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Review -Field Lines from a Point Charge The electric field lines from a point charge extend out radially Physics for Scientists & Engineers 2 For a positive point charge, the field lines point outward Spring Semester 2005 • Terminate at infinity Lecture 5 For a negative charge, the field lines point inward • Originate at infinity January 19, 2005 Physics for Scientists&Engineers 2 1 Electric Field from a Point Charge January 19, 2005 Physics for Scientists&Engineers 2 Electric Field from Three Point Charges Suppose we have two charges, q and q0 , separated by a distance r, the electric force between the two charges is 1 qq0 F= 4!" 0 r 2 Consider three charges We can consider q0 to be a test charge, and we can define the electric field from charge q as The three charges are placed at q1 : ( a, 0 ) q2 : ( 0, 0 ) q3 : ( 0,b ) What is the electric field at point P? E= q1 = 1.50 µC q2 = 2.50 µC q3 = !3.50 µC F 1 q = q0 4!" 0 r 2 The electric field is a vector and to add electric fields we must add the components separately January 19, 2005 Physics for Scientists&Engineers 2 2 3 January 19, 2005 Physics for Scientists&Engineers 2 4 1 Electric Field from Three Point Charges (2) Electric Field from Three Point Charges (3) Now we add the components The electric field at P due to q1 is ! q ! E1 = k 12 ex b Ex = k The electric field at P due to q3 is q " kq sin ! % " q sin ! q3 % Ey = $ 22 2 ' + k 32 = k $ 22 + # a +b & # a + b 2 a 2 '& a ! q ! E3 = k 32 ey a # a& # 8.00 m & ! = tan "1 % ( = tan "1 % = 53.1° $ b' $ 6.00 m (' The electric field at P due to q2 is a 2 + b 2 = ( 8.00 m ) + ( 6.00 m ) = 100 m 2 2 ! " kq cos! % ! " kq sin ! % ! E2 = $ 22 2 ' ex + $ 22 2 ' ey # a +b & # a +b & 2 ! is given by tan -1 ( a / b ) January 19, 2005 # ( 2.50 µC ) sin ( 53.1° ) "3.50 µC & Ey = 8.99 !10 9 % + = "311 N/C 100 m 2 ( 8.00 m )2 (' $ Physics for Scientists&Engineers 2 5 Now we use the components to get the magnitude ( 509 N/C) Physics for Scientists&Engineers 2 6 A system of two oppositely charged point particles is called an electric dipole The vector sum of the electric field from the two charges gives the electric field of the dipole E = Ex2 + Ey2 E= January 19, 2005 Electric Field from an Electric Dipole Electric Field from Three Point Charges (4) 2 2 " 1.50 µC ( 2.50 µC ) cos ( 53.1° ) % Ex = 8.99 !10 9 $ 2 + ' = 509 N/C 100 m 2 # ( 6.00 m ) & a + b is the distance from q2 to point P squared 2 q1 " kq2 cos! % " q1 q2 cos! % +$ ' = k $# 2 + 2 ' b2 # a2 + b2 & b a + b2 & + ( !311 N/C ) = 597 N/C 2 We have shown the electric field lines from a dipole Now the direction # Ey & # "311 N/C & ! = tan "1 % ( = tan "1 % $ 509 N/C (' $ Ex ' ! = "31.5° electric field points to the right and down January 19, 2005 Physics for Scientists&Engineers 2 7 January 19, 2005 Physics for Scientists&Engineers 2 8 2 Electric Field from an Electric Dipole (2) Electric Field from an Electric Dipole (3) Let’s get an expression for the electric field of a dipole along a line including both charges Let’s put both charges on the x-axis a distance d apart We will derive a general expression good anywhere along the dashed line and then get an expression for the electric field a long distance from the dipole January 19, 2005 Physics for Scientists&Engineers 2 • Put -q at x = -d/2 • Put +q at x = +d/2 Calculate the electric field at a point P a distance x from the origin 9 January 19, 2005 Physics for Scientists&Engineers 2 Electric Field from an Electric Dipole (4) Electric Field from an Electric Dipole (5) The principle of superposition tells us that the electric field at any point x is the sum of the electric field from +q and -q Rearranging we get E = E+ + E! = E= January 19, 2005 Let’s look at this equation far away (x >> d) 1 q 1 q 2 # 2 4!" 0 $ 4!" 0 $ 1 ' 1 ' &% x # d )( &% x + d )( 2 2 Physics for Scientists&Engineers 2 #2 #2 q $$ d ' d ' ' $ & 1 # )( # &% 1 + )( ) 4!" 0 x 2 &% % 2x 2x ( This equation gives the electric field everywhere on the x axis except for x = ±d 1 q 1 !q + 4"# 0 r+2 4"# 0 r!2 Replacing r+ and r- we get E= 10 E! q %% d d ( % (( ' 1 + $ ...*) $ '& 1 $ + ...*) *) 4"# 0 x 2 '& & x x E! q $ 2d ' qd & )= 4"# 0 x 2 % x ( 2"# 0 x 3 or 11 January 19, 2005 Physics for Scientists&Engineers 2 12 3 Definition of Electric Dipole Example - Electric Dipole Moment of Water We can define the vector electric dipole moment as a vector that points from the negative charge to the positive charge ! ! p = qd • q is the magnitude of one of the opposite charges • d is the distance between the charges Using this definition we can write the electric field far away from an electric dipole as p E= 2!" 0 x 3 Physics for Scientists&Engineers 2 The charge distributions of the atoms are approximately spherical and the hydrogen atoms are arranged such that they make an angle of 105° with each other Suppose we approximate the water molecules as two positive charges located at the center of the hydrogen atoms and two negative charges located at the center of the oxygen atom. What is the electric dipole moment of a water molecule? • p is the magnitude of the dipole moment January 19, 2005 Everyone knows that water is H2O 13 January 19, 2005 Physics for Scientists&Engineers 2 14 Electric Dipole Moment of Water (2) We assume that the center of mass of the two hydrogen atoms is halfway between the two atoms and that the two positive charges are effectively located there The distance between the these two positive charges and the two negative charges assumed at the center of the oxygen atom is d = (10 !10 m)cos 52.5° = 0.6 "10 !10 m Our result for the electric dipole moment of water is then p = 2ed = (3.2 !10 "19 C)(0.6 !10 "10 m) = 2 !10 "29 C m This oversimplified result is close to the known value of 6.2⋅ 10-30 C ⋅ m • Assumed charge distribution not what actually happens January 19, 2005 Physics for Scientists&Engineers 2 15 4