Electric charges and fields, Assignments of Physics

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PHYSICS >> Multiple Choice Questions (MCQs) \\\\ —— »>>> DIRECTIONS : This section contains multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which only one is correct. 1. Two small spheres each having the charge +O are suspended by insulating threads of length L from a hook. This arrangement is taken in space where there is no gravitational ettect, then the angle between the two suspensions and the tension is _ in each thread. 1 @ - 1 0 (a) 0,2) 90", — & 4né9 (21) 4neq L 2 2 a, 1 i - | OQ () 180°, (d) 180°, Ane, 24° 4m&, [ 2. Two equal point charges each of 3uC are separated by a certain distance in metres. If they are located at (i+ 7 +k) and (2 +3) + k). then the electrostatic force between them is (a) 9x 109N (b) 16x 103N (c) 10°7°N (d) 9x107°N 3. A bodyis positively charged, it implies that (a) there is only positive charge in the body (b) there is positive as well as negative charge in the body but the positive charge is more than negative charge (c) there is equal positive and negative charge in the body but the positive charge lies in the outer regions (d) negative charge is displaced from its position 4. On rubbing, when one body gets positively charged and other negatively charged, the electrons transferred from positively charged body to negatively charged body are (a) valence electrons only (b) electrons of inner shells (c) both valence electrons and electrons of inner shell (d) yet to be established 5. Three charges +g, +2g and +4g are connected by strings as shown in the figure. What is ratio of tensions in the strings AB and BC? Electric Charges and Fields A B Ce OS) +24 +4q +q (a) 1:2 (b) 1:3 (c) 2:1 (d) 3:1 Two charge g and —3q are placed fixed on x-axis separated by distance @. Where should a third charge 2q be placed such that it will not experience any force? q —3q A-———<$<—<—E, (b) Ey, Angle rotated 0—~> © @ | . T _ e. é - Angle rotated 0-—> Angle rotated 0—> The electric intensity due toa dipole of length 10 cm and having a charge of 500 uC, at a point on the axis ata distance 20 cm from one of the charges in air, is (a) 625*10’N/C (b) 9.28= 107 N/C (c) 13.1« 10! N/C (d) 20.5 10’N/C A rod of length 2.4 m and radius 4.6 mm carries a negative charge of 4.2 x 10-7 C spread uniformly over it surface. The electric field near the mid-point of the rod, at a point on its surface is (a) -8.6* 1ONC'! (b) 86x 10°NC! (c) -6.7x 1OONC! (d) 6.7 x 104*NC! The total electric flux emanating from a closed surface enclosing an o.-particle is (e-electronic charge) ze e Ege (a) (ob) —— (c) ef (d) £9 £0 4 A square surface of side meter in the plane of the paper is placed in a uniform electric field £ (volt/m) acting along the same plane at an angle 9 with the horizontal side of the square as shown in Figure. The electric flux linked to the surface, in units of volt. m, is E (a) EL (b) EL? cos (c) EL?sing (d) zero va The E-rcurve for an infinite linear charge distribution will be ; a (a) E b) F > > @ i «@ t | i od T+ 35. 36. a7 38. 39. 40. 41. 42. Figure shows electric field lines in which an electric dipole P is placed as shown. Which of the following statements is correct? —<—<—_ _ Pe —_s+p a _ (a) The dipole will not experience any force (b) Thedipole will experience a force towards right (c) Thedipole will experience a force towards lett (d) The dipole will experience a force upwards A point charge + q is placed at a distance d from an isolated conducting plane. The field ata point P on the other side of the plane is (a) directed perpendicular to the plane and away from the plane (b) directed perpendicular to the plane but towards the plane (c) directed radially away from the point charge (d) directed radially towards the point charge. Ahemisphere is uniformely charged positively. The electric field at a point on a diameter away from the centre is directed (a) perpendicular to the diameter (b) parallel to the diameter (c) atan angle tilted towards the diameter (d) atan angle tilted away from the diameter Among two dises A and B, first have radius 10 cm and charge 10-° 1C and second have radius 30 cm and charge 10°5C. When they are touched, charge on both q a and gq, respectively will, be (a) y= 2.75pC, gg=3.15 pC (b) qy=1.09nC, qp = 1.53 pC (C) a= dg =5-5ye (d) None of these Figure shows some of the electric field lines corresponding to an electric field. The figure suggests that (a) Ea Ep Eo ee (b) E, =Ep=E a E gf ae q (c) E,y=Eo>Eg NE The surface density on the copper sphere is c. The electric field strength on the surface of the sphere is (a) o (b) o/2 (c) o/2e, (dd) Q/é, When a body is charged by induction, then the body (a) becomes neutral (b) does not lose any charge (c) loses whole of the charge on it (d) loses part of the charge on it Quantisation of charge implies (a) charge cannot be destroyed (b) charge exists on particles (d) £,=D. mass of B (c) massofA m) (c) charged as well as uncharged particles (d) all the distances In annihilation process, in which an electron anda positron transform into two gamma rays, which property of electric charge is displayed? (a) Additivity of charge (b) Quantisation of charge (c) Conservation of charge (d) Attraction and repulsion Which of the following statements is incorrect? (a) The charge g on a body is always given by g = ne, where #7 is any integer, positive or negative. (b) By convention, the charge on an electron is taken to be negative. (c) The fact that electric charge is always an integral multiple of e is termed as quantisation of charge. (d) The quatisation of charge was experimentally demonstrated by Newton in 1912. Which of the following statements 1s incorrect? Study of charges, by scientists, concludes that (a) there are two kinds of electric charges. (b) bodies like plastic, fur acquire elecrtic charge on rubbing. (c) like charges attract, unlike charges repel each other. (d) the property which differentiates two kinds of charges is called the polarity of the charge. 51. 52; SS. 34. 55. 57. 58. 60. What happens when some charge is placed on a soap bubble? (a) Its radius decreases (b) Its radius increases (c) The bubble collapses (d) None of these Consider a neutral conducting sphere. A positive point charge is placed outside the sphere. The net charge on the sphere is then (a) negative and distributed uniformly over the surface of the sphere (b) negative and appears only at the point on the sphere closest to the point charge (c) negative and distributed non-uniformly over the entire surface of the sphere (d) zero A charged particle is free to move in an electric field. It will travel (a) always along a line of force (b) alonga line of force, if its initial velocity is zero (c) along a line of force, ifit has some initial velocity in the direction of an acute angle with the line of force (d) none of the above If one penetrates a uniformly charged spherical cloud, electric field strength (a) decreases directly as the distance from the centre (b) increases directly as the distance from the centre (c) remains constant (d) None of these Electric lines of force about a negative point charge are (a) circular anticlockwise (b) circular clockwise (c) radial, inwards (d) radial, outwards Electric lines of force (a) exist everywhere (b) exist only in the immediate vicinity of electric charges (c) exist only when both positive and negative charges are near one another (d) are imaginary A region surrounding a stationary electric dipoles has (a) magnetic field only (b) electric field only (c) both electric and magnetic fields (d) no electric and magnetic fields The electric field at a point on equatorial line of a dipole and direction of the dipole moment (a) will be parallel (b) will be in opposite direction (c) will be perpendicular (d) are not related An electric dipole is placed at an angle of 30° to a non-uniform electric field. The dipole will experience (a) atranslational force only in the direction of the field (b) atranslational force only in the direction normal to the direction of the field (c) atorque as well as a translational force (d) a torque only The spatial distribution of electric field due to charges (A, B) is shown in figure. Which one of the following statements 1s correct ? 61. 62. 64. 65. 66. 67. (a) Ais +ve and B-ve, |A| > |B) (b) Ais—ve and B +ve, |A| =|B| (c) Bothare+ve butA>B (d) Bothare —ve butA>B n the figure, charge q is placed at origin O. When the charge q is displaced from its position the electric field at point P changes (a) atthe same time when q is displaced. OP (b) atatime after a where c is the speed of light. OPcos8 a OP sin@ c Anelectric dipole is placed in a uniform electric field. The dipole will experience (a) aforce that will displace it in the direction of the field (b) a force that will displace it in a direction opposite to the field. (c) atorque which will rotate it without displacement (d) atorque which will rotate it and a force that will displace it On decreasing the distance between the two charges of a dipole which is perpendicular to electric field and decreasing the angle between the dipole and electric field, the torque on the dipole (a) increases (c) remains same (c) atatime after (d) atatime after (b) decreases (d) cannot be predicted. A sphere of radius R has uniform distribution of electric charge in its volume. At a distance x from its centre for x Assertion & Reason \\\ —— Find the net electric flux through the surface is (a) (b) (c) (d) +360Nm2/C Electric field at point A depends on (a) qy A (b) qo (c) both q, and q, (d) The surface density on the copper sphere is o. The electric field strength on the surface of the sphere is (a) o (b) 3/2 (c) of/2e, (d) o/s, A charge q is placed at the centre of the open end of a cylindrical vessel. The flux of the electric field through the surface of the vessel is (a) (b) (c) q/2e, (d) 2q/s, At the centre of a cubical box + Q charge is placed. The value of total flux that is coming out a wall is (a) Q/e, (6b) Q/3e, © Q/4e, (d) Q/ be, @ @ Cc — 670 Nm?2/C +670 Nm2/C Q —360Nm7/C charge. eb None of these Zero q q/e ta) »>>»> DIRECTIONS : Each of these questions contains an assertion followed by reason. Read them carefully and answer the question on the basis of following options. You have to select the one that best describes the two statements. (a) (b) (c) (d) 102. 103. Ifboth Assertion and Reason are correct and the Reason is the correct explanation of the Assertion. Ifboth Assertion and Reason are correct but Reason is not the correct explanation of the Assertion. Ifthe Assertion is correct but Reason is incorrect. Ifthe Assertion is incorrect but the Reason is correct. Assertion : When bodies are charged through friction, there is a transfer of electric charge from one body to another, but no creation or destruction of charge. Reason: This follows from conservation of electric charges. Assertion : Coulomb force and gravitational force follow the same inverse-square law. Reason : Both laws are same in all aspects. 104. 105. 106. 107. 108. 109. 110. 111. 112. Assertion : The property that the force with which two charges attract or repel each other are not affected by the presence ofa third charge. Reason : Force on any charge due to a number of other charge is the vector sum of all the forces on that charge due to other charges, taken one at a time. Assertion ; Consider two identical charges placed distance 2d apart, along x-axis. The equilibrium of a positive i test charge placed at the point -——>B O midway between them is Q O Q stable for displacements along K 7d >| the x-axis. Reason: Force on test charge is zero. Assertion ; If a proton and an electron are placed in the same uniform electric field. They experience different acceleration. Reason : Electric force on a test charge is dependent of its mass. Assertion : A metallic shield in form ofa hollow shell may be built to block an electric field. Reason : In a hollow spherical shield, the electric field inside it 1s zero at every point. Assertion : A point charge is brought in an electric field, the field at a nearby point will increase or decrease, depending on the nature of charge. Reason : The electric field is independent ofthe nature of charge. Assertion : Electric field is always normal to equipotential surfaces and along the direction of decreasing order of potential. Reason : Negative gradient of electric potential is electric field. [CBSE Sample 2021] Assertion : On disturbing an electric dipole in stable equillibrium in an electric field, it returns back to its stable equillibrium orientation. Reason : A restoring torque acts on the dipole on being disturbed from its stable equillibrium. Assertion : On going away from a point charge or a small electric dipole, electric field decreases at the same rate in both the cases. Reason : Electric field is inversely proportional to square of distance from the charge or an electric dipole. Assertion : Four point charges q,, q,. q3 and q, are as shown in figure. The flux over the shown Gaussian surface depends only on charges q, and qj. @q, Gaussian surface bade Reason : Electric field at all points on Gaussian surface depends only on charges q, and q). = saya ANSWER KEY & SOLUTIONS (a) In the absence of gravitational force, the only force acts on the spheres is electrostatic repulsion and so the angle between two suspension becomes 180°. So force between the sphere 1 @ 4m &y (2L)° F= (b) Here, 1 =i+j+k>p =21+3j+k f—b-F = (21+3)+k)— G+j+k) =1+2) 7 EV)? +2)? =V5 By Coulomb's law, pe! 9192 _ 9x10? x3x10° x3x10~° Angy yr? 5 y = x107 N, Nearest answer is 16 10-3 N, (b) When we say that a body is charged, we always mean that the body is having excess of electrons (negatively charged) or is of deficient of electrons (positively charged). (a) Valence electrons are outermost electrons these can get transferred on rubbing. Se ee d? (2 d)* d k- i 2q . kq: = _ Ske? Tre = 5 5 a (ay d? 2q q —3q 0) — cs d sf Let a charge 2g be placed at P, at a distance / from A where charge q 1s placed, as shown in figure. The charge 2g will not experience any force, when force, when force of repulsion on it due to g is balanced by force of attraction on it due to —-3g at B where AB =d (2qq) __ @q-39) Ane (7 4neg(f+dy (C+ dP =30 or 20-2d-P=0 pa 2dtvad* +20" _d , V3d 2 4 2° a d+3d a (c) Let be the number of electrons missing. 2 1 og _—"s - = q=\4negd7F F =ne Ane) Fd" ” 2 é Se 90 10. 11. 12. 14. 15. 16. @) (b) Force on charge q, due to q, is N92 Fo =k be Force on charge g, due to q, is Fy =k a’ The X - component of the force (F.) on q, is Fin + 3 sin 8 1142 4192 .. Poh +k sinO . h? ae ; 13 . aoe + Sin 0 ) a zg g . @ O,40=O ...(i) and F=k2= (ii) 3 k = From (i) and (ii) F = 100-0) ; dF O For F to be maximum dO, QO, = Q> 5 (a) Figure indicates the presence of some positive charge to the left of A. Ex > Ep (et, direction of dipole moment. B . (a) —eE=mg y ee. 31 pe — = -5.6x10EN/C 1.6x10— (d) Electric field lines do not form closed loop. This follows from the conservative nature of electric field. (b) Given: Dipole moment of the dipole= / and uniform electric field = £ . We know that dipole moment (p) = q.a (where q is the charge and a is dipole length). And when a dipole of dipole moment ? is placed in uniform electric field E , then Torque (t) = Either force x perpendicular distance between the two forces = ga E sin @ or t = pE sin 8 or 7=pe E (vector form) 17. 18. 19. 20. 21. 22; 23. 24. 2kp kp _ (b) We have EF = Ks and E, “3° ~ Ba 2b, (a) (a) Given: Length ofthe dipole (2/)= 10cm = 0.1m or /= 0.05m Charge on the dipole (q) = 500 uC = 500 x 10°® C and distance of the point on the axis from the mid-point of the dipole (r) =20+ 5 =25 cm=0.25 m. We know that the electric field intensity due to dipole on the given point (E) ] , 2(q.2/)r 7 ATE 9 (r? es 2(500x10-® x 0.1) x 0.25 [(0.25)? -(0.05)" |? ~9x10" x = 6.25x10' N/C (k=1 forair) (c) Here, 6 =2.4 m, r=4.6 mm = 4.6 « 103 m g=—42 £10-7C Linear charge density, A = — -4.2x10 7 = 1.75 x10-’ pA 24 a Electric field, E = 2TSgr 7 -1.75«107' 2x 3.14x8.854x 107? x4.6x107 =-6.7 x 10°N C! (a) According to Gauss’s law total electric flux through a es bo. _— closed surface is — times the total charge inside that € surface. q Electric flux, 6, = = 0 Charge on o-particle=2e .. 0,.= — (d) Electric flux, 6= £A cos 9, where 6 = angle between £ and normal to the surface. Here 0-5 = =0 (c) The field due to infinite linear charge distribution 1 dq l E= => = nso! r Boe So curve is hyperbolic. (d) According to Gauss’ Law fE. fy = Qenclosed by closed surface Fay Go so total flux = Q/e, Since cube has six face, so flux coming out through one wall or one face is Q/6€,. 25. 26. pas 28. 29. 30. 31. 32, (a) Gaussian surface cannot pass through any discrete charge because electric field due to a system of discrete charges is not well defined at the location of the charges. But the Gaussian surface can pass through a continuous charge distribution. (d) Here, E=274+3j+k NC! $=10i m? Electric flux, = E.S =(21+3j+kKNC7!).(10i m’) =20Nm2c~! (d) Since electric field Ff decreases inside water, therefore flux 6= E+ A also decreases. (a) The flux is zero according to Gauss’ Law because it is a open surface which enclosed a charge q. (d) (a) According to Gauss’s theorem 2X So, net charge enclosed by the surface is zero if the net electric flux through a closed surface is zero. (a) The force on q, depend on the force acting between q, and q, and q, and q, so that the net force acting on q, by q, and q, by q, is along the + x-direction, so the force acting between q,, qy and q,, q; is attractive force as shown in figure: aA +, q,< > x +q, ye The attractive force between these charges states that q, is a negative charge (since, q, and q, are positive). Then the force acting between q, and charge Q (positive) is also know as attractive force and then the net force on q, by q,, q3 and Q arealong the same direction as shown in the figure. The figure shows that the force on q, shall increase along the positive x-axis due to the positive charge Q. (a) Ifa positive point charge is brought near an isolated conducting sphere without touching the sphere, then the free electrons in the sphere are attracted towards the positive Charge and electric field passes through a charged body. This leaves an excess of positive charge on the (right) surface of sphere due to the induction process. 54. 58. 60. 61. 62. 64. 65. 66. 69. 70. 71. 72. 73. 74, (a) 55. (c) 56. (d) 57. (b) (b) The direction of electric field at equatorial point A or B will be in opposite direction, as that of direction of dipole moment. — eo (c) The electric field will be different at the location of force on the two charges. Therefore the two charges will be unequal. This will result in a force as well as torque. (a) Since lines of force starts from A and ends at B, soA is +ve and B is —ve. Lines of forces are more crowded near A, soA>B. (b) The electric field around a charge propagates with the speed of light away from the charge. Therefore the distance _ OP speed G required time = (c) When a dipole is placed in a uniform electric field, two equal and opposite forces act on it. Therefore, a torque acts which rotates the dipole. (b) Since t= p£ sin 6 on decreasing the distance between the two charges, and on decreasing angle @ between the dipole and electric field, sin 8 decreases therefore torque decreases. (c) Since x < R, that is the point is inside the sphere electric field Eocx.as E= Lo R? (d) Electric displacement vector, D =cE —- As, €=e)K . D =e9KE (a) 67. (c) 68. (d) (a) Gaussian surface cannot pass through any discrete charge because electric field due to a system of discrete charges is not well defined at the location of the charges. But the Gaussian surface can pass through a continuous charge distribution. (b) (c) pEdd =(), represents charge inside close surtace is zero. Electric field as any point on the surface may be zero. (b) Total flux coming out from unit charge (a) Atthe interior point of a hollow sphere, the electric field is zero. (a) By Gauss theorem Total charge inside cube gq Total electric flux = aes = Sy 73. 76. 77. 78. 79. 80. $1. 82. 83. $4. 85. 86. 88. 89. 90. 91. 92. 93. 94. (b) (c) Ifelectric dipole, the flux coming out from positive charge is equal to the flux coming in at negative charge i.e. total charge on sphere= 0. From Gauss law, total flux passing through the sphere = 0. (c) Electric field near the conductor surface is given by Go wa) 3 — and it is perpendicular to surface. e) (d) (a) Here, g=1C, &)=8.85 « 10-1? CN! m2 Number of lines of force = Electric force q et Ey 885x107 = 1.13 x 10!! (c) (a) Flux(@) = E- AA=EAA cos 60° = EV? /2 (b) Charging by induction involves transfer of charges from one part to the other of the body. No loss of charge is involved. (c) Oncharging by conduction, body may gain mass, if it acquires negative charge. It may lose mass, ifit acquires positive charge. (c) Positive charge is due to deficiency of electrons. (c) Net charge acquired by induction is zero, as there is only transfer of electrons from one part of body to the other. (a) 87. (b) i (b) The direction of electric field at equatorial point A or B will be in Opposite direction, as that of direction of dipole moment. (c) The electric field will be different at the location of force on the two charges. Therefore the two charges will be unequal. This will result in a force as well as torque. (c) Intensity of electric field due to a Dipole 2 ] Aneyr? V3cos70+1 > “ “73 (b) Since t=p£ sin 6 on decreasing the distance between the two charges, and on decreasing angle 9 between the dipole and electric field, sin @ decreases therefore torque decreases. E= (c) > E.d4=0, represents charge inside close surface is zero. Electric field as any point on the surface may be zero. (c) =E(ds) cos@ = E(2nr’)cos0°=2nr E. (a) For the curved surface, 0 = 90° “. O=E ds cos90°=0. 95, 96. 97. 98. 99. 100. 101. 102. 103. 104. 105. 106. 107. 108. 109. 110. 111. (d) (c) Flux does not depend on the size and shape of the close surface, and so, it remains same. Zq_ -5.9x10” @) g=—t- = & 8.85xl0 * (c) Electric field at any point depends on all the charges. (d) According to Gauss's theorem, Ef ds =| Here fas = 4nR? | =— 670 Nm?2/C & /4nR? | Bat. fe g/4nR? =6] & or E=0/€, (a) The flux is zero according to Gauss’ Law because it 1s a open surface which enclosed a charge q. (d) According to Gauss’ Law = flux f Eds = Q enclosed by closed surface Eo so total flux = Q/e, Since cube has six face, so flux coming out through one wall or one face is Q/6é,. (a) Conservation of electric charge states that the total charge of an isolated system remains unchanged with time. (c) Coulomb force and gravitational force follow the same inverse-square law. But gravitational force is always attractive force, while coulomb force can be of both force attractive and repulsive. (b) The individual force are unaffected due to presence of other charges. This is the principal of superposition of charges. 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. (b) [f+vecharge is displaced along x-axis, then net force will always act in a direction opposite to that of displacement and the test charge will always come back to its original position. (c) Electron and proton have same amount of charge so they have same coulomb force. They have different acceleration because they have different masses. (a) The electrostatic shielding is possible by metallic conductor. (c) The electric field will increase if positive charge is brought in an electric field. (b) b=-— (a) The restoring torque brings it back to its stable equilibrium. (d) The rate of decrease of electric field is different in the two cases. In case of a point charge, it decreases as 1/r2 but in the case of electric dipole it decreases more rapidly, as BE x 1/77. 112. 113. 114. 115. 116. 118. 119. 120. 121. 122. 123. 124. 125. 126. (d) Electric field at any point depends on presence ofall charges. 2 (b) “=A [Letr, andr, be two different radii] q2 1 2 2 go, Fi 480%? _. ad 2 Ey, 4megrf = q2 yj soE, =E, (a) The electric field due to disc is superposition of electric field due to its constituent ring as given in Reason. (b) Though the net charge on the conductor is still zero but due to induction negatively charged region is nearer to the rod as compared to the positively charged region. That is why the conductor gets attracted towards the rod. (a) 117. (b) _ eA 12 12 = ~6x1l0 (62 10") B" /éxtoe" (90°) (3 x 10° x ¢)) By Gauss law, we know that > =— Here, Net electric flux, 6=4,-6, £9 2 q | =9 165-62 10a" - 3 q=36 105 a, =O 0 (680) Given, Total surface charge carried by earth q=? According to Gauss’s law. o=1=BA = = 8.85 x 10-1 x 150 x (6.37 x 10®)% = 680 Ke As electric field directed inward hence q =— 680 Ke or, q= €gEA= Ee Ear’. (100) Here, E must be perpendicular to Y-Z plane, i.e., area must be parallel to X-plane, so ds = 207 units . electric flux= E.ds = (57 + 47 + 9k).(207) = 100 units (False) No. of lines entering the surface = No. of lines leaving the surface. (True) (True) The individual force are unaffected due to the presence of other charges. ( “ True) 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.