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1/44, , Pradeep's Fundamental Physics (XI) ReTay, , , , a QUANTIZATION OF CHARGE, , Formula used. ¢ = + ne, Units used, ¢ and e are in coulomb, n is a number, Standard value. ¢ = 1-6 x 10-!? coulomb, , , , {examptefl] Which is bigger, a coulomb or, charge on an electron ? How many electronic, charges form one coulomb of charge ?, , (Ph. Board 2011), , Solution. A coulomb of charge is bigger than, the charge on an electron., , Magnitude of charge on one electron,, , e = 1-6 x 107! coulomb, Number of electronic charges in one coulomb,, a, e 116x107!, , Note that one coulomb is too big a unit of charge., , f PF] How much positive and, negative charge is there in a cup of water ?, , NCERT Solved Example, , = 0-625 x 10!, , , , EE ee, Solution. Suppose the cup contains 250 cc of, water (H,0)., Mass of 250 cm? of water = 250 g., Molecular weight of water = 2 + 16 = 18, Number of molecules in 18 g of water, = 6.023 x 10%, Number of molecules in 250 g of water, _ 6.023 x 1073 x 250, 7 18, As each molecule of water contains 10 electrons,, therefore, total number of electrons, ., _ 10x 6.023 x10 x 250, a 18, As q=ne, therefore,, q = 8365 x 10% x 1-6 x 10-9, = 1338 x 107C, lexampteb) If a body gives out 10°, , n = 8-365 x 107, , electrons every second, how much time is required ., , to get a total charge of 1 C from it?, NCERT Solved Example, eS, , Charge given/sec, ia, qzne= 10? x 16x 10-7 C, , =16x 101°C, Total charge, Q=1C, , . Time required -< = Texio- sec, 9. 6:25 x10?, , = 625 x10" 8 = F699 394x365, = 198-18 year, jexampie[] Two bodies A and B carry, charges - 300 #C and — 0:44 .C. How many, electrons should be transferred from A toB so that, they acquire equal charges?, , Solution. Here, q, = — 3-00 #C, , and Qn = - 0-44 pC, , Let n electrons be transferred from A to B, when, , A and B would carry same charge., Charge on A = Charge on B, — 3-00 + ne =— 0:44 — ne, 2 ne = 3-00 — 0-44 = 2-56 (uC), , , , , , n=—, , 2e, e=16x 1019 C= 16x 10°73 pc, , ee 3, ~axnexio — 08% 10, , Taking, , =8x 102, , COULOMB’S LAW, , Formulae used. Fy = sls Fquthaail + ;, Ane, me, , el g'lal_ 1 igtle!, 4m ge, r, , Units used. 4), 42 are in coulomb, F in newton and, rin metre, i, , f 1, Standard Values, ——— = 20-2, tne, 910° Nm? C~, €9= 8-85 x 10712. C2 Ne! m2, , Scanned with CamScanner

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ELECTROSTATIC CHARGES AND FIELDS, , example fy A free pith ball P of 10 g, , carries a positive charge of 5 x 10-* C, What must, , be the nature and magnitude of charge that should, be given to another pith ball Q fixed 7 cm below, the former ball, so that upper ball is stationary ?, (Hr, Board 2001), Solution. Here, m = 10 g = 10 x 10 kg,, 1 =5x108C,r=7com=7x 10-2 m, Second ball Q must carry positive charge so that, force of repulsion balances the weight of ball P., When ball remains stationary, Fig. 1.50,, F=mg, ! Nae =mg, 4me) r°, , , , , , FIGURE 1.50, F, , I © °, , 7cm mg, , Te., , 9x10 x5x10-8 xq,, (7x10-?)?, , , , , , , , = 10x 103 x 9.8, , _ 10x1073 x9.8x(7x10-2)2, 2 9x109x5x10°8, = 1-067 x 10% C, \Exampie[9 A particle carrying charge, +q is held at the centre of a square of each side, one metre. It is surrounded by eight charges, , arranged on the square as shown in Fig. 1.51. If q, =2uC, what is the net force on the particle?, , FIGURE 1.51, -2q -3q, , , , , , , , , , , , , , 1/45, , Solution, As is clear from Fig. 1.51, forces on, the particle at O due to (- 2 %-29);(-34,-3 qQ), and (+ 4 q, +4 q) are equal and opposite. They cancel, out in pairs. However, forces due to +7 qand-7q, add up. Therefore, net force on the particle at O is, , rot, 09@+74(q), 4ne, (1/2)2, — 910° x14 (2x1076)2, = AEX (2x10")*, 14, =36x 14x 4x 103N, , =2:016N, , , , ‘Example f] Coulomb’s law for electrostatic force between two point charges and, Newton’s law for gravitational force between two, Stationary point masses, both have inverse square, dependence on the distance between the charges/, masses (a) compare the strength of these forces, by determining the ratio of their magnitudes, (@ for an electron and a proton (ii) for two protons, (b) estimate the accelerations for electron and, proton due to electrical force of their mutual, attraction when they are 1A apart., , NCERT Solved Example, Solution. (a) (i) For an electron and proton, 1 exe., 4ne, P, , , , LE I=, , Gm,.m, , =e?, [Als 2, , IF | 1 e, , \F, | 4ne, Gm,m,, , , , , , , , 9x 109 (1-6 x 107!9)?, * 667 «10-1 x9 x 10-3! x 1.66% 10-27, , = 2:3 x 10?, , (ii) Similarly, for two protons,, |F1 1 e, , | Fy 4ne, Gm, m,, , , , 9x 10° (1-6 x 10719)?, © 667X107! x (1-66 x 10°27 )2, , = 1:3 x 1056, , Scanned with CamScanner

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1/46, , (b) Force of mutual attraction between an, electron and a proton,, , re! e _ 9x10? 1-6 x10-)?, ~ 7 (10-12, , Ane, re, =23x10°N, , 231078, Acceleration of electron = ue 2 fen, m, — 9x10-3!, , = 2:5 x 1022 m/s?, , ; F 2:3x10°8 ., Acceleration of proton = —- =——, m 1-66 x 10-27, , = 1:3 x 10! m/s?, , jExamptef] A charged metallic sphere A, is suspended by a nylon thread. Another charged, metallic sphere B carried by an insulating handle, is brought close to A such that the distance between, their centres is 10 cm as shown in Fig. 1.52(a). The, resulting repulsion of A is noted (for example, by, shining a beam of light and measuring the, deflection of its shadow on a calibrated screen)., , , , FIGURE 1.52, _—_ —_, + >, BS °, k—— 10cm ——1, —_——, , , , , , ge", , ded:, , — 55cm —, , , , , , , , Pradeeg's Fundamental Physics (XID Pay, , Spheres A and B are touched by uncharged spheres, Cand D respectively, as shown in Fig. 1,52(6).C, and D are then removed and B is brought closer, to A to a distance of 5-0 cm between their centres,, as shown in Fig. 1.52(c). What is the expected, repulsion of A on the basis of Coulomb’s Law ?, Spheres A and C and spheres B and D have, identical sizes. Ignore the sizes of A and B in, comparison to separation between their centres,, NCERT Solved Example, , Solution. Let the original repulsive force, between A and B be, kay, , 2, , As A and C have same size, charges are shared, equally. Again, as B and D have same size, their, charges are also shared equally., , As charges on A and B are halved, and distance, between them is also halved from 10 cm to 5 cm,, therefore,, , F=, , F’ = —__+*_ = —,- =F, , (Example) Two electrons and a positive, charge q are held along a straight line. At what, position and for what value of q will the system be, in equilibrium ? Check whether it is stable,, unstable or neutral equilibrium., , Solution. Let two electrons of charges — e each, be held at A and B. The third charge + q must be, placed at the centre O of AB. The forces on + q, due, , to two electrons being equal and opposite, cancel, eachother and it is in equilibrium., , ; For the charge (— e) at A to be in equilibrium,, Fig. 1.53, force on charge at A due to —e charge at, , B + force on charge at A di =, ze g ue to + q charge at O =, , , , FIGURE 1.53, , , , , , , , Lee, Tee ES 2 0, 0 x 4meq (x/2)2, , 1, , 4, TE,, , or, , , , 2, , &__ 1 glex4, Peg 4neE, x2, e, , = =4q or qz=el4, , Scanned with CamScanner

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ELECTROSTATIC CHARGES AND FIELDS, , , , If charge at O is moved slightly towards A, it, Jd not return to O on its own and shall continue, a towards A. Hence equilibrium is unstable., jo, example [) Two point charges 4 1C and, 1 pC are separated by a distance of 2 m in air., ma the point on the line joining the charges at, which net electric field of the system is zero., ICBSE 2017(C)], Solution. Here, gq, = 4 WC =4 x 10°C, =1HC=10°C, d=2m, , FIGURE 1.54, , , , , , Let electric field of the system be zero at P,, where AP =r, Fig. 1.54., At P,E, =E,, | axl q, x1, , 4ne, 1? 4ne,(2-r)?, , , , 4 1 2 1, —e or ==—_, rr (2-r)? r 2-r, r=4-2r, 3r=4, r=43m, , ai SUPERPOSITION PRINCIPLE, , ee ne, Formula used, Fo = Fo + Font Fost, , Units used. Force is in newton, when charges are, in coulomb and distance is in metre,, , _ Example MH ten Positively charged particles are kept fixed on x-axis at points x = 10 cm,, 20 cm, 30 cm, ...., 100 cm. The first particle has, charge 1-0 x 10-8 C, second 8 x 10-8 C, third 27 x, 10 C, and so on. Tenth particle will have charge, 1000 x 10-8 C. Find the magnitude of electric force, acting on a 1C charge placed at the origin., , Solution. By superposition principle, force on, charge 1¢ placed at origin, Fo= Foy + Foy +... + Foro, , , , 1/47, =f 4 By +e], 4ne, i 5 is, 10x10 8x10-8 2710-8, 0.10)? (0-20) -30)2, =9x109 x1] O10" — 20)" (030), 1000x1078, Posse, (1-00)?, , =9x 10? x 10% [1+24+3+4.... +10}, =9x 103 x 55 = 4-95 x 105N, , (exampte fH] Find the magnitude of the, resultant force on a charge of 1 C held at P due, to two charges of + 2 x 10° C and - 10-8 C at A, and B respectively., , Given AP = 10 cm and BP =5 cm., , ZAPB = 90°, Fig. 1.55., , , , FIGURE 1.55, , , , , , , , , , Solution. Here, F = ?,, Charge at P,g=1uC=10°C, Charge at A, q,=+2x 10°C, Charge at B, q.=- 10°C, AP = 10 cm =0-1 m, BP=5 cm=0-05 m, ZAPB = 90°, Force at P due to q, charge at A,, , , , 1 nF, = | > along AP produced, i an <, AP? long AP produce, , _ 9x10? x2x10-8 x10, - (1)?, Force at P due to q charge at B,, 1, , 9249, r= —,;. along PB, 24g, BP?’ "8, , =18x 103N, , , , _ 9x10? x 108 x 10-6, , = -3, (G05)? =36 x 103N, , > >, As angle between F, and F, is 90°,, , Scanned with CamScanner

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1/48, , , = /pe4 2, -. Resultant force, F = F? + F?, = 418x107)? +. (36x 1079)2, , = 18 x 1079 x 2.236, =40x102N, , JExamplel§] Consider three charges 4},, 42» 73 each equal to q at the vertices of an, equilateral triangle of side J. What is the force on, a charge Q (with the same sign as q) placed at the, centroid of the triangle ?, NCERT Solved Example, , Sa See eee, Solution. As shown in Fig. 1.56, draw, AD 1 BC., , , , AD = AB cos 30° =, FIGURE 1.56, A(q=a), , , , , , , , , , ~. Force on Q at O due to charge g, =q at A, -_! Qq_ __3Qq, , 4ne, (1/ ¥3)2 4n€, 12’ along AO, Similarly, force on Q due to charge Q=qaB, , = 324, 2 4ne,/? along BO, , , , , , A, , and force on Q due to charge q, = q at C, 304, , ~ 2’ along CO, 4negl 8, , Angle between forces Fy and F3 = 120°, By parallelogram law, resultant of F, and, , Fy = 304 along OA, , 4ne,l?, , 3, , , , Pradeep’s Fundamental Physics (X11), , 3Qq__, , 3Qq ., 4negl? 4n€,, , +. Total force on Q = p, lexam pei) Consider the charges 4, 7 and, , -qplaced at the vertices of an oe hae, of each side /. What is the force on eac ?, , NCERT solved Example, —__—__—, Solution. As is clear from Fig. 1.57, force on, q=qatA, , , , FIGURE 1.57, ¢ (a3=-4), , , , , , , , , , where F = —44__, 4m Pe, , a ., and 7, = unit vector along BC, , Force on (q3 = q) at B,, > 3? U8 A, Fy = Fy + Fy, =Fr,, , A, where —_r, = unit vector along AC, , Force on q,=~q at C,, , — > >, Fy = Fit Fy, , = FP + F2 +2F F, cos60°] 4, =V3Fin, , a, where ” =unit vector along the direction, , bisecting ZBCA,, We can show that, , ~ >» @, F\+F,+F,=0, , Scanned with CamScanner