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2, Electric charges and fields, , 1. Electric charge:, An electric charge is a fundamental physical quantity due to which electrical and other related, effects are produced in the matter., Charges are of two types, • Charge is a scalar quantity., 1. Positive charges (vitreous), • Dimensional formulae of charge is [AT]., 2. Negative charges (resinous), ➢ Electricity: The study of charges., Note2: Dr. Gilbert (A personal doctor of queen, ➢ Electrostatics: The study of charges at rest. Elizabeth I of England) led to the conclusion that, ➢ Current electricity: The study of charges in many substances like amber aquire this property.), motion., Note3: Benjamin Franklin (American) introduced, Note1: Thales (Greek philosopher-600BC) the concept of +ve and –ve charges., observed that amber (elektron) rubbed with, woolen cloth attracts light objects like feather, Note4: The name Electricity is coined from the, Greek word elektron meaning amber., pieces of paper etc., Any substance rubbed with any other under suitable conditions will become charged., , 2. Methods of Charging:, 1. Charging by Conduction, The phenomenon of charging the bodies when, Charging an uncharged body with actual they are rubbed with each other is called, physical contact of another charged body is Electrification by friction. The charge thus, called charging by conduction., developed is called Frictional electricity., 2. Charging by Induction, Name of the body which acquires, Charging of an uncharged body in the, Positive charge, Negative charge, presence of a charged body without, making a physical contact with it, is called, 1. Glass rod, 1. Silk cloth, charging by induction, 2. Woollen cloth, 2. Amber, 3. Electrification by friction, 3. Woollen cloth, 3. Rubber, When a glass rod is rubbed with silk, the, 4. Woollen cloth, 4. Plastic, glass rod becomes positively charged and the silk, 5. Flannel(cat skin), 5. Ebonite, becomes negatively charged., 6. Dry hair, 6. Plastic comb, Earthing or Grounding:, The process of Charge flow from a charged, A body in its normal state is electrically neutral., body to earth, when it is in contact with earth., Questions:, 1. When a glass rod is rubbed with silk, it acquires positive charge. How it becomes positively charged?, [Glass becomes positively charged by the transfer of electrons from glass rod to silk.], 2. When a polythene piece is rubbed with wool, it acquires negative charge. Is there a transfer of mass, from wool to polythene?, [Yes, since electron has mass, there is a transfer of mass from wool to polythene.], 3. A comb run through one’s dry hair attracts small bits of paper. Why? What happens if the hair is wet, or if it is a rainy day?, [When a comb run through one’s dry hair, it gets charged due to friction. Hence it attracts small bits of, paper. If hair is wet, the comb does not get charged due to the absence of friction.], , HSPTA KANNUR
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3, Electric charges and fields, , 3. Properties of charges, 1. Like charges repel and unlike charges attract., 3. Charge is conserved., 2. A charged body attracts (light) uncharged 4. Charge is quantized., bodies., 5. Charges add up like real numbers., −𝟏𝟗, Charge on an electron (e=1.6 x 𝟏𝟎 c) is considered as the smallest possible charge., 𝟏, 𝟐, However quarks have charges ± 𝟑 e and ± 𝟑 e., , 4. Law of conservation of charge, “For an isolated system, the net charge always e.g. 2. In the reaction,, remains constant”, 236, 141, 92, 1, 1, e.g. 1., When a glass rod is rubbed with silk, the 235, 92U+ 0n→ 92U→ 56Ba + 36Kr+3 0n+Energy, glass rod becomes +vely charged and the silk Net charge before reaction=0+92=92, becomes –vely charged., Net charge after reaction=56+36=92, Thus the net charge before and after rubbing, are zero., Thus the charge is conserved., 5. Quantisation of charge, “Charge on a body is always an integral, multiple of the smallest unit of charge ‘e’.”, q = ±ne, Problems:, 1. A body is found to have a negative charge of, electrons transferred., 𝐪, 𝐞, , [q = ±ne; n= =, , 3.2x10−19, 1.6 x 10−19, , 𝐪, , 16x10−19, 1.6 x 10−19, , Any charged body cannot possess charge which, is a fraction of ±e., 3.2 x 10−19 C on rubbing. Calculate the number of, , =𝟐, , 2. A body is found to have a negative charge of, electrons transferred., [n=𝐞 =, , Where ‘n’ is an integer and e=1.6 x 𝟏𝟎−𝟏𝟗 C, , 𝟐], 16 x 10−19 C on rubbing. Calculate the number of, , = 𝟏𝟎, , 𝟏0 ], , 3. A polythene piece is rubbed with wool is found to have a negative charge of 3x10−7C. Calculate the, number of electrons transferred from wool to polythene., 𝐪, , [n=𝐞 =, 4., , 3x10−7, 1.6 x 10−19, , =, , 𝟏. 𝟖𝟕𝟓 𝐱 𝟏𝟎𝟏𝟐 ], , Can a body have a charge of 2.4 x 10−19C? Justify your answer., 𝐪, , [n=𝐞 =, , 2.4 x10−19, 1.6 x 10−19, , = 𝟏. 𝟓, , Fraction of charge is not possible. ], , 5. Find the number of electrons in 1C., 𝐪, , [n=𝐞 =, , 1C, 1.6 x 10−19, , =, , 0.625 x𝟏𝟎𝟏𝟗 ], , Questions:, 1. Vehicles carrying inflammable materials have metallic ropes touching the ground. Why?, 2. Tyres of aircrafts are made slightly conducting. Why?, [Charges accumulated on the vehicle due to air friction may produce spark., Metallic conductors and conducting tyres allow the charges to flow to earth. ], , HSPTA KANNUR
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6, Electric charges and fields, , ELECTRIC FIELD, 10. Electric field, “The region of space where an electric charge Note: The test charge should be vanishingly small., experiences electrostatic force is called electric, (If the test charge is large, it will disturb the, field.”, source charge and hence measured electric field, The strength of the electric field is defined as will not be accurate.), the force per unit test charge placed at that point., ⃗ = 𝐋𝐭 𝐅, ∴, 𝐄, 𝐪𝟎 →𝟎 𝐪𝟎, , N, , ⃗= 𝐅, 𝐄, 𝐪, , V, , Unit: C or (E = d; unit:, , 𝟎, , Dimensions:, , volt, ), m, −3 −1, , MLT−2, =MLT, IT, , I, , ➢ If the strength of electric field is the same at every point, it is called Uniform electric field., ➢ If its strength is different at every point, it is called Non-Uniform electric field., , 11. Electric Field Lines:, “The path along which a unit positive, charge would move if it is free to do so.”, Note 1: It is a convenient way of representing, electric field., Note 2: It is a concept for visualasing the, electric field introduced by Faraday., , 4. Two field lines never intersect., 5. Two field lines have a tendency to repel, each other. That is they can exert a, lateral pressure., 6. They are always perpendicular to the, surface of the charged conductor., 7. Number of field lines is proportional to, the magnitude of the charge., 8. They do not form any closed loops., , Properties of Electric field lines:, 1. Diverge out from a positive charge and Electric field due to two positive charges., converge at a negative charge., +, , +, , Positive charge, , -, , negative charge, , +, , Electric field due to two opposite charges., , +, , -, , 2. The tangent drawn at any point on the, electric field line gives the direction of the, electric field., Qn: Two field lines never intersect. Why?, 3. The number field lines crossing per unit, If two lines of force intersect, there will be two, area (normal to the lines of force) at any directions for electric field at the point of, intersection. It is impossible., point give the electric field at that point., , HSPTA KANNUR
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15, Electric charges and fields, Problems, Problem1. Pick the odd one out based on the net flux., , (i), , (ii), , (iii), , (iv), , [Ans: Case (i), , Net flux is not zero], , Problem2. Show that the entire charge given to a conductor with a cavity always resides on its outer, surface., [Ans:, , ⃗⃗⃗⃗ =, ⃗ . ds, Electric field inside a conductor is zero. ( ∮ E, , Q, , 𝟏, 𝛜𝟎, , q), , Hence there cannot be charge inside the Guassian surface.], Problem3. In the above conductor, another conductor with a charge ‘q’ is inserted into the cavity which, is insulated from the first conductor. Show that the total charge on the first conductor is (Q+q)., [Ans:, cavity, , Q +q, , +q produces an induced –ve charge on the inner wall of the, leaving +q charge on the outer surface, Hence the total charge on the first conductor is (Q+q)], , Problem 4: NCERT EXE 2.4 A spherical conductor of radius 12 cm has a charge of 1.6 × 10–7C, distributed uniformly on its surface., (b) just outside the sphere, , What is the electric field, , (c) at a point 18 cm from the centre of the sphere?, , 1.6 × 10–7, , [(a) Zero, , (a) inside the sphere, , 𝛔 𝟒 𝐱 𝟑.𝟏𝟒 𝐱( 𝟏𝟐 𝐱𝟏𝟎−𝟐 )𝟐, (b) E= =, =9.99 x, 𝛜𝟎, 8.854 x 10−12, , 𝛔 𝐑𝟐, =105, 𝛜𝟎 𝐫 𝟐, , 104 =𝟏𝟎𝟓 N/C (c) E=, , x, , ( 12 x10−2 )2, =, ( 18 x10−2 )2, , 4.4x𝟏𝟎𝟒 ], , Problem 5: NCERT e.g 1.1 How can you charge a metal sphere positively without touching it?, [Ans:-, , (a), , (b), , (d), , (c), , (e), , (a) an uncharged metallic sphere. (b) Bring a negatively charged rod close to the sphere, the near end, becomes positively charged. (c) Earth the sphere. The electrons will flow to the ground while the, positive charges at the near end will remain held there due to the negative charges on the rod., (d) Disconnect the sphere from the ground. The positive charge continues to be held at the near end., (e)Remove the negatively charged rod, the positive charge will spread uniformly over the sphere.], [By bringing a positively charged rod, the metal sphere can be negatively charged by induction], Problem 6: NCERT e.g 1.2 If 109 electrons move out of a body to another body every second, how much, time is required to get a total charge of 1 C on the other body?, q, , 1, , [q = ±ne, n=e=1.6 x 10−19=0.625 x1019 :, , HSPTA KANNUR, , t=, , 0.625 x1019, =, 109, , 𝟎. 𝟔𝟐𝟓 𝐱𝟏𝟎𝟏𝟎 s]
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16, Electric charges and fields, Problem 7: NCERT e.g 1.3 How much positive and negative charge is there in a cup of water (assume, mass= 250 g)?, 6.02 × 1023, 18, , [Mass of 6.02 × 1023 molecules of water = 18 g;, , the number of molecules in1g of water =, , The number of molecules in one cup of water =250 x, , 6.02 × 1023, 18, , ;, , Each molecule of H2 O contains 2 hydrogen atoms and 1 oxygen atom, i.e., 10 electrons and 10 protons., Hence the total positive ( negative) charge =10 x 250 x, , 6.02 × 1023, x, 18, , 1.6 × 10–19 C, , =1.34 × 107 C], , Problem 8: NCERT e.g 1.4 Coulomb’s law for electrostatic force between two point charges and, Newton’s law for gravitational force between two stationary point masses, both have inverse-square, dependence onthe distance between the charges/masses. (a) Compare the strength of these forces by, determining the ratio of their magnitudes (i) for an electron and a proton and (ii) for two protons. (b), Estimate the accelerations of electron and proton due to the electrical force of their mutual attraction, when they are 1 Å (= 10-10 m) apart? (mp = 1.67 × 10–27 kg, me = 9.11 × 10–31 kg), 𝟏 𝐪𝟏 𝐪𝟐, , [(a) (i) for electron- proton:, , Fe 𝟒𝛑𝛜𝟎 𝐫𝟐, = m1 m2, Fg, G 2, , 9x109 𝐪𝟏 𝐪𝟐, , =, , r, , 𝟏 𝐪𝟏 𝐪𝟐, , (ii) For two protons, , 𝟐, F, 𝟒𝛑𝛜, : e= m𝟎1m𝐫 2, Fg, G 2, , =, , r, , G m1 m2, , 9x109 𝐪𝟏 𝐪𝟐, G m1 m2, , (b), , Acceleration of, , F, proton: a=m, , =, , 𝟏 𝐪𝟏 𝐪𝟐, 𝟒𝛑𝛜𝟎 𝐫𝟐, , m, , =, , =2.27 x𝟏𝟎𝟑𝟗, , 9x109 x1.6 x 10−19 x1.6 x 10−19, 6.67 x 10−11 x1.67 × 10–27 x1.67 × 10–27, , =1.23 x𝟏𝟎𝟑𝟔, , =, , 𝟏 𝐪𝟏 𝐪𝟐, , 𝟐, F 𝟒𝛑𝛜, Acceleration of electron: a= = 𝟎 𝐫, m, m, , 9x109 x1.6 x 10−19 x1.6 x 10−19, , =6.67 x 10−11 x1.67 × 10–27 x9.11 × 10–31, , =, , 9x109 x1.6 x 10−19 x1.6 x 10−19, 9.11 × 10−31 x(10−10 )2, , 9x109 x1.6 x 10−19 x1.6 x 10−19, 1.67 × 10−27 x(10−10 )2, , =2.52 x 𝟏𝟎𝟐𝟐 m𝐬 −𝟏, , =1.37 x 𝟏𝟎𝟏𝟗 m𝐬 −𝟏 ], , Problem 9: NCERT e.g 1.5 A charged metallic sphere A is suspended by a nylon thread. Another charged, metallic sphere B held by an insulating handle is brought close to A such that the distance between their, centres is 10 cm, as shown in Fig. 1.7(a). The resulting repulsion of A is noted Spheres A and B are, touched by uncharged spheres C and D respectively, as shown in Fig. 1.7(b). C and D are then removed, and B is brought closer to A to a distance of 5.0 cm between their centres, as shown in Fig. 1.7(c). What, is the expected repulsion of A on the basis of Coulomb’s law? Spheres A and C and spheres B and D have, identical sizes. Ignore the sizes of A and B in comparison to the separation between their centres., , [Initial force F=, , 𝟏, 𝟒𝛑𝛜𝟎, , 𝐪𝐪′, , 𝟏, , 𝐪 𝐪′, ⁄𝟐 ⁄𝟐, 𝟏, 𝐪𝐪′, 𝟏 𝐪𝐪′, =, =, ------(2), 𝟓𝟐, 𝟒𝛑𝛜𝟎 𝟒 𝐱 𝟐𝟓 𝟒𝛑𝛜𝟎 𝟏𝟎𝟎, 𝟎, , ---(1) : New force F= 𝟒𝛑𝛜, 𝟏𝟎𝟐, , Thus the electrostatic force on A, due to B, remains unaltered], XXXXXXXXXXXXXX, , HSPTA KANNUR
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17, Electric charges and fields, , Additional exercises:, 1. AIPMT 1994: The given figure gives electric field lines due to two charges q1 and q2. What are the, signs of the two charges?, , (a) q1 is positive but q2 is negative., (c) both are negative., , (b) q1 is negative but q2 is positive., (d) both are positive., , [Ans:(c), , (c) both are negative ], , 2. AIPMT 1999: When air is replaced by a dielectric medium of constant K, the maximum force of, attraction between two charges separated by a distance, (a) increases K times, , (b) remains unchanged, , (c)decreases K times, , (d) increases K –1 times, , F, K, , [Ans: (c) using the formula F= 0;, , (c)decreases K times], , 3. AIPMT 2011: A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is, doubled, then the outward electric flux will, (a) increase four times (b) be reduced to half, (c) remain the same (d) be doubled, [Ans: (c)], (c) remain the same], 4. NEET 2019: Two point charges A and B, having charges +Q and –Q respectively, are placed at certain, distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force, between the charges., [Ans: (c)][𝐐𝐀 = Q -, , Q, 𝟑, =𝟒Q, 4, , (a) 4F, ;, , 𝐐𝐁 = -, , (b) F, , (c), , 9, F, 16, , 15, 16, , (d) F, 3, , 3, , F2 4 X 4, = 1, F1, , Q, 𝟑, Q + 4 =− 𝟒 Q:, , 9, , =16, , 𝟗, 𝟏𝟔, , 𝐅], , 5. NEET 2016 An electric dipole is placed at an angle of 30° with an electric field intensity 2 × 10 5 N/C. It, experiences a torque equal to 4 Nm. The charge on the dipole, if the dipole length is 2 cm, is, (a) 8 mC, (b) 2 mC, (c) 5 mC, (d) 7 µC, τ, , 4, , [Ans: (b)][𝛕=pE𝐬𝐢𝐧 𝛉= q(2l) Esin θ; q=(2l) E sin θ=(2x 10−2 )x 2 × 105, , sin 30, , =2x 𝟏𝟎−𝟑 C], , 6. NEET 2017: Suppose the charge of a proton and an electron differ slightly. One of them is - e, the, other is (e + ∆e). If the net of electrostatic force and gravitational force between two hydrogen atoms, placed at a distance d (much greater than atomic size) apart is zero, then ∆e is of the order of [Given, mass of hydrogen mh = 1.67 × 10-27 kg], (a) 10-23 C, (b) 10-37 C, (c) 10-47 C, (d) 10-20 C], [Ans:(b)][ K, , (∆e)2, =, r2, , m2, , 6.67 x 10−11, 9 x 109, , G r2 ; ∆e =1.67𝑥10−27 √, , = 1.44 x 10−37C, , XXXXXXXXXXXXX, , HSPTA KANNUR, , 𝟏𝟎−𝟑𝟕 C]
