 π ε0 l 2 EXAMPLE The total force F2 on charge q at B is thus F2 = F ˆr 2, where ˆr 2 is a unit vector along AC. 1.6 13 Reprint 2025-26 Physics Similarly the total force on charge –q at C is F3 = 3 F ˆn , where ˆn is the unit vector along the direction bisecting the ÐBCA. It is interesting to see that the sum of the forces on the three charges 1.6 is zero, i.e., F1 + F2 + F3 = 0 The result is not at all surprising. It follows straight from the fact that Coulomb’s law is consistent with Newton’s third law. The proof EXAMPLE is left to you as an exercise. 1.7 ELECTRIC FIELD Let us consider a point charge Q placed in vacuum, at the origin O. If we place another point charge q at a point P, where OP = r, then the charge Q will exert a force on q as per Coulomb’s law. We may ask the question: If charge q is removed, then what is left in the surrounding? Is there nothing? If there is nothing at the point P, then how does a force act when we place the charge q at P. In order to answer such questions, the early scientists introduced the concept of field. According to this, we say that the charge Q produces an electric field everywhere in the surrounding. When another charge q is brought at some point P, the field there acts on it and produces a force. The electric field produced by the charge Q at a point r is given as 1 Q 1 Q rˆ rˆ = E ( r ) = (1.6) 4 πε0 r 2 4 πε0 r 2 where rˆ = r/r, is a unit vector from the origin to the point r. Thus, Eq.(1.6) specifies the value of the electric field for each value of the position vector r. The word “field” signifies how some distributed quantity (which could be a scalar or a vector) varies with position. The effect of the charge has been incorporated in the existence of the electric field. We obtain the force F exerted by a charge Q on a charge q, as 1 Qq F = 2 rˆ (1.7) 4 πε0 r Note that the charge q also exerts an equal and opposite force on the charge Q. The electrostatic force between the charges Q and q can be looked upon as an interaction between charge q and the electric field of Q and vice versa. If we denote the position of charge q by the vector r, it experiences a force F equal to the charge q multiplied by the electric field E at the location of q. Thus, F(r) = q E(r) (1.8) Equation (1.8) defines the SI unit of electric field as N/C*. Some important remarks may be made here: (i) From Eq. (1.8), we can infer that if q is unity, the electric field due to FIGURE 1.8 Electric a charge Q is numerically equal to the force exerted by it. Thus, the field (a) due to a electric field due to a charge Q at a point in space may be defined charge Q, (b) due to a as the force that a unit positive charge would experience if placed charge –Q. 14 * An alternate unit V/m will be introduced in the next chapter. Reprint 2025-26 Electric Charges and Fields at that point. The charge Q, which is producing the electric field, is called a source charge and the charge q, which tests the effect of a source charge, is called a test charge. Note that the source charge Q must remain at its original location. However, if a charge q is brought at any point around Q, Q itself is bound to experience an electrical force due to q and will tend to move. A way out of this difficulty is to make q negligibly small. The force F is then negligibly small but the ratio F/q is finite and defines the electric field:  F  E = lim (1.9) q → 0  q  A practical way to get around the problem (of keeping Q undisturbed in the presence of q) is to hold Q to its location by unspecified forces! This may look strange but actually this is what happens in practice. When we are considering the electric force on a test charge q due to a charged planar sheet (Section 1.14), the charges on the sheet are held to their locations by the forces due to the unspecified charged constituents inside the sheet. (ii) Note that the electric field E due to Q, though defined operationally in terms of some test charge q, is independent of q. This is because F is proportional to q, so the ratio F/q does not depend on q. The force F on the charge q due to the charge Q depends on the particular location of charge q which may take any value in the space around the charge Q. Thus, the electric field E due to Q is also dependent on the space coordinate r. For different positions of the charge q all over the space, we get different values of electric field E. The field exists at every point in three-dimensional space. (iii) For a positive charge, the electric field will be directed radially outwards from the charge. On the other hand, if the source charge is negative, the electric field vector, at each point, points radially inwards. (iv) Since the magnitude of the force F on charge q due to charge Q depends only on the distance r of the charge q from charge Q, the magnitude of the electric field E will also depend only on the distance r. Thus at equal distances from the charge Q, the magnitude of its electric field E is same. The magnitude of electric field E due to a point charge is thus same on a sphere with the point charge at its centre; in other words, it has a spherical symmetry. 1.7.1 Electric field due to a system of charges Consider a system of charges q1, q2, ..., qn with position vectors r1, r2, ..., rn relative to some origin O. Like the electric field at a point in space due to a single charge, electric field at a point in space due to the system of charges is defined to be the force experienced by a unit test charge placed at that point, without disturbing the original positions of charges q1, q2, ..., qn. We can use Coulomb’s law and the superposition principle to determine this field at a point P denoted by 15position vector r. Reprint 2025-26 Physics Electric field E1 at r due to q1 at r1 is given by 1 q1 E1 = 2 ˆr1P 4 πε0 r1 P where ˆr1P is a unit vector in the direction from q1 to P, and r1P is the distance between q1 and P. In the same manner, electric field E2 at r due to q2 at r2 is 1 q 2 E2 = 2 ˆr2P 4 πε0 r2 P where ˆr2P is a unit vector in the direction from q2 to P FIGURE 1.9 Electric field at a point and r2P is the distance between q2 and P. Similar due to a system of charges is the expressions hold good for fields E3, E4, ..., En due to vector sum of the electric fields at charges q3, q4, ..., qn. the point due to individual charges. By the superposition principle, the electric field E at r due to the system of charges is (as shown in Fig. 1.9) E(r) = E1 (r) + E2 (r) + … + En(r) 1 q1 1 q 2 1 q n = 2 rˆ1P + 2 rˆ2 P + ... + 2 rˆn P 4 πε0 r1P 4 πε0 r2 P 4 πε0 rn P 1 n q i E(r) = 2 ˆri P (1.10) 4 πε0 =∑i 1 ri P E is a vector quantity that varies from one point to another point in space and is determined from the positions of the source charges. 1.7.2 Physical significance of electric field You may wonder why the notion of electric field has been introduced here at all. After all, for any system of charges, the measurable quantity is the force on a charge which can be directly determined using Coulomb’s law and the superposition principle [Eq. (1.5)]. Why then introduce this intermediate quantity called the electric field? For electrostatics, the concept of electric field is convenient, but not really necessary. Electric field is an elegant way of characterising the electrical environment of a system of charges. Electric field at a point in the space around a system of charges tells you the force a unit positive test charge would experience if placed at that point (without disturbing the system). Electric field is a characteristic of the system of charges and is independent of the test charge that you place at a point to determine the field. The term field in physics generally refers to a quantity that is defined at every point in space and may vary from point to point. Electric field is a vector field, since force is a vector quantity. The true physical significance of the concept of electric field, however, emerges only when we go beyond electrostatics and deal with time- dependent electromagnetic phenomena. Suppose we consider the force between two distant charges q1, q2 in accelerated motion. Now the greatest speed with which a signal or information can go from one point to another 16 is c, the speed of light. Thus, the effect of any motion of q1 on q2 cannot Reprint 2025-26 Electric Charges and Fields arise instantaneously. There will be some time delay between the effect (force on q2) and the cause (motion of q1). It is precisely here that the notion of electric field (strictly, electromagnetic field) is natural and very useful. The field picture is this: the accelerated motion of charge q1 produces electromagnetic waves, which then propagate with the speed c, reach q2 and cause a force on q2. The notion of field elegantly accounts for the time delay. Thus, even though electric and magnetic fields can be detected only by their effects (forces) on charges, they are regarded as physical entities, not merely mathematical constructs. They have an independent dynamics of their own, i.e., they evolve according to laws of their own. They can also transport energy. Thus, a source of time- dependent electromagnetic fields, turned on for a short interval of time and then switched off, leaves behind propagating electromagnetic fields transporting energy. The concept of field was first introduced by Faraday and is now among the central concepts in physics. Example 1.7 An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude 2.0 × 104 N C–1 [Fig. 1.10(a)]. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [Fig. 1.10(b)]. Compute the time of fall in each case. Contrast the situation with that of ‘free fall under gravity’. FIGURE 1.10 Solution In Fig. 1.10(a) the field is upward, so the negatively charged electron experiences a downward force of magnitude eE where E is the magnitude of the electric field. The acceleration of the electron is ae = eE/me where me is the mass of the electron. Starting from rest, the time required by the electron to fall through a 2 h 2 h m e distance h is given by t e = = a e e E For e = 1.6 × 10–19C, me = 9.11 × 10–31 kg, E = 2.0 × 104 N C–1, h = 1.5 × 10–2 m, te = 2.9 × 10–9s In Fig. 1.10 (b), the field is downward, and the positively charged proton experiences a downward force of magnitude eE. The acceleration of the proton is ap = eE/mp EXAMPLE where mp is the mass of the proton; mp = 1.67 × 10–27 kg. The time of fall for the proton is 1.7 17 Reprint 2025-26 Physics 2 h 2 h m p –7 t p = = = 1. 3 × 10 s a p e E Thus, the heavier particle (proton) takes a greater time to fall through the same distance. This is in basic contrast to the situation of ‘free fall under gravity’ where the time of fall is independent of the mass of the body. Note that in this example we have ignored the acceleration due to gravity in calculating the time of fall. To see if this is justified, let us calculate the acceleration of the proton in the given electric field: e E a p = m p (1 . 6 × 10 − 19 C) × (2. 0 × 10 4 N C −1 ) = −27 1 .67 × 10 kg 12 –2 1.7 = 1 .9 × 10 m s which is enormous compared to the value of g (9.8 m s–2), the acceleration due to gravity. The acceleration of the electron is even greater. Thus, the effect of acceleration due to gravity can be ignored EXAMPLE in this example. Example 1.8 Two point charges q1 and q2, of magnitude +10–8 C and –10–8 C, respectively, are placed 0.1 m apart. Calculate the electric fields at points A, B and C shown in Fig. 1.11. FIGURE 1.11 Solution The electric field vector E1A at A due to the positive charge q1 points towards the right and has a magnitude (9 × 10 9 Nm 2 C -2 ) × (10 −8 C) E1A = 2 = 3.6 × 104 N C–1 (0.05m) 1.8 The electric field vector E2A at A due to the negative charge q2 points towards the right and has the same magnitude. Hence the magnitude of the total electric field EA at A is EA = E1A + E2A = 7.2 × 104 N C–1 EXAMPLE EA is directed toward the right.18 Reprint 2025-26 Electric Charges and Fields The electric field vector E1B at B due to the positive charge q1 points towards the left and has a magnitude (9 × 10 9 Nm 2 C –2 ) × (10 −8 C) E1B = 2 = 3.6 × 104 N C–1 (0.05 m) The electric field vector E2B at B due to the negative charge q2 points towards the right and has a magnitude (9 × 10 9 Nm 2 C –2 ) × (10 −8 C) E 2B = 2 = 4 × 103 N C–1 (0.15 m) The magnitude of the total electric field at B is EB = E1B – E2B = 3.2 × 104 N C–1 EB is directed towards the left. The magnitude of each electric field vector at point C, due to charge q1 and q2 is (9 × 10 9 Nm 2 C –2 ) × (10 − 8 C) E1C = E 2C = 2 = 9 × 103 N C–1 (0.10 m) The directions in which these two vectors point are indicated in Fig. 1.11. The resultant of these two vectors is π π EXAMPLE E C = E1c cos + E 2 c cos = 9 × 103 N C–1 3 3 1.8 EC points towards the right.

14.3 Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated 341 Reprint 2025-26 Physics by energy band gap respectively equal to (Eg)C, (Eg)Si and (Eg)Ge. Which of the following statements is true? (a) (Eg)Si < (Eg)Ge < (Eg)C (b) (Eg)C < (Eg)Ge > (Eg)Si (c) (Eg)C > (Eg)Si > (Eg)Ge (d) (Eg)C = (Eg)Si = (Eg)Ge 14.4 In an unbiased p-n junction, holes diffuse from the p-region to n-region because (a) free electrons in the n-region attract them. (b) they move across the junction by the potential difference. (c) hole concentration in p-region is more as compared to n-region. (d) All the above. 14.5 When a forward bias is applied to a p-n junction, it (a) raises the potential barrier. (b) reduces the majority carrier current to zero. (c) lowers the potential barrier. (d) None of the above. 14.6 In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency. Reprint 2025-26 Notes Reprint 2025-26 Physics APPENDICES APPENDIX A 1 THE GREEK ALPHABET APPENDIX A 2 COMMON SI PREFIXES AND SYMBOLS FOR MULTIPLES AND SUB-MULTIPLES Reprint 2025-26 AppendicesAnswers APPENDIX A 3 SOME IMPORTANT CONSTANTS OTHER USEFUL CONSTANTS 345 Reprint 2025-26 Physics ANSWERS CHAPTER 9 9.1 v = –54 cm. The image is real, inverted and magnified. The size of the image is 5.0 cm. As u ® f, v ® ¥; for u < f, image is virtual. 9.2 v = 6.7 cm. Magnification = 5/9, i.e., the size of the image is 2.5 cm. As u ® ¥; v ® f (but never beyond) while m ® 0. 9.3 1.33; 1.7 cm 9.4 nga = 1.51; nwa = 1.32; ngw = 1.144; which gives sin r = 0.6181 i.e., r ~ 38°. 9.5 r = 0.8 × tan ic and sin ci = 1/1.33 ≅ 0.75 , where r is the radius (in m) of the largest circle from which light comes out and ic is the critical angle for water-air interface, Area = 2.6 m2 9.6 n ≅ 1.53 and Dm for prism in water ≅ 10° 9.7 R = 22 cm 9.8 Here the object is virtual and the image is real. u = +12 cm (object on right; virtual) (a) f = +20 cm. Image is real and at 7.5 cm from the lens on its right side. (b) f = –16 cm. Image is real and at 48 cm from the lens on its right side. 9.9 v = 8.4 cm, image is erect and virtual. It is diminished to a size 1.8 cm. As u ® ¥, v ® f (but never beyond f while m ® 0). Note that when the object is placed at the focus of the concave lens (21 cm), the image is located at 10.5 cm (not at infinity as one might wrongly think). 9.10 A diverging lens of focal length 60 cm 9.11 (a) ve = –25 cm and fe = 6.25 cm give ue = –5 cm; vO = (15 – 5) cm = 10 cm, fO = uO = – 2.5 cm; Magnifying power = 20 (b) uO = – 2.59 cm. Magnifying power = 13.5. 9.12 Angular magnification of the eye-piece for image at 25 cm 25 25   1  11; | u e |= cm = 2 .27cm ; vO = 7.2 cm 2.5 11 Separation = 9.47 cm; Magnifying power = 88 9.13 24; 150 cm 9.14 (a) Angular magnification = 1500 346 (b) Diameter of the image = 13.7 cm. Reprint 2025-26 Answers