{"id":118348,"date":"2022-04-25T18:02:24","date_gmt":"2022-04-25T12:32:24","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=118348"},"modified":"2022-04-25T18:02:24","modified_gmt":"2022-04-25T12:32:24","slug":"chapter-1-electric-charges-and-fields-questions-and-answers-ncert-solutions-for-class-12-physics","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-1-electric-charges-and-fields-questions-and-answers-ncert-solutions-for-class-12-physics","title":{"rendered":"Chapter 1 – Electric Charges And Fields Questions and Answers: NCERT Solutions for Class 12 Physics"},"content":{"rendered":"

1. What is the force between two small charged spheres of charges
\n2\u00d710\u22127C2\u00d710\u22127C and 3\u00d710\u22127C3\u00d710\u22127C placed 30cm30cm apart in air?<\/h2>\n

Ans: We are given the following information:
\nRepulsive force of magnitude,F=6\u00d710\u22123NF=6\u00d710\u22123N
\nCharge on the first sphere, q1=2\u00d710\u22127Cq1=2\u00d710\u22127C
\nCharge on the second sphere, q2=3\u00d710\u22127Cq2=3\u00d710\u22127C
\nDistance between the two spheres, r=30cm=0.3mr=30cm=0.3m
\nElectrostatic force between the two spheres is given by Coulomb\u2019s law as,
\nF=14\u03c0\u03b50q1q2r2F=14\u03c0\u03b50q1q2r2
\nWhere, \u03b50\u03b50is the permittivity of free space and,
\n14\u03c0\u03b50=9\u00d7109Nm2C\u2212214\u03c0\u03b50=9\u00d7109Nm2C\u22122
\nNow on substituting the given values, Coulomb\u2019s law becomes,
\n$\\begin{align}
\n& F=\\frac{9\\times {{10}^{9}}\\times 2\\times {{10}^{-7}}\\times 3\\times {{10}^{-7}}}{{{\\left( 0.3 \\right)}^{2}}} \\\\
\n& \\therefore F=6\\times {{10}^{-3}}N \\\\
\n\\end{align}$
\nTherefore, we found the electrostatic force between the given charged spheres to be F=6\u00d710\u22123NF=6\u00d710\u22123N. Since the charges are of the same nature, we could say that the force is repulsive.<\/h3>\n

2. The electrostatic force on a small sphere of charge 0.4\u03bcC0.4\u03bcC due to another small sphere of charge \u22120.8\u03bcC\u22120.8\u03bcC in air is 0.2N.
\na) What is the distance between the two spheres?<\/h2>\n

Ans: Electrostatic force on the first sphere is given to be, F=0.2NF=0.2N
\nCharge of the first sphere is, q1=0.4\u03bcC=0.4\u00d710\u22126Cq1=0.4\u03bcC=0.4\u00d710\u22126C
\nCharge of the second sphere is, q2=\u22120.8\u03bcC=\u22120.8\u00d710\u22126Cq2=\u22120.8\u03bcC=\u22120.8\u00d710\u22126C
\nWe have the electrostatic force given by Coulomb\u2019s law as,
\nF=14\u03c0\u03b50q1q2r2F=14\u03c0\u03b50q1q2r2
\n\u21d2r=q1q24\u03c0\u03b50F\u2212\u2212\u2212\u2212\u221a\u21d2r=q1q24\u03c0\u03b50F
\nSubstituting the given values in the above equation, we get,
\n\u21d2r=0.4\u00d710\u22126\u00d78\u00d710\u22126\u00d79\u00d71090.2\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221a\u21d2r=0.4\u00d710\u22126\u00d78\u00d710\u22126\u00d79\u00d71090.2
\n\u21d2r=144\u00d710\u22124\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221a\u21d2r=144\u00d710\u22124
\n\u2234r=0.12m\u2234r=0.12m
\nTherefore, we found the distance between charged spheres to be r=0.12mr=0.12m.<\/h3>\n

b) What is the force on the second sphere due to the first?<\/h2>\n

Ans: From Newton\u2019s third law of motion, we know that every action has an equal and opposite reaction.
\nThus, we could say that the given two spheres would attract each other with the same force.
\nSo, the force on the second sphere due to the first sphere will be 0.2N0.2N.<\/h3>\n

3. Check whether the ratio ke2Gmempke2Gmempis dimensionless. Look up a table of physical constants and hence determine the value of the given ratio. What does the ratio signify?<\/h2>\n

Ans: We are given the ratio, ke2Gmempke2Gmemp.
\nHere, G is the gravitational constant which has its unit Nm2kg\u22122Nm2kg\u22122;
\nmemeand mpmp are the masses of electron and proton in kgkg respectively;
\nee is the electric charge in CC;
\nkk is a constant given by k=14\u03c0\u03b50k=14\u03c0\u03b50
\nIn the expression for k, \u03b50\u03b50 is the permittivity of free space which has its unit Nm2C\u22122Nm2C\u22122.
\nNow, we could find the dimension of the given ratio by considering their units as follows:
\nke2Gmemp=[Nm2C\u22122][C]2[Nm2kg\u22122][kg][kg]=M0L0T0ke2Gmemp=[Nm2C\u22122][C]2[Nm2kg\u22122][kg][kg]=M0L0T0
\nClearly, it is understood that the given ratio is dimensionless.
\nNow, we know the values for the given physical quantities as,
\ne=1.6\u00d710\u221219Ce=1.6\u00d710\u221219C
\nG=6.67\u00d710\u221211Nm2kg\u22122G=6.67\u00d710\u221211Nm2kg\u22122
\nme=9.1\u00d710\u221231kgme=9.1\u00d710\u221231kg
\nmp=1.66\u00d710\u221227kgmp=1.66\u00d710\u221227kg
\nSubstituting these values into the required ratio, we get,
\nke2Gmemp=9\u00d7109\u00d7(1.6\u00d710\u221219)26.67\u00d710\u221211\u00d79.1\u00d710\u22123\u00d71.67\u00d710\u221222ke2Gmemp=9\u00d7109\u00d7(1.6\u00d710\u221219)26.67\u00d710\u221211\u00d79.1\u00d710\u22123\u00d71.67\u00d710\u221222
\n\u21d2ke2Gmemp\u22482.3\u00d71039\u21d2ke2Gmemp\u22482.3\u00d71039
\nWe could infer that the given ratio is the ratio of electrical force to the gravitational force between a proton and an electron when the distance between them is kept constant.<\/h3>\n

4.
\na) Explain the meaning of the statements \u2018electric charge of a body is quantized\u2019.<\/h2>\n

Ans: The given statement \u2018Electric charge of a body is quantized\u2019 means that only the integral number (1,2,3,…,n)(1,2,3,…,n) of electrons can be transferred from one body to another.
\nThat is, charges cannot be transferred from one body to another in fraction.
\nb) Why can one ignore quantization of electric charge when dealing with macroscopic i.e., large scale charges?
\nAns: On a macroscopic scale or large-scale, the number of charges is as large as the magnitude of an electric charge.
\nSo, quantization is considered insignificant at a macroscopic scale for an electric charge and electric charges are considered continuous.<\/h3>\n

5. When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.<\/h2>\n

Ans: Rubbing two objects would produce charges that are equal in magnitude and opposite in nature on the two bodies.
\nThis happens due to the reason that charges are created in pairs. This phenomenon is called charging by friction.
\nThe net charge of the system however remains zero as the opposite charges equal in magnitude annihilate each other.
\nSo, rubbing a glass rod with a silk cloth creates opposite charges of equal magnitude on both of them and this observation is found to be consistent with the law of conservation of charge.<\/h3>\n

6. Four point charges qA=2\u03bcCqA=2\u03bcC, qB=\u22125\u03bcCqB=\u22125\u03bcC, qC=2\u03bcCqC=2\u03bcCand qD=\u22125\u03bcCqD=\u22125\u03bcC are located at the corners of a square ABCD with side 10cm. What is the force on the 1\u03bcC1\u03bcC charge placed at the centre of this square?<\/h2>\n

Ans: Consider the square of side length 10cm10cm given below with four charges at its corners and let O be its centre.
\nFrom the figure we find the diagonals to be,
\nAC=BD=102\u2013\u221acmAC=BD=102cm
\n\u21d2AO=OC=DO=OB=52\u2013\u221acm\u21d2AO=OC=DO=OB=52cm
\nNow the repulsive force at O due to charge at A,
\nFAO=kqAqOOA2=k(+2\u03bcC)(1\u03bcC)(52\u221a)2FAO=kqAqOOA2=k(+2\u03bcC)(1\u03bcC)(52)2\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (1)
\nAnd the repulsive force at O due to charge at D,
\nFDO=kqDqOOD2=k(+2\u03bcC)(1\u03bcC)(52\u221a)2FDO=kqDqOOD2=k(+2\u03bcC)(1\u03bcC)(52)2\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (2)
\nAnd the attractive force at O due to charge at B,
\nFBO=kqBqOOB2=k(\u22125\u03bcC)(1\u03bcC)(52\u221a)2FBO=kqBqOOB2=k(\u22125\u03bcC)(1\u03bcC)(52)2\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (3)
\nAnd the attractive force at O due to charge at C,
\nFCO=kqCqOOC2=k(\u22125\u03bcC)(1\u03bcC)(52\u221a)2FCO=kqCqOOC2=k(\u22125\u03bcC)(1\u03bcC)(52)2\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (4)
\nWe find that (1) and (2) are of same magnitude but they act in the opposite direction and hence they cancel out each other.
\nSimilarly, (3) and (4) are of the same magnitude but in the opposite direction and hence they cancel out each other too.
\nHence, the net force on charge at centre O is found to be zero.<\/h3>\n

7.
\na) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?<\/h2>\n

Ans: An electrostatic field line is a continuous curve as the charge experiences a continuous force on being placed in an electric field.
\nAs the charge doesn\u2019t jump from one point to the other, field lines will not have sudden breaks.<\/h3>\n

b) Explain why two field lines never cross each other at any point?<\/h2>\n

Ans: If two field lines are seen to cross each other at a point, it would imply that the electric field intensity has two different directions at that point, as two different tangents (representing the direction of electric field intensity at that point) can be drawn at the point of intersection.
\nThis is however impossible and thus, two field lines never cross each other.<\/h3>\n

8. Two point charges qA=3\u03bcCqA=3\u03bcCand qB=\u22123\u03bcCqB=\u22123\u03bcCare located 20cm apart in vacuum.
\na) What is the electric field at the midpoint O of the line AB joining the two charges?<\/h2>\n

Ans: The situation could be represented in the following figure. Let O be the midpoint of line AB.
\n(Image Will Be Updated Soon)
\nWe are given:
\nAB=20cmAB=20cm
\nAO=OB=10cmAO=OB=10cm
\nTake E to be the electric field at point O, then,
\nThe electric field at point O due to charge +3\u03bcC+3\u03bcCwould be,
\nE1=3\u00d710\u221264\u03c0\u03b50(AO)2=3\u00d710\u221264\u03c0\u03b50(10\u00d710\u22122)2NC\u22121E1=3\u00d710\u221264\u03c0\u03b50(AO)2=3\u00d710\u221264\u03c0\u03b50(10\u00d710\u22122)2NC\u22121along OB
\nThe electric field at point O due to charge \u22123\u03bcC\u22123\u03bcCwould be,
\nE2=\u2223\u22233\u00d710\u221264\u03c0\u03b50(OB)2\u2223\u2223=3\u00d710\u221264\u03c0\u03b50(10\u00d710\u22122)2NC\u22121E2=|3\u00d710\u221264\u03c0\u03b50(OB)2|=3\u00d710\u221264\u03c0\u03b50(10\u00d710\u22122)2NC\u22121along OB
\nThe net electric field,
\n\u21d2E=E1+E2\u21d2E=E1+E2
\n\u21d2E=2\u00d79\u00d7109\u00d73\u00d710\u22126(10\u00d710\u22122)2\u21d2E=2\u00d79\u00d7109\u00d73\u00d710\u22126(10\u00d710\u22122)2
\n\u21d2E=5.4\u00d7106NC\u22121\u21d2E=5.4\u00d7106NC\u22121
\nTherefore, the electric field at mid-point O is E=5.4\u00d7106NC\u22121E=5.4\u00d7106NC\u22121 along OB.<\/h3>\n

b) If a negative test charge of magnitude 1.5\u00d710\u221219C1.5\u00d710\u221219C is placed at this point, what is the force experienced by the test charge?<\/h2>\n

Ans: We have a test charge of magnitude 1.5\u00d710\u22129C1.5\u00d710\u22129C placed at mid-point O and we found the electric field at this point to be E=5.4\u00d7106NC\u22121E=5.4\u00d7106NC\u22121.
\nSo, the force experienced by the test charge would be F,
\n\u21d2F=qE\u21d2F=qE
\n\u21d2F=1.5\u00d710\u22129\u00d75.4\u00d7106\u21d2F=1.5\u00d710\u22129\u00d75.4\u00d7106
\n\u21d2F=8.1\u00d710\u22123N\u21d2F=8.1\u00d710\u22123N
\nThis force will be directed along OA since like charges repel and unlike charges attract.<\/h3>\n

9. A system has two charges qA=2.5\u00d710\u22127CqA=2.5\u00d710\u22127Cand qB=\u22122.5\u00d710\u22127CqB=\u22122.5\u00d710\u22127Clocated at points A:(0,0,\u221215cm)A:(0,0,\u221215cm) and B:(0,0,+15cm)B:(0,0,+15cm) respectively. What are the total charge and electric dipole moment of the system?<\/h2>\n

Ans: The figure given below represents the system mentioned in the question:
\nThe charge at point A, qA=2.5\u00d710\u22127CqA=2.5\u00d710\u22127C
\nThe charge at point B, qB=\u22122.5\u00d710\u22127CqB=\u22122.5\u00d710\u22127C
\nThen, the net charge would be, q=qA+qB=2.5\u00d710\u22127C\u22122.5\u00d710\u22127C=0q=qA+qB=2.5\u00d710\u22127C\u22122.5\u00d710\u22127C=0
\nThe distance between two charges at A and B would be,
\nd=15+15=30cmd=15+15=30cm
\nd=0.3md=0.3m
\nThe electric dipole moment of the system could be given by,
\nP=qA\u00d7d=qB\u00d7dP=qA\u00d7d=qB\u00d7d
\n\u21d2P=2.5\u00d710\u22127\u00d70.3\u21d2P=2.5\u00d710\u22127\u00d70.3
\n\u2234P=7.5\u00d710\u22128Cm\u2234P=7.5\u00d710\u22128Cm along the
\n+z+z
\naxis.
\nTherefore, the electric dipole moment of the system is found to be 7.5\u00d710\u22128Cm7.5\u00d710\u22128Cm and it is directed along the positive
\nzz
\n-axis.<\/h3>\n

10. An electric dipole with dipole moment 4\u00d710\u22129Cm4\u00d710\u22129Cm is aligned at 30\u221830\u2218 with direction of a uniform electric field of magnitude 5\u00d7104NC\u221215\u00d7104NC\u22121. Calculate the magnitude of the torque acting on the dipole.<\/h2>\n

Ans: We are given the following:
\nElectric dipole moment, p\u2192=4\u00d710\u22129Cmp\u2192=4\u00d710\u22129Cm
\nAngle made by p\u2192p\u2192 with uniform electric field, \u03b8=30\u2218\u03b8=30\u2218
\nElectric field, E\u2192=5\u00d7104NC\u22121E\u2192=5\u00d7104NC\u22121
\nTorque acting on the dipole is given by
\n\u03c4=pEsin\u03b8\u03c4=pEsin\u2061\u03b8
\nSubstituting the given values we get,
\n\u21d2\u03c4=4\u00d710\u22129\u00d75\u00d7104\u00d7sin30\u2218\u21d2\u03c4=4\u00d710\u22129\u00d75\u00d7104\u00d7sin\u206130\u2218
\n\u21d2\u03c4=20\u00d710\u22125\u00d712\u21d2\u03c4=20\u00d710\u22125\u00d712
\n\u2234\u03c4=10\u22124Nm\u2234\u03c4=10\u22124Nm
\nThus, the magnitude of the torque acting on the dipole is found to be 10\u22124Nm10\u22124Nm.<\/h3>\n

11. A polythene piece rubbed with wool is found to have a negative charge of 3\u00d710\u22127C3\u00d710\u22127C
\na) Estimate the number of electrons transferred (from which to which?)<\/h2>\n

Ans: When polythene is rubbed against wool, a certain number of electrons get transferred from wool to polythene.
\nAs a result of which wool becomes positively charged on losing electrons and polythene becomes negatively charged on gaining them.
\nWe are given:
\nCharge on the polythene piece, q=\u22123\u00d710\u22127Cq=\u22123\u00d710\u22127C
\nCharge of an electron, e=\u22121.6\u00d710\u221219Ce=\u22121.6\u00d710\u221219C
\nLet n be the number of electrons transferred from wool to polythene, then, from the property of quantization we have,
\nq=neq=ne
\n\u21d2n=qe\u21d2n=qe
\nNow, on substituting the given values, we get,
\n\u21d2n=\u22123\u00d710\u22127\u22121.6\u00d710\u221219\u21d2n=\u22123\u00d710\u22127\u22121.6\u00d710\u221219
\n\u2234n=1.87\u00d71012\u2234n=1.87\u00d71012
\nTherefore, the number of electrons transferred from wool to polythene would be1.87\u00d710121.87\u00d71012.<\/h3>\n

b) Is there a transfer of mass from wool to polythene?<\/h2>\n

Ans: Yes, during the transfer of electrons from wool to polythene, along with charge, mass is transferred too.
\nLet mm be the mass being transferred in the given case and meme be the mass of the electron, then,
\nm=me\u00d7nm=me\u00d7n
\n\u21d2m=9.1\u00d710\u221231\u00d71.85\u00d71012\u21d2m=9.1\u00d710\u221231\u00d71.85\u00d71012
\n\u2234m=1.706\u00d710\u221218kg\u2234m=1.706\u00d710\u221218kg
\nThus, we found that a negligible amount of mass does get transferred from wool to polythene.<\/h3>\n

12.
\na) Two insulated charged copper spheres AA and BB have their centres separated by a distance of 50cm50cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5\u00d710\u22127C6.5\u00d710\u22127C? The radii of A and B are negligible compared to the distance of separation.<\/h2>\n

Ans: We are given:
\nCharges on spheres AA and BB are equal,
\nqA=qB=6.5\u00d710\u22127CqA=qB=6.5\u00d710\u22127C
\nDistance between the centres of the spheres is given as,
\nr=50cm=0.5mr=50cm=0.5m
\nIt is known that the force of repulsion between the two spheres would be given by Coulomb\u2019s law as,
\nF=qAqB4\u03c0\u03b50r2F=qAqB4\u03c0\u03b50r2
\nWhere, \u03b5o\u03b5o is the permittivity of the free space
\nSubstituting the known values into the above expression, we get,
\nF=9\u00d7109\u00d7(6.5\u00d710\u22127)2(0.5)2=1.52\u00d710\u22122NF=9\u00d7109\u00d7(6.5\u00d710\u22127)2(0.5)2=1.52\u00d710\u22122N
\nThus, the mutual force of electrostatic repulsion between the two spheres is found to beF=1.52\u00d710\u22122NF=1.52\u00d710\u22122N.<\/h3>\n

b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?<\/h2>\n

Ans: It is told that the charges on both the spheres are doubled and the distance between the centres of the spheres is halved. That is,
\nqA\u2032=qB\u2032=2\u00d76.5\u00d710\u22127=13\u00d710\u22127CqA\u2032=qB\u2032=2\u00d76.5\u00d710\u22127=13\u00d710\u22127C
\nr\u2032=12(0.5)=0.25mr\u2032=12(0.5)=0.25m
\nNow, we could substitute these values in Coulomb\u2019s law to get,
\nF\u2032=qA\u2032qB\u20324\u03c0\u03b50r\u20322F\u2032=qA\u2032qB\u20324\u03c0\u03b50r\u20322
\n\u21d2F=9\u00d7109\u00d7(13\u00d710\u22127)2(0.25)2\u21d2F=9\u00d7109\u00d7(13\u00d710\u22127)2(0.25)2
\n\u21d2F=0.243N\u21d2F=0.243N
\nThe new mutual force of electrostatic repulsion between the two spheres is found to be 0.243N0.243N.<\/h3>\n

13. Suppose the spheres AA and BB in question 1212 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between AA and BB?<\/h2>\n

Ans: We are given the following:
\nDistance between the spheres AA and BB is r=0.5mr=0.5m
\nThe charge on each sphere initially is found to be qA=qB=6.5\u00d710\u22127CqA=qB=6.5\u00d710\u22127C
\nNow, when an uncharged sphere CC is made to touch the sphere AA, a certain amount of charge from AA will get transferred to the sphere CC, making both AA and CC to have equal charges in them. So,
\nqA\u2032=qC=12(6.5\u00d710\u22127)=3.25\u00d710\u22127CqA\u2032=qC=12(6.5\u00d710\u22127)=3.25\u00d710\u22127C
\nNow, when the sphere CC is made to touch the sphere BB, there is a similar transfer of charge making both CC and BB to have equal charges in them. So,
\nqC\u2032=qB\u2032=3.25\u00d710\u22127+6.5\u00d710\u221272=4.875\u00d710\u22127CqC\u2032=qB\u2032=3.25\u00d710\u22127+6.5\u00d710\u221272=4.875\u00d710\u22127C
\nThus, the new force of repulsion between the spheres AA and BB would now become,
\nF\u2032=qA\u2032qB\u20324\u03c0\u03b50r2F\u2032=qA\u2032qB\u20324\u03c0\u03b50r2
\n\u21d2F\u2032=9\u00d7109\u00d73.25\u00d710\u22127\u00d74.875\u00d710\u22127(0.5)2\u21d2F\u2032=9\u00d7109\u00d73.25\u00d710\u22127\u00d74.875\u00d710\u22127(0.5)2
\n\u21d2F\u2032=5.703\u00d710\u22123N\u21d2F\u2032=5.703\u00d710\u22123N<\/h3>\n

14. Figure below shows tracks taken by three charged particles in a uniform electrostatic field. Give the signs of the three charges and also mention which particle has the highest charge to mass ratio?<\/h2>\n

Ans: From the known properties of charges, we know that the unlike charges attract and like charges repel each other.
\nSo, the particles 1 and 2 that move towards the positively charged plate while repelling away from the negatively charged plate would be negatively charged and the particle 3 that moves towards the negatively charged plate while repelling away from the positively charged plate would be positively charged.
\nNow, we know that the charge to mass ratio (which is generally known as emf) is directly proportional to the displacement or the amount of deflection for a given velocity.
\nSince the deflection of particle 3 is found to be maximum among the three, it would have the highest charge to mass ratio.<\/h3>\n

15. Consider a uniform electric field E=3\u00d7103i^N\/CE=3\u00d7103i^N\/C.
\na) Find the flux of this field through a square of side 10cm10cmwhose plane is parallel to the y-z plane.<\/h2>\n

Ans: We are given:
\nElectric field intensity, E\u2192=3\u00d7103i^N\/CE\u2192=3\u00d7103i^N\/C
\nMagnitude of electric field intensity, \u2223\u2223\u2223E\u2192\u2223\u2223\u2223=3\u00d7103N\/C|E\u2192|=3\u00d7103N\/C
\nSide of the square, a=10cm=0.1ma=10cm=0.1m
\nArea of the square, A=a2=0.01m2A=a2=0.01m2
\nSince the plane of the square is parallel to the y-z plane, the normal to its plane would be directed in the x direction. So, angle between normal to the plane and the electric field would be, \u03b8=0\u2218\u03b8=0\u2218
\nWe know that the flux through a surface is given by the relation,
\n\u03d5=|E||A|cos\u03b8\u03d5=|E||A|cos\u2061\u03b8
\nSubstituting the given values, we get,
\n\u21d2\u03d5=3\u00d7103\u00d70.01\u00d7cos0\u2218\u21d2\u03d5=3\u00d7103\u00d70.01\u00d7cos\u20610\u2218
\n\u2234\u03d5=30Nm2\/C\u2234\u03d5=30Nm2\/C
\nThus, we found the net flux through the given surface to be \u03d5=30Nm2\/C\u03d5=30Nm2\/C.<\/h3>\n

b) What would be the flux through the same square if the normal to its plane makes 60\u221860\u2218 angle with the x-axis?<\/h2>\n

Ans: When the plane makes an angle of 60\u221860\u2218 with the x-axis, the flux through the given surface would be,
\n\u03d5=|E||A|cos\u03b8\u03d5=|E||A|cos\u2061\u03b8
\n\u21d2\u03d5=3\u00d7103\u00d70.01\u00d7cos60\u2218\u21d2\u03d5=3\u00d7103\u00d70.01\u00d7cos\u206160\u2218
\n\u21d2\u03d5=30\u00d712\u21d2\u03d5=30\u00d712
\n\u21d2\u03d5=15Nm2\/C\u21d2\u03d5=15Nm2\/C
\nSo, we found the flux in this case to be, \u03d5=15Nm2\/C\u03d5=15Nm2\/C.<\/h3>\n

16. What is the net flux of the uniform electric field of exercise 1.151.15 through a cube of side 20cm20cm oriented so that its faces are parallel to the coordinate planes?<\/h2>\n

Ans: We are given that all the faces of the cube are parallel to the coordinate planes.
\nClearly, the number of field lines entering the cube is equal to the number of field lines entering out of the cube. As a result, the net flux through the cube would be zero.<\/h3>\n

17. Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0\u00d7103Nm2\/C8.0\u00d7103Nm2\/C.
\na) What is the net charge inside the box?<\/h2>\n

Ans: We are given that:
\nNet outward flux through surface of the box,
\n\u03d5=8.0\u00d7103Nm2\/C\u03d5=8.0\u00d7103Nm2\/C
\nFor a body containing of net charge qq, flux could be given by,
\n\u03d5=q\u03b50\u03d5=q\u03b50
\nWhere, \u03b50=8.854\u00d710\u221212N\u22121C2m\u22122=\u03b50=8.854\u00d710\u221212N\u22121C2m\u22122= Permittivity of free space
\nTherefore, the charge qq is given by
\nq=\u03d5\u03b50q=\u03d5\u03b50
\n\u21d2q=8.854\u00d710\u221212\u00d78.0\u00d7103\u21d2q=8.854\u00d710\u221212\u00d78.0\u00d7103
\n\u21d2q=7.08\u00d710\u22128\u21d2q=7.08\u00d710\u22128
\n\u21d2q=0.07\u03bcC\u21d2q=0.07\u03bcC
\nTherefore, the net charge inside the box is found to be 0.07\u03bcC0.07\u03bcC.<\/h3>\n

b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or why not?<\/h2>\n

Ans: No, the net flux entering out through a body depends on the net charge contained within the body according to Gauss\u2019s law.
\nSo, if the net flux is given to be zero, then it can be inferred that the net charge inside the body is zero.
\nHowever, the net charge of the body being zero only implies that the body has equal amount of positive and negative charges and thus, we cannot conclude that there were no charges inside the box.<\/h3>\n

18. A point charge +10\u03bcC+10\u03bcC is a distance 5cm5cm directly above the centre of a square of side 10cm10cm, as shown in Figure below. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10cm10cm)<\/h2>\n

Ans: Consider the square as one face of a cube of edge length 10cm10cm with a charge qq at its centre, according to Gauss’s theorem for a cube, total electric flux is through all its six faces.
\n\u03d5total=q\u03b50\u03d5total=q\u03b50
\nThe electric flux through one face of the cube could be now given by,
\n\u03d5=\u03d5total6\u03d5=\u03d5total6
\n.
\n\u03d5=16q\u03b50\u03d5=16q\u03b50
\n\u03b50=8.854\u00d710\u221212N\u22121C2m\u22122=\u03b50=8.854\u00d710\u221212N\u22121C2m\u22122= Permittivity of free space
\nThe net charge enclosed would be, q=10\u03bcC=10\u00d710\u22126Cq=10\u03bcC=10\u00d710\u22126C
\nSubstituting the values given in the question, we get,
\n\u03d5=16\u00d710\u00d710\u221268.854\u00d710\u221212\u03d5=16\u00d710\u00d710\u221268.854\u00d710\u221212
\n\u2234\u03d5=1.88\u00d7105Nm2C\u22121\u2234\u03d5=1.88\u00d7105Nm2C\u22121
\nTherefore, electric flux through the square is found to be 1.88\u00d7105Nm2C\u221211.88\u00d7105Nm2C\u22121.<\/h3>\n

19. A point charge of 2.0\u03bcC2.0\u03bcC is kept at the centre of a cubic Gaussian surface of edge length 9cm9cm. What is the net electric flux through this surface?<\/h2>\n

Ans: Let us consider one of the faces of the cubical Gaussian surface considered (square).
\nSince a cube has six such square faces in total, we could say that the flux through one surface would be one-sixth the total flux through the gaussian surface considered.
\nThe net flux through the cubical Gaussian surface by Gauss\u2019s law could be given by,
\n\u03d5total=q\u03b50\u03d5total=q\u03b50
\nSo, the electric flux through one face of the cube would be,
\n\u03d5=\u03d5total6\u03d5=\u03d5total6
\n\u21d2\u03d5=16q\u03b50\u21d2\u03d5=16q\u03b50\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (1)
\nBut we have,
\n\u03b50=8.854\u00d710\u221212N\u22121C2m\u22122=\u03b50=8.854\u00d710\u221212N\u22121C2m\u22122= Permittivity of free space
\nCharge enclosed, q=10\u03bcC=10\u00d710\u22126Cq=10\u03bcC=10\u00d710\u22126C
\nSubstituting the given values in (1) we get,
\n\u03d5=16\u00d710\u00d710\u221268.854\u00d710\u221212\u03d5=16\u00d710\u00d710\u221268.854\u00d710\u221212
\n\u21d2\u03d5=1.88\u00d7105Nm2C\u22121\u21d2\u03d5=1.88\u00d7105Nm2C\u22121
\nTherefore, electric flux through the square surface is 1.88\u00d7105Nm2C\u221211.88\u00d7105Nm2C\u22121.<\/h3>\n

20. A point charge causes an electric flux of \u22121.0\u00d7103Nm2\/C\u22121.0\u00d7103Nm2\/C to pass through a spherical Gaussian surface of 10cm10cm radius centred on the charge.
\na) If the radius of the Gaussian surface were doubled, how much flux could pass through the surface?<\/h2>\n

Ans: We are given:
\nElectric flux due to the given point charge, \u03d5=\u22121.0\u00d7103Nm2\/C\u03d5=\u22121.0\u00d7103Nm2\/C
\nRadius of the Gaussian surface enclosing the point charge,r=10.0cmr=10.0cm
\nElectric flux piercing out through a surface depends on the net charge enclosed by the surface according to Gauss\u2019s law and is independent of the dimensions of the arbitrary surface assumed to enclose this charge.
\nHence, if the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i.e., \u2212103Nm2\/C\u2212103Nm2\/C.<\/h3>\n

b) What is the magnitude of the point charge?<\/h2>\n

Ans: Electric flux could be given by the relation,
\n\u03d5total=q\u03b50\u03d5total=q\u03b50
\nWhere,q=q= net charge enclosed by the spherical surface
\n\u03b50=8.854\u00d710\u221212N\u22121C2m\u22122=\u03b50=8.854\u00d710\u221212N\u22121C2m\u22122= Permittivity of free space
\n\u21d2q=\u03d5\u03b50\u21d2q=\u03d5\u03b50
\nSubstituting the given values,
\n\u21d2q=\u22121.0\u00d7103\u00d78.854\u00d710\u221212=\u22128.854\u00d710\u22129C\u21d2q=\u22121.0\u00d7103\u00d78.854\u00d710\u221212=\u22128.854\u00d710\u22129C
\n\u21d2q=\u22128.854nC\u21d2q=\u22128.854nC
\nThus, the value of the point charge is found to be \u22128.854nC\u22128.854nC.<\/h3>\n

21. A conducting sphere of radius 10cm10cm has an unknown charge. If the electric field at a point 20cm20cm from the centre of the sphere of magnitude 1.5\u00d7103N\/C1.5\u00d7103N\/C is directed radially inward, what is the net charge on the sphere?<\/h2>\n

Ans: We have the relation for electric field intensity EE at a distance
\n(d)(d)
\nfrom the centre of a sphere containing net charge qq is given by,
\nE=q4\u03c0\u03b50d2E=q4\u03c0\u03b50d2 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (1)
\nWhere,
\nNet charge, q=1.5\u00d7103N\/Cq=1.5\u00d7103N\/C
\nDistance from the centre, d=20cm=0.2md=20cm=0.2m
\n\u03b50=8.854\u00d710\u221212N\u22121C2m\u22122=\u03b50=8.854\u00d710\u221212N\u22121C2m\u22122= Permittivity of free space
\n14\u03c0\u03b50=9\u00d7109Nm2C\u2212214\u03c0\u03b50=9\u00d7109Nm2C\u22122
\nFrom (1), the unknown charge would be,
\nq=E(4\u03c0\u03b50)d2q=E(4\u03c0\u03b50)d2
\nSubstituting the given values we get,
\n\u21d2q=1.5\u00d7103\u00d7(0.2)29\u00d7109=6.67\u00d710\u22129C\u21d2q=1.5\u00d7103\u00d7(0.2)29\u00d7109=6.67\u00d710\u22129C
\n\u21d2q=6.67nC\u21d2q=6.67nC
\nTherefore, the net charge on the sphere is found to be6.67nC6.67nC.<\/h3>\n

22. A uniformly charged conducting sphere of 2.4m2.4m diameter has a surface charge density of 80.0\u03bcC\/m280.0\u03bcC\/m2.
\na) Find the charge on the sphere.<\/h2>\n

Ans: Given that,
\nDiameter of the sphere, d=2.4md=2.4m.
\nRadius of the sphere, r=1.2mr=1.2m.
\nSurface charge density,
\n\u03c3=80.0\u03bcC\/m2=80\u00d710\u22126C\/m2\u03c3=80.0\u03bcC\/m2=80\u00d710\u22126C\/m2
\nTotal charge on the surface of the sphere,
\nQ=Charge density \u00d7Surface areaQ=Charge density \u00d7 Surface area
\n\u21d2Q=\u03c3\u00d74\u03c0r2=80\u00d710\u22126\u00d74\u00d73.14\u00d7(1.2)2\u21d2Q=\u03c3\u00d74\u03c0r2=80\u00d710\u22126\u00d74\u00d73.14\u00d7(1.2)2
\n\u21d2Q=1.447\u00d710\u22123C\u21d2Q=1.447\u00d710\u22123C
\nTherefore, the charge on the sphere is found to be 1.447\u00d710\u22123C1.447\u00d710\u22123C.<\/h3>\n

b) What is the total electric flux leaving the surface of the sphere?<\/h2>\n

Ans: Total electric flux (\u03d5total)(\u03d5total) leaving out the surface containing net charge QQ is given by Gauss\u2019s law as,
\n\u03d5total=Q\u03b50\u03d5total=Q\u03b50\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (1)
\nWhere, permittivity of free space,
\n\u03b50=8.854\u00d710\u221212N\u22121C2m\u22122\u03b50=8.854\u00d710\u221212N\u22121C2m\u22122
\nWe found the charge on the sphere to be,
\nQ=1.447\u00d710\u22123CQ=1.447\u00d710\u22123C
\nSubstituting these in (1), we get,
\n\u03d5total=1.447\u00d710\u221238.854\u00d710\u221212\u03d5total=1.447\u00d710\u221238.854\u00d710\u221212
\n\u21d2\u03d5total=1.63\u00d710\u22128NC\u22121m2\u21d2\u03d5total=1.63\u00d710\u22128NC\u22121m2
\nTherefore, the total electric flux leaving the surface of the sphere is found to be 1.63\u00d710\u22128NC\u22121m21.63\u00d710\u22128NC\u22121m2.<\/h3>\n

23. An infinite line charge produces a field of magnitude 9\u00d7104N\/C9\u00d7104N\/C at a distance of 2cm2cm. Calculate the linear charge density.<\/h2>\n

Ans: Electric field produced by the given infinite line charge at a distance ddhaving linear charge density\u03bb\u03bb could be given by the relation,
\nE=\u03bb2\u03c0\u03b50dE=\u03bb2\u03c0\u03b50d
\n\u21d2\u03bb=2\u03c0\u03b50Ed\u21d2\u03bb=2\u03c0\u03b50Ed\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (1)
\nWe are given:
\nd=2cm=0.02md=2cm=0.02m
\nE=9\u00d7104N\/CE=9\u00d7104N\/C
\nPermittivity of free space,
\n\u03b50=8.854\u00d710\u221212N\u22121C2m\u22122\u03b50=8.854\u00d710\u221212N\u22121C2m\u22122
\nSubstituting these values in (1) we get,
\n\u21d2\u03bb=2\u03c0(8.854\u00d710\u221212)(9\u00d7104)(0.02)\u21d2\u03bb=2\u03c0(8.854\u00d710\u221212)(9\u00d7104)(0.02)
\n\u21d2\u03bb=10\u00d710\u22128C\/m\u21d2\u03bb=10\u00d710\u22128C\/m
\nTherefore, we found the linear charge density to be 10\u00d710\u22128C\/m10\u00d710\u22128C\/m.<\/h3>\n

24. Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0\u00d710\u221222Cm\u2212217.0\u00d710\u221222Cm\u22122. What is EE in the outer region of the first plate? What is EE in the outer region of the second plate? What is E between the plates?<\/h2>\n

Ans: The given nature of metal plates is represented in the figure below:
\nHere, A and B are two parallel plates kept close to each other. The outer region of plate A is denoted as II, outer region of plate B is denoted as IIIIII, and the region between the plates, A and B, is denoted as IIII.
\nIt is given that:
\nCharge density of plate A, \u03c3=17.0\u00d710\u221222C\/m2\u03c3=17.0\u00d710\u221222C\/m2
\nCharge density of plate B, \u03c3=\u221217.0\u00d710\u221222C\/m2\u03c3=\u221217.0\u00d710\u221222C\/m2
\nIn the regions IIandIIIIII, electric field E is zero. This is because the charge is not enclosed within the respective plates.
\nNow, the electric field EE in the region IIII is given by
\nE=|\u03c3|\u03b50E=|\u03c3|\u03b50
\nWhere,
\nPermittivity of free space \u03b50=8.854\u00d710\u221212N\u22121C2m\u22122\u03b50=8.854\u00d710\u221212N\u22121C2m\u22122
\nClearly,
\nE=17.0\u00d710\u2212228.854\u00d710\u221212E=17.0\u00d710\u2212228.854\u00d710\u221212
\n\u21d2E=1.92\u00d710\u221210N\/C\u21d2E=1.92\u00d710\u221210N\/C
\nThus, the electric field between the plates is 1.92\u00d710\u221210N\/C1.92\u00d710\u221210N\/C.<\/h3>\n

25. An oil drop of 1212 excess electrons is held stationary under a constant electric field of 2.55\u00d7104NC\u221212.55\u00d7104NC\u22121 in Millikan’s oil drop experiment. The density of the oil is 1.26gm\/cm31.26gm\/cm3. Estimate the radius of the drop. (g=9.81ms\u22122,e=1.60\u00d710\u221219C)(g=9.81ms\u22122,e=1.60\u00d710\u221219C).<\/h2>\n

Ans: It is given that:
\nThe number of excess electrons on the oil drop,
\nn=12n=12
\nElectric field intensity, E=2.55\u00d7104NC\u22121E=2.55\u00d7104NC\u22121
\nThe density of oil, \u03c1=1.26gm\/cm3=1.26\u00d7103kg\/m3\u03c1=1.26gm\/cm3=1.26\u00d7103kg\/m3
\nAcceleration due to gravity, g=9.81ms\u22122g=9.81ms\u22122
\nCharge on an electron e=1.60\u00d710\u221219Ce=1.60\u00d710\u221219C
\nRadius of the oil drop =r=r
\nHere, the force (F) due to electric field E is equal to the weight of the oil drop (W).
\nClearly,
\nF=WF=W
\n\u21d2Eq=mg\u21d2Eq=mg
\n\u21d2Ene=43\u03c0r2\u03c1\u00d7g\u21d2Ene=43\u03c0r2\u03c1\u00d7g
\nWhere,
\nqq is the net charge on the oil drop =ne=ne
\nmm is the mass of the oil drop =Volume of the oil drop\u00d7Density of oil=Volume of the oil drop\u00d7Density of oil
\n=43\u03c0r3\u00d7p=43\u03c0r3\u00d7p
\nTherefore, radius of the oil drop can be calculated as
\nr=3Ene4\u03c0\u03c1g\u2212\u2212\u2212\u2212\u2212\u221ar=3Ene4\u03c0\u03c1g
\n\u21d2r=3\u00d72.55\u00d7104\u00d712\u00d71.6\u00d710\u2212194\u00d73.14\u00d71.26\u00d7103\u00d79.81\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221a\u21d2r=3\u00d72.55\u00d7104\u00d712\u00d71.6\u00d710\u2212194\u00d73.14\u00d71.26\u00d7103\u00d79.81
\n\u21d2r=946.09\u00d710\u221221\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221a\u21d2r=946.09\u00d710\u221221
\n\u21d2r=9.72\u00d710\u221210m\u21d2r=9.72\u00d710\u221210m
\nTherefore, the radius of the oil drop is 9.72\u00d710\u221210m9.72\u00d710\u221210m.<\/h3>\n

26. Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?<\/h2>\n

a)<\/h2>\n

Ans: The field lines shown in (a) do not represent electrostatic field lines because field lines must be normal to the surface of the conductor which is a characterizing property of electric field lines.<\/h3>\n

b)<\/h2>\n

Ans: The lines shown in (b) do not represent electrostatic field lines because field lines cannot emerge from a negative charge and cannot terminate at a positive charge since the direction of the electric field is from positive to negative charge.<\/h3>\n

c)<\/h2>\n

Ans: The field lines shown in (c) do represent electrostatic field lines as they are directed outwards from positive charge in accordance with the property of electric field.<\/h3>\n

d)<\/h2>\n

Ans: The field lines shown in (d) do not represent electrostatic field lines because electric field lines should not intersect each other.<\/h3>\n

e)<\/h2>\n

Ans: The field lines shown in (e) do not represent electrostatic field lines because electric field lines do not form closed loops<\/h3>\n

27. In a certain region of space, the electric field is along the z-direction throughout. The magnitude of the electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 105NC\u22121105NC\u22121 per meter. What are the force and torque experienced by a system having a total dipole moment equal to 10\u22127Cm10\u22127Cm in the negative z-direction?<\/h2>\n

Ans: We know that the dipole moment of the system, P=q\u00d7dl=\u221210\u22127CmP=q\u00d7dl=\u221210\u22127Cm.
\nAlso, the rate of increase of electric field per unit length is given as
\ndEdl=105NC\u22121dEdl=105NC\u22121
\nNow, the force (F) experienced by the system is given by F=qEF=qE
\nF=qdEdl\u00d7dlF=qdEdl\u00d7dl
\n\u21d2F=PdEdl\u21d2F=PdEdl
\n\u21d2F=\u221210\u22127\u00d7105\u21d2F=\u221210\u22127\u00d7105
\n\u21d2F=\u221210\u22122N\u21d2F=\u221210\u22122N
\nClearly, the force is equal to \u221210\u22122N\u221210\u22122N in the negative z-direction i.e., it is opposite to the direction of the electric field.
\nThus, the angle between the electric field and dipole moment is equal to
\n180\u2218180\u2218
\nNow, the torque is given by \u03c4=PEsin\u03b8\u03c4=PEsin\u2061\u03b8
\n\u03c4=PEsin180\u2218=0\u03c4=PEsin\u2061180\u2218=0
\nTherefore, it can be concluded that the torque experienced by the system is zero.<\/h3>\n

28.
\na) A conductor A with a cavity as shown in the Fig. 1.36(a) is given a charge QQ. Show that the entire charge must appear on the outer surface of the conductor.
\n(Image Will Be Updated Soon)<\/h2>\n

Ans: Firstly, let us consider a Gaussian surface that is lying within a conductor as a whole and enclosing the cavity. Clearly, the electric field intensity E inside the charged conductor is zero.
\nNow, let
\nqq
\nbe the charge inside the conductor and \u03b50\u03b50, the permittivity of free space.
\nAccording to Gauss’s law,
\nFlux is given by
\n\u03d5=E\u2192.ds=q\u03b50\u03d5=E\u2192.ds=q\u03b50
\nHere, \u03d5=0\u03d5=0 as E=0E=0 inside the conductor
\nClearly,
\n0=q8.854\u00d710\u2212120=q8.854\u00d710\u221212
\n\u21d2q=0\u21d2q=0
\nTherefore, the charge inside the conductor is zero.
\nAnd hence, the entire charge QQ appears on the outer surface of the conductor.<\/h3>\n

b) Another conductor B with charge qq is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q+qQ+q Fig.1.36(b)Fig.1.36(b).<\/h2>\n

Ans: The outer surface of conductor A has a charge of QQ.
\nIt is given that another conductor B, having a charge
\n+q+q
\nis kept inside conductor A and is insulated from conductor A.
\nClearly, a charge of
\n\u2212q\u2212q
\nwill get induced in the inner surface of conductor A and a charge of
\n+q+q
\nwill get induced on the outer surface of conductor A.
\nTherefore, the total charge on the outer surface of conductor A amounts to Q+qQ+q.<\/h3>\n

c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.<\/h2>\n

Ans: A sensitive instrument can be shielded from a strong electrostatic field in its environment by enclosing it fully inside a metallic envelope.
\nSuch a closed metallic body provides hindrance to electrostatic fields and thus can be used as a shield.<\/h3>\n

29. A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is [\u03c32\u03b50]n\u2227[\u03c32\u03b50]n\u2227, where n\u2227n\u2227 is the unit vector in the outward normal direction, and \u03c3\u03c3 is the surface charge density near the hole.<\/h2>\n

Ans: Firstly, let us consider a conductor with a cavity or a hole as shown in the diagram below. It is known that the electric field inside the cavity is zero.
\nLet us assume E to be the electric field just outside the conductor, qq be the electric charge, \u03c3\u03c3 be the charge density, and \u03b50\u03b50, the permittivity of free space.
\nWe know that charge |q|=\u03c3\u00d7d|q|=\u03c3\u00d7d
\nNow, according to Gauss’s law,
\n\u03d5=E.ds=|q|\u03b50\u03d5=E.ds=|q|\u03b50
\n\u21d2E.ds=\u03c3\u00d7d\u03b50\u21d2E.ds=\u03c3\u00d7d\u03b50
\n\u21d2E=\u03c3\u03b50n\u2227\u21d2E=\u03c3\u03b50n\u2227
\nwhere n\u2227n\u2227 is the unit vector in the outward normal direction.
\nThus, the electric field just outside the conductor is \u03c3\u03b50n\u2227\u03c3\u03b50n\u2227.
\nNow, this field is actually a superposition of the field due to the cavity E1E1 and the field due to the rest of the charged conductor E2E2.
\nThese electric fields are equal and opposite inside the conductor whereas equal in magnitude as well as direction outside the conductor.
\nClearly,
\nE1+E2=EE1+E2=E
\n\u21d2E1=E2=E2=\u03c32\u03b50n\u2227\u21d2E1=E2=E2=\u03c32\u03b50n\u2227
\nTherefore, the electric field in the hole is \u03c32\u03b50n\u2227\u03c32\u03b50n\u2227.
\nHence, proved.<\/h3>\n

30. Obtain the formula for the electric field due to a long thin wire of uniform linear charge density \u03bb\u03bb without using Gauss’s law. Hint:UseCoulomb\u2032slawdirectlyandevaluatethenecessaryintegralHint:UseCoulomb\u2032slawdirectlyandevaluatethenecessaryintegral<\/h2>\n

Ans: Firstly, let us take a long thin wire XY as shown in the figure below. This wire is of uniform linear charge density \u03bb\u03bb.
\n(Image Will Be Updated Soon)
\nNow, consider a point A at a perpendicular distance l from the mid-point O of the wire as shown in the figure below:
\n(Image Will Be Updated Soon)
\nConsider E to be the electric field at point A due to the wire.
\nAlso consider a small length element dxdx on the wire section with OZ=xOZ=x as shown.
\nLet qq be the charge on this element.
\nClearly, q=\u03bbdxq=\u03bbdx
\nNow, the electric field due to this small element can be given as
\ndE=14\u03c0\u03b50\u03bbdx(AZ)2dE=14\u03c0\u03b50\u03bbdx(AZ)2
\nHowever, AZ=12+x2\u2212\u2212\u2212\u2212\u2212\u2212\u221aAZ=12+x2
\n\u21d2dE=\u03bbdx4\u03c0\u03b50(12+x2)\u21d2dE=\u03bbdx4\u03c0\u03b50(12+x2)
\nNow, let us resolve the electric field into two rectangular components. Doing so, dEcos\u03b8dEcos\u2061\u03b8 is the perpendicular component and dEsin\u03b8dEsin\u2061\u03b8 is the parallel component.
\nWhen the whole wire is considered, the component dEsin\u03b8dEsin\u2061\u03b8 gets cancelled and only the perpendicular component dEcos\u03b8dEcos\u2061\u03b8 affects the point A.
\nThus, the effective electric field at point A due to the element dxdx can be written as
\ndE1=\u03bbdxcos\u03b84\u03c0\u03b50(l2+x2)dE1=\u03bbdxcos\u2061\u03b84\u03c0\u03b50(l2+x2) ….(1)
\nNow, in \u0394AZO\u0394AZO, we have
\ntan\u03b8=xltan\u2061\u03b8=xl
\nx=ltan\u03b8……(2)x=ltan\u2061\u03b8……(2)
\nOn differentiating equation (2), we obtain
\ndx=lsec2d\u03b8……(3)dx=lsec2d\u03b8……(3)
\nFrom equation (2)
\nx2+l2=l2+l2tan2\u03b8x2+l2=l2+l2tan2\u03b8
\n\u21d2l2(1+tan2\u03b8)=l2sec2\u03b8\u21d2l2(1+tan2\u03b8)=l2sec2\u03b8
\n\u21d2x2+l2=l2sec2\u03b8…..(4)\u21d2x2+l2=l2sec2\u03b8…..(4)
\nPutting equations (3) and (4) in equation (1), we obtain
\ndE1=\u03bblsec2d\u03b84\u03c0\u03b50(l2sec2\u03b8)cos\u03b8dE1=\u03bblsec2d\u03b84\u03c0\u03b50(l2sec2\u03b8)cos\u2061\u03b8
\n\u21d2dE1=\u03bbcos\u03b8d\u03b84\u03c0\u03b50l…..(5)\u21d2dE1=\u03bbcos\u2061\u03b8d\u03b84\u03c0\u03b50l…..(5)
\nNow, the wire is taken so long that it ends from \u2212\u03c02\u2212\u03c02 to +\u03c02+\u03c02.
\nTherefore, by integrating equation (5), we obtain the value of field E1E1 as
\n\u222b\u2212\u03c02\u03c02dE1=\u222b\u2212\u03c02\u03c02\u03bb4\u03c0\u03b50lcos\u03b8d\u03b8\u222b\u2212\u03c02\u03c02dE1=\u222b\u2212\u03c02\u03c02\u03bb4\u03c0\u03b50lcos\u2061\u03b8d\u03b8
\n\u21d2E1=\u03bb4\u03c0\u03b50l\u00d72\u21d2E1=\u03bb4\u03c0\u03b50l\u00d72
\n\u21d2E1=\u03bb2\u03c0\u03b50l\u21d2E1=\u03bb2\u03c0\u03b50l
\nThus, the electric field due to the long wire is derived to be equal to
\n\u03bb2\u03c0\u03b50l\u03bb2\u03c0\u03b50l<\/h3>\n

31. It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up quark (denoted by uu)\u00a0 of charge (+12)e(+12)e and the ‘down’ quark (denoted by dd) of charge \u2212(13)e\u2212(13)e together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.<\/h2>\n

Ans: It is known that a proton has three quarks. Let us consider
\nnn
\nup quarks in a proton, each having a charge of +(23e)+(23e).
\nNow, the charge due to
\nnn
\nup quarks =(23e)n=(23e)n
\nThe number of down quarks in a proton =3\u2212n=3\u2212n
\nAlso, each down quark has a charge of \u221213e\u221213e
\nTherefore, the charge due to (3\u2212n)(3\u2212n) down quarks =(\u221213e)(3\u2212n)=(\u221213e)(3\u2212n)
\nWe know that the total charge on a proton =+e=+e
\nTherefore,
\ne=(23e)n+(\u221213e)(3\u2212n)e=(23e)n+(\u221213e)(3\u2212n)
\n\u21d2e=(2ne3)\u2212e+ne3\u21d2e=(2ne3)\u2212e+ne3
\n\u21d22e=ne\u21d22e=ne
\n\u21d2n=2\u21d2n=2
\nClearly, the number of up quarks in a proton, n=2n=2
\nThus, the number of down quarks in a proton =3\u2212n=3\u22122=1=3\u2212n=3\u22122=1
\nTherefore, a proton can be represented as uuduud.
\nA neutron is also said to have three quarks. Let us consider
\nnn
\nup quarks in a neutron, each having a charge of +(23e)+(23e) .
\nIt is given that the charge on a neutron due to
\nnn
\nup quarks =(+32e)n=(+32e)n
\nAlso, the number of down quarks is (3\u2212n)(3\u2212n), each having a charge of =(\u221232)e=(\u221232)e
\nThus, the charge on a neutron due to (3\u2212n)(3\u2212n) down quarks =(\u221213e)(3\u2212n)=(\u221213e)(3\u2212n)
\nNow, we know that the total charge on a neutron =0=0
\nThus,
\n0=(23e)n+(\u221213e)(3\u2212n)0=(23e)n+(\u221213e)(3\u2212n)
\n\u21d20=(2ne3)\u2212e+ne3\u21d20=(2ne3)\u2212e+ne3
\n\u21d2e=ne\u21d2e=ne
\n\u21d2n=1\u21d2n=1
\nClearly, the number of up quarks in a neutron, n=1n=1
\nThus, the number of down quarks in a neutron =3\u2212n=2=3\u2212n=2
\nTherefore, a neutron can be represented as uddudd.<\/h3>\n

32.
\na) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where
\nE=0E=0
\n) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.<\/h2>\n

Ans: Firstly, let us assume that the small test charge placed at the null point of the given setup is in stable equilibrium.
\nBy stable equilibrium, it means that even a slight displacement of the test charge in any direction will cause the charge to return to the null point as there will be strong restoring forces acting around it.
\nThis further suggests that all the electric lines of force around the null point act inwards and towards the given null point.
\nBut by Gauss law, we know that the net electric flux through a chargeless enclosing surface is equal to zero. This truth contradicts the assumption which we had started with. Therefore, it can be concluded that the equilibrium of the test charge is necessarily unstable.<\/h3>\n

b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed at a certain distance apart.<\/h2>\n

Ans: When we consider this configuration setup with two charges of the same magnitude and sign placed at a certain distance apart, the null point happens to be at the midpoint of the line joining these two charges.
\nAs per the previous assumption, the test charge, when placed at this mid-point will experience strong restoring forces when it tries to displace itself.
\nBut when the test charge tries to displace in a direction normal to the line joining the two charges, the test charge gets pulled off as there is no restoring force along the normal to the line considered.
\nSince stable equilibrium prioritizes restoring force in all directions, the assumption in this case also gets contradicted.<\/h3>\n

33. A particle of mass mm and charge (\u2212q)(\u2212q) enters the region between the two charged plates initially moving along x- axis with speed vxvx (like particle 1 in Fig 1.33). The length of plate is LL and a uniform electric field EE is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL22mvx2qEL22mvx2.
\nCompare this motion with motion of a projectile in the gravitational field discussed in section 4.10 of class XI textbook of Physics.<\/h2>\n

Ans: It is given that:
\nThe charge on a particle of mass m=\u2212qm=\u2212q
\nVelocity of the particle =vx=vx
\nLength of the plates =L=L
\nMagnitude of the uniform electric field between the plates =E=E
\nMechanical force, F= Mass (m)\u00d7Acceleration (a)F= Mass (m)\u00d7Acceleration (a)
\nThus, acceleration, a=Fma=Fm
\nHowever, electric force, F=qEF=qE
\nTherefore, acceleration, =qEm=qEm………(1)
\nHere, the time taken by the particle to cross the field of length LL is given by,
\nt=Length of the plateVelocity of the plate=Lvxt=Length of the plateVelocity of the plate=Lvx ……(2)
\nIn the vertical direction, we know that the initial velocity, u=0u=0
\nNow, according to the third equation of motion, vertical deflection ss of the particle can be derived as
\ns=ut+12at2s=ut+12at2
\n\u21d2s=0+12(qEm)(Lvx)2\u21d2s=0+12(qEm)(Lvx)2
\n\u21d2s=qEL22mvx2\u21d2s=qEL22mvx2 …..(3)
\nThus, the vertical deflection of the particle at the far edge of the plate is qEL22mvx2qEL22mvx2.
\nIn comparison, we can see that this is similar to the motion of horizontal projectiles under gravity.<\/h3>\n

34. Suppose that the particle in Exercise in 1.33 is an electron projected with velocity vx=2.0\u00d7106ms\u22121vx=2.0\u00d7106ms\u22121. If EEbetween the plates separated by 0.5cm0.5cm is 9.1\u00d7102N\/C9.1\u00d7102N\/C, where will the electron strike the upper plate? (|e|=1.6\u00d710\u221219C,me=9.1\u00d710\u221231kg|e|=1.6\u00d710\u221219C,me=9.1\u00d710\u221231kg )<\/h2>\n

Ans: We are given the velocity of the particle, vx=2.0\u00d7106ms\u22121vx=2.0\u00d7106ms\u22121.
\nSeparation between the two plates, d=0.5cm=0.005md=0.5cm=0.005m
\nElectric field between the two plates, E=9.1\u00d7102N\/CE=9.1\u00d7102N\/C
\nCharge on an electron, e=1.6\u00d710\u221219Ce=1.6\u00d710\u221219C
\nmass of an electron, me=9.1\u00d710\u221231kgme=9.1\u00d710\u221231kg
\nLetssbe the deflection when the electron strikes the upper plate at the end of the plate LL, then, we have the deflection given by,
\ns=qEL22mvxs=qEL22mvx
\n\u21d2L=2dmvxqE\u2212\u2212\u2212\u2212\u2212\u221a\u21d2L=2dmvxqE
\nSubstituting the given values,
\n\u21d2L=2\u00d70.005\u00d79.1\u00d710\u221231\u00d7(2.0\u00d7106)21.6\u00d710\u221219\u00d79.1\u00d7102\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221a=0.025\u00d710\u22122\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221a=2.5\u00d710\u22124\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221a\u21d2L=2\u00d70.005\u00d79.1\u00d710\u221231\u00d7(2.0\u00d7106)21.6\u00d710\u221219\u00d79.1\u00d7102=0.025\u00d710\u22122=2.5\u00d710\u22124
\n\u21d2L=1.6\u00d710\u22122=1.6cm\u21d2L=1.6\u00d710\u22122=1.6cm
\nTherefore, we found that the electron will strike the upper plate after travelling a distance of 1.6cm1.6cm.<\/h3>\n","protected":false},"excerpt":{"rendered":"

Class 12 Physics NCERT book solutions for Chapter 1 – Electric Charges And Fields Questions and Answers.<\/p>\n","protected":false},"author":21830,"featured_media":118347,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7],"tags":[],"class_list":{"0":"post-118348","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-education"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/118348","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/users\/21830"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/comments?post=118348"}],"version-history":[{"count":1,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/118348\/revisions"}],"predecessor-version":[{"id":118350,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/118348\/revisions\/118350"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media\/118347"}],"wp:attachment":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media?parent=118348"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/categories?post=118348"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/tags?post=118348"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}