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Transcript
Chapter 4 Energy and Potential 4.1 Energy to move a point charge through a Field • Force on Q due to an electric field FE QE • Differential work done by an external source moving Q dW QE dL • Work required to move a charge a finite distance 4.2 Line Integral • Work expression without using vectors W Q final E L dL initial EL is the component of E in the dL direction • Uniform electric field density W QE L BA Example y E( x y ) x 2 .8 A .6 1 Q 2 1 B 0 1 2 x y Path: 2 1 z 1 Calculate the work to cary the charge from point B to point A. W A A A 0 1 2 0 1 2 Q E( x y ) d x Q E( x y ) d y Q E( x y ) d z 0 1 2 B B B Plug path in for x and y in E(x,y) A A A 0 1 2 2 2 W Q E 0 1 x 0 d x Q E 1 y 0 1 d y Q E( 0 0) d z 2 B B 0 1 B2 W 0.96 Example y E( x y ) x 2 Q 2 .8 A .6 1 1 B 0 1 Path: 3( x 1) z y 1 (straight line) Calculate the work to cary the charge from point B to point A. W A A A 0 1 2 0 1 2 Q E( x y ) d x Q E( x y ) d y Q E( x y ) d z 0 1 2 B B B Plug path in for x and y in E(x,y) A 0 W Q E[ 0 3( x 1) ] d x Q 0 B 0 A1 B1 y E 1 0 1 d y Q 3 A2 E( 0 0) d z W 0.96 B2 • Same amount of work with a different path • Line integrals are path independent 2 4.3 Potential Difference • Potential Difference • Using radial distances from the point charge 4.3 Potential • Measure potential difference between a point and something which has zero potential “ground” V AB VA VB Example – D4.4 6x2 E( x y z) 6y 4 a) Find Vmn 2 M 6 1 3 N 3 2 M0 VMN N 0 M1 6x d x N 2 1 M2 6y d y N 4 dz 2 b) Find Vm if V=0 at Q(4,-2,-35) 4 Q 2 35 M0 VM Q 0 M1 6x d x Q 2 1 M2 6y d y Q 4 dz VM 120 2 c) Find Vn if V=2 at P(1, 2, -4) 1 P 2 4 N0 VN P 0 N1 2 6x d x P 1 N2 6y d y P 4 dz 2 2 VN 19 VMN 139 4.4 Potential Field of a Point Charge • Let V=0 at infinity • Equipotential surface: – A surface composed of all points having the same potential Example – D4.5 2 P1 3 1 9 Q 15 10 Q is located at the origin 12 0 8.85 10 a) Find V1 if V=0 at (6,5,4) 6 P0 5 4 V1 1 1 P0 4 0 P1 Q V1 20.677 b) Find V1 if V=0 at infinity V1 Q 1 V1 36.047 4 0 P1 c) Find V1 if V=5 at (2,0,4) 2 P5 0 4 V1 1 1 5 P5 4 0 P1 Q V1 10.888 Potential field of single point charge Q1 V ( r) 4 0 r r1 A |r - r1| Q1 Move A from infinity Potential due to two charges V( r) Q1 4 0 r r1 Q2 4 0 r r2 A |r - r1| Q1 |r - r2| Q2 Move A from infinity Potential due to n point charges Continue adding charges V( r) Q1 4 0 r r1 Q2 4 0 r r2 n V( r) m1 .... Qm 4 0 r r m Qn 4 0 r r n Potential as point charges become infinite Volume of charge Line of charge Surface of charge V( r) v r prime d v prime 4 r r 0 prime V( r) L r prime d L prime 4 r r 0 prime V( r) S r prime d S prime 4 r r 0 prime Example Find V on the z axis for a uniform line charge L in the form of a ring V( r) L r prime d L prime 4 r r 0 prime Conservative field No work is done (energy is conserved) around a closed path KVL is an application of this 4.6 Potential gradient Relationship between potential and electric field intensity V=- Ed dL Two characteristics of relationship: 1. The magnitude of the electric field intensity is given by the maximum value of the rate of change of potential with distance 2. This maximum value is obtained when the direction of E is opposite to the direction in which the potential is increasing the most rapidly Gradient • The gradient of a scalar is a vector • The gradient shows the maximum space rate of change of a scalar quantity and the direction in which the maximum occurs • The operation on V by which -E is obtained E = - grad V = - V Gradients in different coordinate systems The following equations are found on page 104 and inside the back cover of the text: gradV gradV gradV V x V V r a x a a r V y a y V z a z 1 V V a a z z 1 V 1 V a a r r sin Cartesian Cylindrical Spherical Example 4.3 Given the potential field, V = 2x2y - 5z, and a point P(-4, 3, 6), find the following: potential V, electric field intensity E potential VP = 2(-4)2(3) - 5(6) = 66 V electric field intensity - use gradient operation E = -4xyax - 2x2ay + 5az EP = 48ax - 32ay + 5az Dipole The name given to two point charges of equal magnitude and opposite sign, separated by a distance which is small compared to the distance to the point P, at which we want to know the electric and potential fields Potential To approximate the potential of a dipole, assume R1 and R2 are parallel since the point P is very distant V V Q 4 0 1 R1 Q d cos 2 4 0 r R2 1 Dipole moment The dipole moment is assigned the symbol p and is equal to the product of charge and separation p = Q*d The dipole moment expression simplifies the potential field equation Example An electric dipole located at the origin in free space has a moment p = 3*ax - 2*ay + az nC*m. Find V at the points (2, 3, 4) and (2.5, 30°, 40°). 3 10 9 p 2 10 9 9 1 10 12 0 8.854 10 2 P 3 4 V p 4 0 P 2.5 30 Pspherical 180 40 180 2 P P Transform this into rectangular coordinates 2.5 sin 30 cos 40 180 180 sin 40 Prectangular 2.5 sin 30 180 180 2.5 cos 30 180 V p 4 0 Prectangular V 0.23 2 Prectangular Prectangular 0.958 Prectangular 0.803 2.165 V 1.973 Potential energy Bringing a positive charge from infinity into the field of another positive charge requires work. The work is done by the external source that moves the charge into position. If the source released its hold on the charge, the charge would accelerate, turning its potential energy into kinetic energy. The potential energy of a system is found by finding the work done by an external source in positioning the charge. Empty universe Positioning the first charge, Q1, requires no work (no field present) Positioning more charges does take work Total positioning work = potential energy of field = WE = Q2V2,1 + Q3V3,1 + Q3V3,2 + Q4V4,1 + Q4V4,2 + Q4V4,3 + ... Manipulate this expression to get WE = 0.5(Q1V 1 + Q2V2 + Q3V3 + …) Where is energy stored? The location of potential energy cannot be precisely pinned down in terms of physical location - in the molecules of the pencil, the gravitational field, etc? So where is the energy in a capacitor stored? Electromagnetic theory makes it easy to believe that the energy is stored in the field itself