XII Physics Chapter 1 - Electric Charges and Fields PDF

CHAPTER 1 Column I Column II Glass Silk ELECTRIC CHARGES AND FIELDS Wool Amber, ebonite, plastic Ebonite Polythene...

CHAPTER 1 Column I Column II Glass Silk ELECTRIC CHARGES AND FIELDS Wool Amber, ebonite, plastic Ebonite Polythene 1. What is meant by electrostatics? Dry hair Comb Ans: Electrostatics is the branch of physics which deals with charges at 7. How does a body get charged? rest. Ans: A body gets charged by the transfer of electrons. The body which 2. Which are the three methods of loses electrons gets positively charged charging a body? and the body which gains electrons Ans: The three methods are: gets negatively charged. a) Rubbing (charging by friction) b) Conduction 8. Is the mass of a body affected by c) Induction charging? Ans: Yes. A positively charged body 3. What is the method to charge an loses electrons. Therefore, its mass insulator? decreases. A negatively charged body Ans: Rubbing gains electrons. So its mass increases. (Electron has a definite mass of 9.1× 4. What are the methods to charge a 10−31 Kg) conductor? Ans: Conduction and induction 9. Distinguish between conductors and insulators. Give examples for 5. What is frictional electricity? Give both. an example. Ans: The materials which allow Ans: The charge obtained by a body on electricity to pass through them easily rubbing with another body is called are called conductors. frictional electricity. Examples: Metals Example: When a glass rod is rubbed The materials which offer a high with silk, the glass rod gets positively resistance to the passage of electricity charged and silk gets negatively are called insulators. charged. Examples: Most of the non-metals like glass, porcelain, plastic, nylon, 6. Give some examples for wood are insulators. substances which get charge on rubbing. 10. Conductors cannot be charged Ans: The substances in column I when by rubbing but insulators can. Why? rubbed with substances in column II, acquire positive charge while substances in column II acquire negative charge. SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut 1 Ans: This is because, when some charge is transferred to a conductor, it readily gets distributed over the entire surface of the conductor. But if some charge is put on an insulator, it stays at the same place. 11. What is meant by charging by conduction? Ans: When a charged body is brought in to contact with an uncharged conductor, charge flows from the 13. What is the use of an charged body to the uncharged body. electroscope? Ans: It is a device used to detect the 12. What is meant by charging by charge on a body. induction? Ans: When a charged body is 14. Briefly explain the working of a brought near to an uncharged gold leaf electroscope. conductor (without touching), that end Ans: A gold leaf electroscope consists of the uncharged conductor which is of a vertical metal rod fixed in a box, near to the charged body gets with two thin gold leaves attached to oppositely charged and the farther end its end. is charged with the same type of charge. When a charged object touches the metal knob at the top of the rod, charge flows on to the leaves. Since both the leaves are charged by the same type of charge, they diverge due to electrostatic repulsion. The separation between the leaves gives a rough measure of the amount of charge. SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut 2  b) Conservation Electric Charge Charge can neither be created nor be destroyed but can be transferred from one body to another. OR 15. State whether the following The total charge of an isolated statement is true or false. system is always conserved. ‘During charging by induction, new c) Quantization of electric charges are created in the body’ charge Ans: False. During induction only a According to quantization of rearrangement of charges takes place. electric charge “in the universe, No new charges are created in the body. charge of any body is an integer multiple of e” 16. Repulsion is the sure test of electrification. Explain Q = ne, where n is an integer and e = 1.6  10-19 C Ans: A charged body can attract That is, charges like 1e, 2e, 3e, ------ another oppositely charged body as well as an uncharged body. But a are possible but a charge like 1.5e is charged body can repel only similar not possible. charged bodies. 19[P]. A polythene piece rubbed with 17. Can a body attract a similar charged body in any case? wool is found to have a negative Ans: Yes. If the charge on one body is charge of 3 x 10-7 C. much greater than the charge on the (a) Estimate the number of electrons other body, it can induce opposite charges on the other body. Then the transferred (from which to which?) attraction can dominate the repulsion. (b) Is there a transfer of mass from wool to polythene? 18. Explain the properties electric charges. Ans: The basic properties electric COULOMB’S LAW charges are: State Coulomb’s inverse square a) Additive property law in electrostatics. If a system contains ‘n’ Ans: Coulomb’s law states that charges q1, q2, q3, ------, qn, then the “the electrostatic force total charge of the system is q1 + q2 + between two stationary point q3 + ------- +qn. SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut 3 charges is directly proportional permittivity of the to the product of the medium to the magnitudes of the charges and permittivity of vacuum. inversely proportional to the square of the distance between 22. If the air medium between two them.” charges is replaced by water, what change you expect in the electrostatic force and why? Ans: Fair 1 q1q2 Fmed  F r 4 r 2 Є is called the permittivity of the The force decreases by  r times. medium. If the charges are placed in vacuum or 23. Write Coulomb’s law in vector air, є = є0, where є0 is the permittivity form. of vacuum or air. Ans: Case(i) When the force is 1 q1q 2 attractive Then, F= 4Πε 0 r 2 є0= 8.854 x 10-12 C2/Nm2 1 =9x109Nm2/C2 4  0 For any other medium є =є0єr ,where ε Force on q1 due to q2 εr = is called the relative 1 q1q2 ε0 F 12  r12 permittivity (or dielectric constant)of 4 0 r 2 the medium with respect to vacuum or Force on q2 due to q1 air. 1 q1q2 F 21  r 21 Fmed  1 q1q2 4 0 r 2 4 0 r r 2 Case(ii) When the force is repulsive Fair Fmed  r 21. Define dielectric constant of a Force on q1 due to q2 medium. 1 q1q2 F 12  r 21 4 0 r 2 Ans: Dielectric constant of a medium is the ratio of SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut 4 Force on q2 due to q1 double the above amount, and the 1 q1q2 F 21  r 21 distance between them is halved? 4 0 r 2 24. Define one coulomb. 27[P]. Suppose the spheres A and B in Ans: According to Coulomb’s law, the above problem have identical 1 q1q 2 F= sizes. A third sphere of the same size 4Πε 0 r 2 Put q1 = q2 = 1C and r = 1m, then but uncharged is bought in contact 11 with the first, then brought in contact F = 9  109 2  9  109 N 1 with the second, and finally removed “One coulomb is defined as that from both. What is the new force of charge which when placed in repulsion between A and B? free space at a distance of 1m 28[P]. Two point charges +16μC and with an equal and similar -9μC are placed 8cm apart in air. charge, will repel with a force You are asked to place a +10 μC charge at a third position such that of 9  109 N.” the net force on +10 μC charge is 25[P]. Four point charges qA=2µC, zero. Where will you place the qB=-5µC, qC=2µC and qD=-5µC are charge? Make necessary calculations. located at the corners of a square 29. State superposition principle. ABCD of side 10cm. What is the force Ans: If there are a number of on a charge of 1µC placed at the charges q1, q2, - - - -qn around centre of the square? a charge ‘q’, then according to 26[P]. (a) Two insulated charged super position principle “the total force acting on ‘q’ is the copper spheres A and B have their vector sum of the forces on ‘q’ centres separated by a distance of due to individual charges”. 50cm. What is the mutual force of Total force, F = F1 + F2 + - - - - +Fn electrostatic repulsion if the charge on 1 qq1 -7 Where F 1  r1 , each is 6.5 x 10 C? The radii of A 4 0 r12 and B are negligible compared to the F2  1 qq2 r2 4 0 r2 2 distance of separation. ------------------------------- (b) What is the force of 1 qqn Fn  rn repulsion if, each sphere is charged 4 0 rn 2 SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut 5 30. What is the use of superposition Ans: principle? Ans: It is used to find the force on a charge due to more than two charges. ELECTRIC FIELD 31. Define electric field. ‘V’is the voltage and ‘d’ is the distance Ans: It is the space around an electric charge, where an electrostatic force is 34. Electric field intensity at a point experienced by another charge. F is defined as E  lim qo q 32. Define electric field intensity at a point? Ans: Electric field intensity at a here what does imply? point is defined as the force Ans: Here q is the test charge which is experienced by unit positive to be placed at the point where the field charge placed at that point. is to be determined. means that this test charge must be very small, otherwise it will produce its own Let a small test charge ‘δq’ is field so that the field at that point will placed at P. Then the force experienced be changed. by this test charge is given by 35. Define an electric line of force or 1 q q electric field line. F= 4Πε 0 r2 Ans: It is defined as “the path Therefore, the force experienced by along which a unit positive F unit positive charge is E  lim charge would move if it is free q 0 q 1 q to do so.” E= 4Πε 0 r 2 36. Draw the electric field lines due 33. Write the equations for electric to (i) an isolated positive charge field intensity. (ii) an isolated negative charge (iii) an electric dipole (iv) Two positive charges Ans: SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut 6  electric field lines start or end at infinity. (ii) In a charge free region electric field lines have no break or they are continuous. (iii) Two electric field lines never intersect each other. (iv) In a uniform electric field, electric field lines are parallel. (v) The number density of electric field lines at a point gives the strength (intensity) of electric field at that point. (vi) The tangent drawn to the electric field line at a point gives the direction of the electric field at that point. (vii) Electric field lines do not form closed loops. 38. Two electric field lines never intersect. Why? Ans: If two electric field lines intersect at a point, then there will be two directions for electric field at that 37. Write the properties of electric point. But this is not possible. So two field lines. electric field lines never intersect each Ans: The properties of electric field other. lines are: (i) Electric field lines start at positive charge and end at negative charge. For an isolated single charge SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut 7 39. Figure below shows the electric 43. What is the SI unit of electric dipole moment? field lines for two point charges Ans: coulomb- meter (Cm) separated by a small distance. 44[P]. A system has two charges qA=2.5 x 10-7 C and qB=-2.5 x 10-7 C located at points A (0,0,-15cm) and B ( 0,0,+15cm), respectively. What are the total charge and electric dipole q1 (a) Determine the ratio moment of the system? q2 (b) What are the signs of q1 and q2? 45. Derive the expression for the torque acting on an electric dipole placed in a uniform electric field. ELECTRIC DIPOLE Ans: Consider an electric dipole of dipole moment p  q  2ap , placed in a 40. What is an electric dipole? uniform electric field. Ans: Two equal and opposite charges separated by a small distance is called an electric dipole. 41. Define electric dipole moment. Ans: Electric dipole moment is defined as the product of Because of the two equal and opposite forces acting at the two ends magnitude of one of the charges of the dipole, a torque is experienced and length of the dipole. by the dipole. So the dipole will rotate till it becomes parallel to the electric Electric dipole moment p  q  2a p field. Electric dipole moment is a vector quantity. 42. What is the direction of electric dipole moment? Ans: Electric dipole moment is directed from negative charge to the positive charge. SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut 8 Torque,    Force  r dis tan ce What happens when an electric  qE  BC dipole is placed in a non- uniform  qE  2a sin  electric field?  (q  2a)E sin  Ans: The dipole will have both rotational and translational motion.  pE sin  The rotational motion will stop, when In vector form,   p  E, The direction of the dipole becomes parallel to the this torque is given by right hand rule. electric field.  51[P]. An electric dipole with dipole What is the maximum value of moment 4 x 10-9 C m is aligned at 300 torque acting on the dipole? Ans: with the direction of a uniform When   90 electric field of magnitude 5x104NC-1. torque,   pE sin 90  pE  1  pE Calculate the magnitude of the torque This is the max imum value of torque. acting on the dipole. t what orientations is the 52. Derive an expression for the dipole placed in a uniform electric electric field at a point on the axial field in the (i) the stable line of an electric dipole equilibrium? Ans: (ii) unstable equilibrium? Ans: (i) stable equilibrium When   0 torque,   pE sin 0  pE  0  0 This is the orientation of stable equilibrium. Consider an electric dipole of dipole (ii) unstable equilibrium moment p  q  2ap.We have to find When   180 the electric field intensity at a point ‘P’ torque,   pE sin180  pE  0  0 on the axial line of the dipole distant ‘r’ This is the orientation of unstable equilibrium. from the midpoint of the dipole. Electric field at P due to the +q charge at A, 48. What is the total charge of an 1 q electric dipole? E  (-p)  40 (r  a)2 Ans: zero (q+ -q = 0) Similarly the electric field at P due to the –q charge at B, 49. What is the total force on an 1 q electric dipole placed in a uniform E  (p) electric field? 40 (r  a) 2 Ans: Zero ( qE + -qE =0) SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut 9 The resultant electric field at P, point ‘P’ on the equatorial line of the E = E+ + E- dipole distant ‘r’ from the midpoint of 1 q 1 q the dipole.  (-p)  (p) 40 (r  a) 2 40 (r  a) 2 Electric field at P due to the +q qp  1 1  charge at A,    2  1 q 40  (r  a) (r  a)  2 E  40 r  a 2 2 qp  (r  a) 2  (r  a) 2   Similarly the electric field at P due to 40  (r  a) 2 (r  a) 2  the –q charge at B, qp  (r 2  2ra  a 2 )  (r 2  2ra  a 2 )  1 q  E  40  [(r  a)(r  a)]2   40 r 2  a 2 qp  4ra  E+ can be split in to two components   2 2 2  40  (r  a )  E+cosθ and E+sinθ. Similarly E- can 1 2(q  2a)rp be split in to components E-cosθ and  E-sinθ. The E+sinθ and E-sinθ 40 (r 2  a 2 ) 2 components cancel each other being 1 2prp  equal and opposite. The E+cosθ and 40 (r  a 2 ) 2 2 E-cosθ components add together. If r 2 a 2 , a 2 can be neglected,then Resultant electric field at P is given 1 2p by, E= p 40 r 3 E=E+cosθ + E-cosθ 1 q 1 q E cos   cos  40 r  a 2 2 40 r  a 2 2 a From figure, cos   1 53. Derive an expression for the r2  a2 2 electric field at a point on the 1 q a E  2 equatorial line of an electric dipole. 40 r  a  r 2  a 2 1/ 2 2 2 Ans: 1 q  2a  40  r 2  a 2 3/ 2 1 p E 40  r 2  a 2 3/ 2 If r 2 a 2 ,a 2 can be neglected,then 1 p E 4 0 r 3 n vector form, 1 p E (p) ˆ Consider an electric dipole of 40 r 3 dipole moment p  q  2ap.We have to find the electric field intensity at a SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut 10 Compare the electric fields on flux of this field through a square of the axial and equatorial lines of an 10cm on a side whose plane is parallel electric dipole. Ans: (i) Eaxial = 2×Eequatorial to the yz plane? (b) What is the flux (ii) Both the fields are inversely through the same square if the normal proportional to r3 to its plane makes a 600 angle with the (iii) The direction of Eaxial is parallel to electric dipole moment and that of x-axis? Eequatorial is antiparallel to electric 58[P]. What is the net flux of the dipole moment. uniform electric field of above ELECTRIC FLUX problem through a cube of side 20 cm 55. Define electric flux. oriented so that its faces are parallel to Ans: Electric flux is defined as the the coordinate planes? total number of electric field lines 59. Define the three charge passing normally through a densities. surface. Ans: The three different charge Electric flux through small area ds is densities are: defined as, (i) Linear charge d  E  dS density(λ) It is the charge per unit q length.   SI unit is C/m (ii) Surface charge The total electric flux through the density(σ) surface S is given It is the charge per unit area by q    E  dS A If the surface S is a plane surface, then SI unit is C/m2 the total electric flux is (iii) Volume charge   E S density(ρ) =ES cos It is the charge per unit  q volume.    Give the SI unit of electric flux. V Ans: Nm2/C or Vm SI unit is C/m3 57[P]. Consider a uniform electric 60. A spherical conducting shell of field E=3x103 i N/C. (a) what is the inner radius r1 and outer radius r2 has SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut 11 a charge ‘Q’. A charge ‘q’ is placed at  E  dS the centre of the shell.   EdS cos 0 ( E dS ) (a) What is the surface charge density =  EdS on the (i) inner surface, (ii) outer  E  dS surface of the shell?  E  4r 2 At the surface of the sphere, (b) Write the expression for the 1 q E= electric field at a point x>r2 from the 4 0 r 2 1 q centre of the shell.   E  dS   4r 2 4 0 r 2 q  GAUSS’S THEOREM 0   What is a Gaussian surface? State Gauss’s theorem in Ans: An imaginary surface enclosing electrostatics. a charge is called a Gaussian surface. Ans: Gauss’s theorem states that “ the A Gaussian surface can be a surface of any shape. total electric flux over a closed surface enclosing a charge is 64. Figure shows three point equal to 1/є0 times the net charges,+2q, -q and +3q.Two charges charge enclosed.” Mathematically Gauss’s theorem can +2q and –q are enclosed in a surface  1 q ‘S’. What is the electric flux due to 0 be stated as  this configuration through the q   E  dS  0 surface ‘S’? Prove Gauss’s theorem. Ans: Consider a point charge q placed at a point. Imagine a sphere of radius r with q as the centre. The total electric flux through the sphere, 65[P]. A point charge +10µC is at a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. What is the magnitude of the electric flux through the SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut 12 square? (Hint: Think of the square as 69. Derive Coulomb’s law from Gauss’s theorem. one face of a cube with edge 10 cm) Ans: By Gauss’s theorem, q  E  dS   0 66[P]. A point charge of 2.0µC is at q the centre of a cubic Gaussian surface  EdScos 0   0 ( E dS ) q 9.0cm on edge. What is the net   EdS   0 electric flux through the surface? q  E  dS  0 67[P]. A point charge causes an q  E  4r 2  0 electric flux of -1.0 x 103 Nm2/C to 1 q pass through a spherical Gaussian  E= 4 0 r 2 surface of 10.0 cm radius centred on this is the electric field at the the charge. (a) If the radius of the surface of the sphere. If we place another ch arg e q on Gaussian surface were doubled, how the surface of the sphere, then force much flux would pass through the acting on it is  surface? 1 qq F This is Coulomb 's law. (b) What is the value of the point 40 r 2 charge?  70. By applying Gauss’s theorem deduce the expression for electric 68[P]. A uniformly charged field due to a spherical shell of conducting sphere of 2.4 m diameter charge (hollow sphere) density σ. Ans: Consider a shell of radius R and has a surface charge density of 80.0 charge density σ. We have to find the µC/m2. (a) Find the charge on the electric field at a point distant r from sphere. (b) What is the total electric the centre of this shell. For this we imagine a Gaussian sphere of radius r, flux leaving the surface of the sphere? concentric with the given shell of charge and passing through P. SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut 13 Case(i): E.F.Outside the shell         The total Electric flux through the In this case the charge enclosed by the Gaussian sphere, Gaussian sphere, q=0  E  dS  Substituting in Gauss 's theorem   EdS cos 0 ( E dS ) 0 E  4r 2  =  EdS 0  E  0.The electric field inside  E  dS a sperical shell of charge is zero.  E  4r 2  he charge enclosed by the Gaussian What is meant by electrostatic sphere, q=A  4R 2 shielding? Applying Gauss’s theorem, Ans: q  E  dS   0 Electric field inside the cavity of a conductor of any shape is zero. This is E  4r 2  4R 2  called electrostatic shielding. R 2 E  0 r 2 Case(ii): E.F. on the shell Put r=R R 2  E  0 R 2 0 72[Q]. Draw the variation of electric  E 0 field intensity of a shell of radius R Case(iii) E.F.inside the shell with distance r. 73. During lightning, a person inside a car is safer than outside. Why? Ans: This is due to electrostatic shielding. Due to electrostatic SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut 14   shielding, electric field inside the car is E dS zero. curved  surface  By applying Gauss’s theorem,  E  2r derive an expression for the electric he charge enclosed by the Gaussian field due to a straight infinitely long cylinder, q   charged wire (line charge) of charge Applying Gauss’s theorem, density λ. q Ans:  E  dS   0  Consider an infinitely long E  2r  0 straight wire of charge density λ. We  have to find the electric field at a point E P distant ‘r’ from this line charge. For 2 0 r this imagine a Gaussian cylinder of radius ‘r’ and length ‘l’ with the line charge as the axis. 74[P]. An infinite line charge produces a field of 9 x104 N/C at a distance of 2 cm. Calculate the linear charge density. 75. Applying Gauss’s theorem find an expression for the electric field he total electric flux, due to an infinitely large plane sheet of charge density ‘σ’  E  dS Ans: Consider an infinitely large plane sheet of charge density λ. We   E  dS   E  dS have to find the electric field at a point curved end surface faces P distant ‘r’ from this plane sheet of   cerved EdScos 0   end EdScos 90 charge. For this imagine a Gaussian cylinder of small area of cross section  surface faces with one end passing through the point   curved EdS   end EdS  0 P, penetrating the sheet and extending surface faces to both sides equally. SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut 15  Ans: Consider two infinitely large plane parallel sheets having charge densities +σ and –σ. The total electric flux through the Gaussian cylinder,  E  dS Electric field due to a single sheet, s  E   E  dS   E  dS 20 curved surface end faces In Region I E I  E  E  0   EdScos 90   EdScos 0 curved surface end faces In Region II E II  E  E   EdS  0   EdS  1 curved end     surface faces =  =2  20 20 20 0 E end  dS faces  E  2A In Region III The charge enclosed by the Gaussian EIII  E  (E)  0 cylinder, q  A Applying Gauss’s theorem, 77[P]. Two charge, thin metal plates q  E  dS   are parallel and close to each other. On their inner faces, the plates have 0 A E  2A  surface charge densities of opposite 0  signs and of magnitude 17.0 x 10-22 E 2 0 C/m2. What is E: (a) in the outer region of the first plate, (b) in the outer region 76. Find the expressions for the of the second plate, and (c) between electric field due to two infinitely large parallel plane sheets of equal the plates? and opposite charge densities. SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut 16