13.4 1.23 351 Reprint 2025-26 Physics 13.5 (i) Q = –4.03 MeV; endothermic (ii) Q = 4.62 MeV; exothermic 56 – 2m 28 Al = 26.90 MeV; not possible. 13.6 Q = m ( 26 Fe ) ( 13 ) 13.7 4.536 × 1026 MeV 13.8 About 4.9 × 104 y 13.9 360 KeV CHAPTER 14 14.1 (c) 14.2 (d) 14.3 (c) 14.4 (c) 14.5 (c) 14.6 50 Hz for half-wave, 100 Hz for full-wave Reprint 2025-26 Bibligraphy BIBLIOGRAPHY TEXTBOOKS For additional reading on the topics covered in this book, you may like to consult one or more of the following books. Some of these books however are more advanced and contain many more topics than this book. 1 Ordinary Level Physics, A.F. Abbott, Arnold-Heinemann (1984). 2 Advanced Level Physics, M. Nelkon and P. Parker, 6th Edition, Arnold-Heinemann (1987). 3 Advanced Physics, Tom Duncan, John Murray (2000). 4 Fundamentals of Physics, David Halliday, Robert Resnick and Jearl Walker, 7th Edition John Wily (2004). 5 University Physics (Sears and Zemansky’s), H.D. Young and R.A. Freedman, 11th Edition, Addison—Wesley (2004). 6 Problems in Elementary Physics, B. Bukhovtsa, V. Krivchenkov, G. Myakishev and V. Shalnov, MIR Publishers, (1971). 7 Lectures on Physics (3 volumes), R.P. Feynman, Addision – Wesley (1965). 8 Berkeley Physics Course (5 volumes) McGraw Hill (1965). a. Vol. 1 – Mechanics: (Kittel, Knight and Ruderman) b. Vol. 2 – Electricity and Magnetism (E.M. Purcell) c. Vol. 3 – Waves and Oscillations (Frank S. Crawford) d. Vol. 4 – Quantum Physics (Wichmann) e. Vol. 5 – Statistical Physics (F. Reif ) 9 Fundamental University Physics, M. Alonso and E. J. Finn, Addison – Wesley (1967). 10 College Physics, R.L. Weber, K.V. Manning, M.W. White and G.A. Weygand, Tata McGraw Hill (1977). 11 Physics: Foundations and Frontiers, G. Gamow and J.M. Cleveland, Tata McGraw Hill (1978). 12 Physics for the Inquiring Mind, E.M. Rogers, Princeton University Press (1960). 13 PSSC Physics Course, DC Heath and Co. (1965) Indian Edition, 14 Physics Advanced Level, Jim Breithampt, Stanley Thornes Publishers (2000). 15 Physics, Patrick Fullick, Heinemann (2000). 16 Conceptual Physics, Paul G. Hewitt, Addision—Wesley (1998). 17 College Physics, Raymond A. Serway and Jerry S. Faughn, Harcourt Brace and Co. (1999). 18 University Physics, Harris Benson, John Wiley (1996). 19 University Physics, William P. Crummet and Arthur B. Western, Wm.C. Brown (1994). 20 General Physics, Morton M. Sternheim and Joseph W. Kane, John Wiley (1988). 21 Physics, Hans C. Ohanian, W.W. Norton (1989). Reprint 2025-26 Physics 22 Advanced Physics, Keith Gibbs, Cambridge University Press (1996). 23 Understanding Basic Mechanics, F. Reif, John Wiley (1995). 24 College Physics, Jerry D. Wilson and Anthony J. Buffa, Prentice Hall (1997). 25 Senior Physics, Part – I, I.K. Kikoin and A.K. Kikoin, MIR Publishers (1987). 26 Senior Physics, Part – II, B. Bekhovtsev, MIR Publishers (1988). 27 Understanding Physics, K. Cummings, Patrick J. Cooney, Priscilla W. Laws and Edward F. Redish, John Wiley (2005). 28 Essentials of Physics, John D. Cutnell and Kenneth W. Johnson, John Wiley (2005). GENERAL BOOKS For instructive and entertaining general reading on science, you may like to read some of the following books. Remember however, that many of these books are written at a level far beyond the level of the present book. 1 Mr. Tompkins in paperback, G. Gamow, Cambridge University Press (1967). 2 The Universe and Dr. Einstein, C. Barnett, Time Inc. New York (1962). 3 Thirty years that Shook Physics, G. Gamow, Double Day, New York (1966). 4 Surely You’re Joking, Mr. Feynman, R.P. Feynman, Bantam books (1986). 5 One, Two, Three… Infinity, G. Gamow, Viking Inc. (1961). 6 The Meaning of Relativity, A. Einstein, (Indian Edition) Oxford and IBH Pub. Co. (1965). 7 Atomic Theory and the Description of Nature, Niels Bohr, Cambridge (1934). 8 The Physical Principles of Quantum Theory, W. Heisenberg, University of Chicago Press (1930). 9 The Physics—Astronomy Frontier, F. Hoyle and J.V. Narlikar, W.H. Freeman (1980). 10 The Flying Circus of Physics with Answer, J. Walker, John Wiley and Sons (1977). 11 Physics for Everyone (series), L.D. Landau and A.I. Kitaigorodski, MIR Publisher (1978). Book 1: Physical Bodies Book 2: Molecules Book 3: Electrons Book 4: Photons and Nuclei. 12 Physics can be Fun, Y. Perelman, MIR Publishers (1986). 13 Power of Ten, Philip Morrison and Eames, W.H. Freeman (1985). 14 Physics in your Kitchen Lab., I.K. Kikoin, MIR Publishers (1985). 15 How Things Work: The Physics of Everyday Life, Louis A. Bloomfield, John Wiley (2005). 16 Physics Matters: An Introduction to Conceptual Physics, James Trefil and Robert M. Hazen, John Wiley (2004). 354 Reprint 2025-26

1.3 SIGNIFICANT FIGURES figures in a measurement. This important remark makes most of the followingAs discussed above, every measurement observations clear:involves errors. Thus, the result of (1) For example, the length 2.308 cm has fourmeasurement should be reported in a way that significant figures. But in different units, theindicates the precision of measurement. same value can be written as 0.02308 m or 23.08Normally, the reported result of measurement is a number that includes all digits in the mm or 23080 µm. number that are known reliably plus the first All these numbers have the same number of digit that is u Reprint 2025-26 4 PHYSICS This shows that the location of decimal point is negative exponent (or power) of 10. In order to of no consequence in determining the number get an approximate idea of the number, we may of significant figures. round off the number a to 1 (for a ≤5) and to 10 The example gives the following rules : (for 5<a ≤10). Then the number can be • All the non-zero digits are significant. expressed approximately as 10b in which the • All the zeros between two non-zero digits exponent (or power) b of 10 is called order of magnitude of the physical quantity. When only are significant, no matter where the an estimate is required, the quantity is of the decimal point is, if at all. order of 10b. For example, the diameter of the • If the number is less than 1, the zero(s) earth (1.28×107m) is of the order of 107m with on the right of decimal point but to the the order of magnitude 7. The diameter of left of the first non-zero digit are not hydrogen atom (1.06 ×10–10m) is of the order of significant. [In 0.00 2308, the underlined 10–10m, with the order of magnitude zeroes are not significant]. –10. Thus, the diameter of the earth is 17 orders • The terminal or trailing zero(s) in a of magnitude larger than the hydrogen atom. It is often customary to write the decimal after number without a decimal point are not the first digit. Now the confusion mentioned in significant. (a) above disappears : [Thus 123 m = 12300 cm = 123000 mm has 4.700 m = 4.700 × 102 cm three significant figures, the trailing zero(s) = 4.700 × 103 mm = 4.700 × 10–3 km being not significant.] However, you can also The power of 10 is irrelevant to the see the next observation. determination of significant figures. However, all • The trailing zero(s) in a number with a zeroes appearing in the base number in the decimal point are significant. scientific notation are significant. Each number [The numbers 3.500 or 0.06900 have four in this case has four significant figures. significant figures each.] Thus, in the scientific notation, no confusion (2) There can be some confusion regarding the arises about the trailing zero(s) in the base trailing zero(s). Suppose a length is reported to number a. They are always significant. be 4.700 m. It is evident that the zeroes here (4) The scientific notation is ideal for reporting are meant to convey the precision of measurement. But if this is not adopted, we use measurement and are, therefore, significant. [If the rules adopted in the preceding example : these were not, it would be superfluous to write • For a number greater than 1, without anythem explicitly, the reported measurement decimal, the trailing zero(s) are notwould have been simply 4.7 m]. Now suppose we change units, then significant. • For a number with a decimal, the trailing4.700 m = 470.0 cm = 4700 mm = 0.004700 km zero(s) are significant. Since the last number has trailing zero(s) in a number with no decimal, we would conclude (5) The digit 0 conventionally put on the left of a erroneously from observation (1) above that the decimal for a number less than 1 (like 0.1250) number has two significant figures, while in is never significant. However, the zeroes at the fact, it has four significant figures and a mere end of such number are significant in a change of units cannot change the number of measurement. significant figures. (6) The multiplying or dividing factors which are (3) To remove such ambiguities in neither rounded numbers nor numbers determining the number of significant representing measured values are exact and figures, the best way is to report every have infinite number of significant digits. For measurement in scientific notation (in the dpower of 10). In this notation, every number is example in r = or s = 2πr, the factor 2 is an expressed as a × 10b, where a is a number 2 between 1 and 10, and b is any positive or exact number and it can be written as 2.0, 2.00 Reprint 2025-26 UNITS AND MEASUREMENT 5 decimal place. The final result should, therefore, t or 2.0000 as required. Similarly, in T = , n is be rounded off to 663.8 g. n Similarly, the difference in length can be an exact number. expressed as : 1.3.1 Rules for Arithmetic Operations with 0.307 m – 0.304 m = 0.003 m = 3 ×10–3 m. Significant Figures Note that we should not use the rule (1) applicable The result of a calculation involving approximate for multiplication and division and write 664 g as measured values of quantities (i.e. values with the result in the example of addition and limited number of significant figures) must 3.00 × 10–3 m in the example of subtraction. They reflect the umeasured values. It cannot be more accurate properly. For addition and subtraction, the rule than the original measured values themselves is in terms of decimal places. on which the result is based. In general, the final result should not have more significant 1.3.2 Rounding off the Ufigures than the original data from which it was The result of computation with approximate obtained. Thus, if mass of an object is measured numbers, which contain more than one to be, say, 4.237 g (four significant figures) and uits volume is measured to be 2.51 cm3, then its for rounding off numbers to the appropriate density, by mere arithmetic division, is significant figures are obvious in most cases. A 1.68804780876 g/cm3 upto 11 decimal places. number 2.746 rounded off to three significant It would be clearly absurd and irrelevant to figures is 2.75, while the number 1.743 would record the calculated value of density to such a be 1.74. The rule by convention is that the precision when the measurements on which the preceding digit is raised by 1 if the value is based, have much less precision. The insignificant digit to be dropped (the following rules for arithmetic operations with underlined digit in this case) is more than significant figures ensure that the final result 5, and is left unchanged if the latter is less of a calculation is shown with the precision that than 5. But what if the number is 2.745 in is consistent with the precision of the input which the insignificant digit is 5. Here, themeasured values : convention is that if the preceding digit is(1) In multiplication or division, the final even, the insignificant digit is simplyresult should retain as many significant dropped and, if it is odd, the preceding digitfigures as are there in the original number with the least significant figures. is raised by 1. Then, the number 2.745 rounded Thus, in the example above, density should off to three significant figures becomes 1.74. On be reported to three significant figures. the other hand, the number 2.735 rounded off to three significant figures becomes 1.74 since 4.237g -3 Density = = 1.69 g cm the preceding digit is odd. 3 2.51 cm In any involved or complex multi-step Similarly, if the speed of light is given as calculation, you should retain, in intermediate 3.00 × 108 m s-1 (three significant figure) and steps, one digit more than the significant digits one year (1y = 365.25 d) has 3.1557 × 107 s (five and round off to proper significant figures at the significant figures), the light year is 9.47 × 1015 m end of the calculation. Similarly, a number (three significant figures). known to be within many significant figures, such as in 2.99792458 × 108 m/s for the speed (2) In addition or subtraction, the final result of light in vacuum, is rounded off to anshould retain as many decimal places as are approximate value 3 × 108 m/s , which is oftenthere in the number with the least employed in computations. Finally, rememberdecimal places. that exact numbers that appear in formulae like For example, the sum of the numbers 436.32 g, 227.2 g and 0.301 g by mere arithmetic L addition, is 663.821 g. But the least precise 2 π in T = 2π , have a large (infinite) number measurement (227.2 g) is correct to only one g Reprint 2025-26 6 PHYSICS of significant figures. The value of π = = 16.2 cm ± 0.6 %. 3.1415926.... is known to a large number of significant figures. You may take the value as Similarly, the breadth b may be written as 3.142 or 3.14 for π, with limited number of b = 10.1 ± 0.1 cm significant figures as required in specific = 10.1 cm ± 1 % cases. Then, the error of the product of two (or more)⊳ Example 1.1 Each side of a cube is experimental values, using the combination of measured to be 7.203 m. What are the errors rule, will be total surface area and the volume of the l b = 163.62 cm2 + 1.6% cube to appropriate significant figures? = 163.62 + 2.6 cm2 Answer The number of significant figures in the measured length is 4. The calculated area This leads us to quote the final result as and the volume should therefore be rounded off l b = 164 + 3 cm2 to 4 significant figures. Here 3 cm2 is the uSurface area of the cube = 6(7.203)2 m2 estimation of area of rectangular sheet. = 311.299254 m2 (2) If a set of experimental data is specified = 311.3 m2 to n significant figures, a result obtained by combining the data will also be valid to n Volume of the cube = (7.203)3 m3 significant figures. = 373.714754 m3 However, if data are subtracted, the number of = 373.7 m3 ⊳ significant figures can be reduced. ⊳ Example 1.2 5.74 g of a substance For example, 12.9 g – 7.06 g, both specified to occupies 1.2 cm3. Express its density by three significant figures, cannot properly be keeping the significant figures in view. evaluated as 5.84 g but only as 5.8 g, as u in a different fashion (smallest number ofmeasured mass whereas there are only 2 decimal places rather than the number of significant figures in the measured volume. significant figures in any of the number added Hence the density should be expressed to only or subtracted). 2 significant figures. (3) The relative error of a value of number 5.74 −3 specified to significant figures depends not Density = g cm 1.2 only on n but also on the number itself. = 4.8 g cm--3 . ⊳ For example, the accuracy in measurement of mass 1.02 g is ± 0.01 g whereas another 1.3.3 Rules for Determining the U in the Results of Arithmetic The relative error in 1.02 g is Calculations = (± 0.01/1.02) × 100 % = ± 1%The rules for determining the u Similarly, the relative error in 9.89 g iserror in the number/measured quantity in = (± 0.01/9.89) × 100 %arithmetic operations can be understood from = ± 0.1 %the following examples. Finally, remember that intermediate results in(1) If the length and breadth of a thin rectangular sheet are measured, using a metre a multi-step computation should be scale as 16.2 cm and, 10.1 cm respectively, there calculated to one more significant figure in are three significant figures in each every measurement than the number of measurement. It means that the length l may digits in the least precise measurement. be written as These should be justified by the data and then l = 16.2 ± 0.1 cm the arithmetic operations may be carried out; Reprint 2025-26 UNITS AND MEASUREMENT 7 otherwise rounding errors can build up. For mass, one dimension in length, and –2 example, the reciprocal of 9.58, calculated (after dimensions in time. The dimensions in all other rounding off) to the same number of significant base quantities are zero. figures (three) is 0.104, but the reciprocal of Note that in this type of representation, the magnitudes are not considered. It is the quality0.104 calculated to three significant figures is of the type of the physical quantity that enters. 9.62. However, if we had written 1/9.58 = 0.1044 Thus, a change in velocity, initial velocity, and then taken the reciprocal to three significant average velocity, final velocity, and speed are figures, we would have retrieved the original all equivalent in this context. Since all these value of 9.58. quantities can be expressed as length/time, This example justifies the idea to retain one their dimensions are [L]/[T] or [L T–1]. more extra digit (than the number of digits in the least precise measurement) in intermediate 1.5 DIMENSIONAL FORMULAE AND steps of the complex multi-step calculations in DIMENSIONAL EQUATIONS order to avoid additional errors in the process The expression which shows how and which of of rounding off the numbers. the base quantities represent the dimensions of a physical quantity is called the dimensional

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.