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Ass.Prof. HALA MOUSTAFA Definition of the electric field E Calculating E due to a charged particle To find E for a group of point charges Electric field lines Motion of charge particles in a uniform electric field The electric dipole in electric field The direction of E If Q is +ve the electric field at point p in space is radially outward from Q If Q is -ve the electric field at point p in space is radially inward toward Q يكون اتجاة المجال الكهربائي الناتج عن شحنة موجبة في اتجاة الخروج من الشحنة. يكون اتجاة المجال الكهربائي الناتج عن شحنة سالبة في اتجاة الدخول على الشحنة. قيمة المجال E=Kq/r2 اتجاة المجال في اتجاة الخروج The lines must begin on positive charges (or infinity) خطوط المجال تبدا من الشحنة الموجبة وتنتهي عند السالبة The lines must end on negative charges (or infinity) خطوط المجال تنتهي عند الشحنة السالبة The no of lines leaving a+ve charge (approaching a-ve charge)is proportional to the magnitude of the charge Electric Field lines cannot cross The e field two charges (positive) The Electric Field two charges (negative) These are imaginary lines pointing outward from positive charge and inward to the negative charge Electric Field It is the space around the electric charge in which the electrical effects appear. Electric Field Strength at Apoint It is the force per unit charge at the point.SI unit is )C/N( E=F/Q )N/C( It is defined as the work done to move a positive unit charge from one point to another If WAB is the work done to move a charge Q from point A to point B ,then the potential difference V is given by: VAB = (J/C) or Volts WAB =QVAB (J) The work done ΔW to move a charge Q against an electric field of strength E a distance ΔX is given by: ΔW=F ΔX F=QE ΔW=QE ΔX ΔV= ΔW = QEΔX=EΔX Q Q E= ΔV (V/m) ΔX The electric dipole is positive charge and negative charge of equal magnitude المناظر لألبرة المغناطيسية في الكهربائية بمعنى الشحنة الموجبة والشحنة السالبة متساويتا We start with special case of a point charge in uniform electric field E. The electric field will exert a force on a charged particle is given by F = qE The force will produce acceleration a = F/m where m is the mass of the particle. Then we can write F = qE = ma The acceleration of the particle is therefore given by a = qE/m If the charge is positive, the acceleration will be in the direction of the electric field. If the charge is negative, the acceleration will be in the direction opposite the electric field. F=QE qE عند وضع شحنة كهربائية كتلتها في مجال كهربي فان المجال الكهربائي يبذل قوة مقدارها A charge of 2μC experiences a force of 34 N when placed at a point in an electric field.Calculate the electric field strength at the point? Q = 2μC= 2x 10-6 C F = 34 N E=F/Q (N/C) E=34 N /2x 10-6 C = 68x 10-6 (N/C) The work done in bringing a charge of 4 C from one point to another is 10 J .What is the potential difference between the two points? Q=4C W = 10 J V = W = 10 = 2.5 V Q 4 The potential difference between two points is 12 V.Find the work done to move a charge of 8 C between the two points? V = 12 C Q=8C W=QV = 8x12 = 96 (J) If the potential difference between two parallel plates separated by a distance 1 mm is 400 v,Find the electric field intensity between the two plates.? ΔX = 1mm = 1 x 10-3 m ΔV = 400 (V/m) E = ΔV = 400 = 4 x 105 (V/m) ΔX 1 x 10-3 Acharge of 2µC experiences a force of 34 N when placed at a point in an electric field.calculate the electric field strength at the point? F=34 N q = 2µC E=F/q E = 34 = 1.7x107 ( N/C) 2x10-6 Calcullate the magnitude of the force on charge of 6 µC when placed in an electric field of strength 4 x10-3 NC-1? E=F/q F= Eq F = (4x10-3) (5x10-6) = 2x10-8 N Find the electric field due to electric dipole along x-axis at point p, which is a distance r from the origin, then assume r>>a The electric dipole is positive charge and negative charge of equal magnitude placed a distance 2a apart as shown in figure 3.6 Find the electric field at point p in figure due to the charges shown. نرقم الشحنات-1 نحدد المجال الناتج عن كل شحنة عند النقطة-2 E نوجد Eh