Tiêu đề 1. ELECTRIC CHARGES AND FIELDS.pdf ✅

TOPIC-1 Electric Field and Dipole Concepts Covered:  Electric charge and their properties  Coulomb's law and the principle of superposition  Electric field, field lines and dipole  Electric field due to dipole, torque on a dipole Revision Notes Electric Charge: Electric Charge is the amount of energy or electrons that pass from one body to another by different modes like conduction, induction,etc. There are two types of electric charges, positive charges and negative charges. Properties of Electric Charge Addition of charges  If a system contains n point charges q1, q2, q3, then the total charge of the system is q = q1 + q2 + q3 + .........+ qn Conservation of charges  The sum of positive and negative charges present in an isolated system, always remains constant.  Charge can neither be created nor destroyed but only exists in positive-negative pairs and only can transfer from one body to another. Quantization of charges  Electric charge is always quantized.  Net charge qnet of an object having Ne electrons, Np protons and Nn neutrons is: qnet = – eNe + eNp + 0Nn = e(Np – Ne) = ± ne, where n is an integer Example-1 If 109 electrons move out of a body to another body every second, how much time is required to get a total charge of 1C on the other body? Sol. 109 electron will take 1s So, 1 electron will take 1 109 s Since, 1C= 6.242 × 1018 electrons ∴ 6.242 ×1018 electrons will take = 6 242 10 10 18 9 . × s = 6.242 ×109 s = 6 242 10 60 60 24 365 9 . × × × × years =197.93 years ≈ 198 years If 109 electrons move out of a body to another body every second, then it will take approx. 198 years to get a total charge of 1C on the other body. Coulomb’s Law  The force of attraction or repulsion between two point charges q1 and q2 separated by a distance r is directly proportional to product of magnitude of charges and inversely proportional to square of the distance between charges, written as: F k q q r = ' | || | 1 2 2 where, F = Force of attraction/repulsion between charges q1 and q2. q1, q2= Magnitudes of charges r = Distance between charges q1, q2 k' = Constant whose value depends on medium where charges are kept When the charges are kept in vacuum, then Learning Objectives After going through this chapter, the students will be able to:  Define electric charge and its properties.  Apply Coulombs' law and principle of superposition.  Understand electric field, field lines, and electric dipole and torque on dipole.  Know about Gauss theorem and its applications. 1 CHAPTER ELECTRIC CHARGES AND FIELDS List of Topics Topic-1: Electric Field and Dipole Topic-2 : Gauss's Theorem and its Applications

CBSE Question Bank Chapterwise & Topicwise, PHYSICS, Class-XII Example-3 Consider three charges q1, q2, q3 each equal to q at the vertices of an equilateral triangle of side l. What is the force on a charge Q (with the same sign as q) placed at the centroid of the triangle, as shown in Fig? Sol. As we know F = q q r 1 2 2 As per the problem we have three charges, q1 = q2 = q3 = q, at the vertices of an equilateral triangle of side I and a charge Q which is of same sign as q is placed at the centroid of the triangle. Now as per fig. ABC is an equilateral triangle having side length I, now if we draw a perpendicular line AD to the side BC we will get the angle between line AC and AD is 30°, as it an equilateral triangle its each angle is equals to 60°. Hence, AD can be written as, AD = AC cos 30° We know, cos 30° = 3 2 and AC = l Now, AD = 3 2 l Now in an equilateral triangle the distance of the centroid is two thirds of the perpendicular line passing through it that is AD. Now we can write the distance of AO of the centroid O from A as 2 3 AD. Similarly, AO = BO = CO = 1 3 l Now putting the value of AD from the above calculation we will get, 2 3 AD = 2 3 3 2 l       Cancelling the common terms we will get, 2 3 AD = 1 3 l Now we can write each of the forces acting on the charge Q as F1 be the force on Q due to charge q at A. F1 = k Q (AO) q 2 along AO Now put value of AO we will get, F1 = k Qq l 1 3 2       ⇒ F1 = 3k Qq l 2 F2 be the force on Q due to charge q at B. F2 = k Qq (BO)2 along BO Now put value of BO we will get, F2 = k Qq l 1 3 2       ⇒ F2 = 3k Qq l 2 F3 be the force on Q due to charge q at C. F3 = k Qq (CO)2 along CO Now put value of CO we will get, F3 = k Qq l 1 3 2       ⇒ F3 = 3k Qq l 2 Now using theorem of parallelogram to find the resultant between the two forces F2 and F3. As it is an equilateral triangle the two forces F2 and F3 makes an angle of 120°. Hence the resultant will be, F F F F o 2 2 3 2 2 3 + + 2 cos120 = 3 3 2 3 3 1 2 2 2 2 2 2 2 k q l k q l k q l k q l  Q Q Q Q      +       +             −       Now on further simplification we will get the resultant of force F2 and F3 is 3k Qq l 2 which is along OA. Now the total force acting on Q will be, Fnet = FAO + FOA Where, FAO is the force along AO and FOA is the force along OA. Now putting the respective values we will get, Fnet = Qq l r k Qq l r 2 2 3   + −( ) \ Fnet = 3 2 k Qq l ( ) r r  −
Electric Charges and Fields Where r  is the unit vector along AO. It is clear also by symmetry that the three forces will sum to zero. Electric field The space around a charge in which a test charge experiences an electric force is called electric field. If a test charge q0 is placed at a point where electric field is E, then force on the test charge is F = q0E The electric field strength due to a point source charge ‘q’ at distance ‘r’ is given by: E q r r → → = 1 4 0 3 πε . or E q r = 1 4 0 2 πε . Electric field lines Electric field lines are imaginary lines that originates from the positive charge and terminates at negative charge. The electric field lines never intersect each other. Strength of electric field is encoded in density of field lines. Example-4 Two point charges q1 and q2, of magnitude + 10–8 C and –10–8 C, respectively, are placed 0.1 m apart. Calculate the electric fields at points A, B and C shown in Fig. Sol. The electric field vector E1A at A due to the positive charge q1 points towards the right and has a magnitude EIC EC E2C EM q1 q2 0.1 m 0.1 m C 0.05 m 0.05 m 0.05 m – B + A EA E1A = ( ) ( C) ( . m) 9 10 10 0 05 9 2 8 2 × × − NmC = 3.6104 NC–1 The electric field vector E2A at A due to the negative charge q2 points towards the right and has the same magnitude. Hence the magnitude of the total electric field EA at A is EA = E1A + E2A = 7.2 × 104 NC–1 EA is directed toward the light. The electric field vector E1B at B due to the positive charge q1 points towards the left and has a magnitude E1B = ( ) ( C) ( . m) 9 10 10 0 05 9 2 8 2 × × − − Nm C 2 = 3.6 × 104 NC–1 The electric field vector E2B at B due to the negative charge q2 points towards the right and has a magnitude E1B = ( ) ( C) ( . m) 9 10 10 0 15 9 2 2 8 2 × × − − Nm C = 4 × 103 NC–1 The magnitude of the total electric field at B is EB = E1B – E2B = 3.2 × 104 NC–1 EB is directed towards the left. The directions in which these two vectors point are indicated in Fig. The resultant of these two vectors is EC = E1 cos(p) 3 + E2 cos p 3 = 9 × 10–3 NC–1 Ec points towards the right. Electric Dipole The system formed by two equal and opposite charges separated by a small distance is called an electric dipole. The force on a dipole in a uniform electric field is zero in both stable as well as unstable equilibrium. The potential energy of a dipole in an uniform electric field is minimum for a stable equilibrium and maximum for an unstable equilibrium. Torque on a dipole [Board, 2023]  In a dipole, when the net force on dipole due to electric field is zero and center of mass of dipole remains fixed, the forces on charged ends produce net torque t about its center of mass. t = F dsin q = qE dsin q = pEsin q t = p E ×