Lecture 11 Electrostatic Potential

Transcription

1 Lecture 11 Electrostatic Potential Sections: 4.1, 4.2, 4.3, 4.4 Homework: See homework file

2 differential work Work: Definition W = F L dw = F dl= FdLcosα total work done from to W = F d L, J dl α L F α work is positive if the applied force is in the direction of displacement (α < 90 ) L F W = F L> 0 work is negative if the applied force is opposite to the direction of displacement (α > 90 ) F L W = 0 W = F L< 0 L F work is zero if force is orthogonal LECTURE 11 slide 2 F

3 Work and Energy work results in change of stored energy work and energy are both measured in joules LECTURE 11 slide 3

4 Work: Examples with Gravity (Uniform Motion) F g F m L α F W = F L> m m 0, man does work (expends energy) W = F Lcosα < 0, rock gains energy g g Wm = Wg (uniform motion, no acceleration) F = F cosα m = F, F = F m m g g g F m W = F Lcosα > g m g W = F L< m 0 0, gravity does work (rock loses energy) man provides enough friction to prevent acceleration, gains energy (heat) α F g L LECTURE 11 slide 4

5 Calculating Work: Line Integration 1 we have already studied the scalar line integral (see L06) Q = ρ dl l the work integral is also a line integral but it involves a dot product W = F d L r( u) = xu ( ) a + yu ( ) a + zu ( ) a dl= dr = dxa + dya + dza dw = F dl= F dx + F dy + F dz x y z x y z F= F a + F a + F a x x y y z z x y z W = ( Fxdx + Fydy + Fzdz) x dw z differential work, elementary work r( u) dr LECTURE 11 slide 5 y

6 Calculating Work: Line Integration 2 RCS: W = F x( x, y, z) dx + Fy( x, y, z) dy + Fz( x, y, z) dz CCS: W= Fρ( ρφ,, zd ) ρ+ Fφ( ρφ,, z) ρdφ+ Fz ( ρφ,, zdz ) SCS: W = Fr (, r θφ, ) dr+ Fθ(, r θφ, ) rdθ+ Fφ(, r θφ, )sin r θ dφ use the parametric equations of the line to reduce the line integral to a single-variable integral; for example, in RCS dlξ x= lx( u), y = ly( u), z = lz( u) dξ = du, ξ x, y, z du u dlx dly dlz W = Fx + Fy + Fz du du du du u dla r ρ dla φ Laz d d La dla θ d La φ change of variable LECTURE 11 slide 6

7 Calculating Work: Line Integration 3 Example: Evaluate the integral W F= ( x+ y) a + ( y x) a x y = F d L along the parabola y 2 = x from (1,1) to (4,2). C u dlx dly dlz W = Fx + Fy + Fz du du du du u Let y = u; x = u2 dl x 2 u, dl y 1 du = du = egin and end points: u = 1, u = W = ( u + u)2 u + ( u u ) du 1 1 W = Fx = x+ y Fy = y x F = 0 z y x LECTURE 11 slide 7

8 Calculating Work: Example 2 Repeat the example from the previous slide but this time take the integration along the straight line from (1,1) to (4,2). LECTURE 11 slide 8

9 Work of External Force to Move Charges electric force FE = QE to move a charge a distance dl against the electric force, external (non-electric) force must be applied; its projection along dl is Fext = FE al = QEa L Example charging a battery: external force does work when moving the charge from to (the system of electric charges gains potential energy, i.e. it is being charged ; the work done by the electric field is W E < 0) ext Fext L E L E 0 W d Q d W = = = > F E F ext Q = QE + displacement LECTURE 11 slide 9

10 Work of Electric Field to Move Charges Example discharging a battery: if charge is already at and is free to move, the electric field brings it back to at the expense of its own energy (electric system of charges loses potential energy, work done by electric field W E > 0) E W Q E dl Q E dl = = > W E dl= E dl ext 0 F E = QE + displacement LECTURE 11 slide 10

11 Electrostatic Potential Difference (Voltage) Definition 1: potential difference V is the work done by the electric field in moving a unit test charge from point to point V = E d L, V [V] = [J/C] Definition 2: potential difference V is the work done by the external forces in moving a unit test charge from point to point V the two definitions are equivalent = E d L, V LECTURE 11 slide 11

12 Example: Potential Difference Due to Line Charge Find the voltage V between point (ρ,φ,z ) and point (ρ,φ,z ) in the field of a line charge ρ l along the z axis. Take the path along the line segments: 1) ρ = ρ, φ φ φ, z = z 2) ρ = ρ, φ = φ, z z z 3) ρ ρ ρ, φ = φ, z = z V = E dl = E( ρ, φ, z) aφ ρdφ+ E( ρ, φ, z) a dz + z ρ ρ E( ρφ,, z ) a φ φ ρ 0 z 0 dρ z LECTURE 11 slide 12 1 ρl E= a 2πε ρ ρl ρ note independence of angular position! V = ln 2πε ρ Can we choose ρ or ρ at infinity or at zero? ρ

13 Example: Potential Difference in the Field of Line Charge 2 LECTURE 11 slide 13

14 Electrostatic (bsolute) Potential Definition 1: the potential at a point is the work done by the electric field in moving a unit test charge from point to a reference point 0 at which the potential is assumed equal to zero Definition 2: the potential at a point is the work done by an external force in moving a unit test charge from a reference point 0, where the potential is assumed equal to zero, to the point. 0 V E dl E dl, V = = the electrostatic potential is the potential energy of a unit test charge 0 LECTURE 11 slide 14

15 bsolute Potential: Example The E field is defined as E(x,y,z) = 6x 2 a x + 6ya y + 4a z. (a) Find the voltage V MN if M(2,6, 1) m and N( 3, 3,2) m. (b) Find the absolute potential V M if V = 0 at Q(4, 2, 35) m. (c) Find the absolute potential V N if V = 2 V at P(1,2, 4) m. LECTURE 11 slide 15

16 V = Potential of a Point/Spherical Charge assume potential is zero at infinity integrate along any radial line Q V = E dl= a 2 R a 4 R dr πε R R dl Q 1 4πε R R V = Q 1 4πε R E + R the potential is the same for all points on the sphere of radius R we say that the sphere is an equipotential surface notice the 1/R dependence on distance and compare it with E LECTURE 11 slide 16

17 You have learned: how to calculate work/energy integrals what we mean by positive and negative work what potential difference (voltage) is in field theory what the potential difference is in the field of a line charge what absolute potential is that the absolute potential of a point/spherical charge depends on distance as 1/R unlike E which decreases as 1/R 2 LECTURE 11 slide 17