University:
California State University, Northridge
Course:
PHYS 220B | Electricity and Magnetism
Academic year:

2008
Views:

215

Pages:

22
Author:

talesofmutiesx1k

20-1 Electric Potential Energy and the Electric Potential Gravitation field Electric field Figure 20-1 Change in Electric Potential Energy In figure 20-1, a positive charge q0 is placed in the field. If the charge moves upward a distance d, the work done by the electric force is negative W = - Fd = - q0Ed The change of the potential energy is ΔU = - W = q0E d (20-1) Definition of Electric potential, V ΔU −W ΔV = = q0 q0 SI Unit: joule/Coulomb = volt, V Note: 1) ΔU = q0 ΔV 2) 1 V=1 J/C 3) Electron volt (eV): potential energy 1 eV=1.6x10-19 J (20 − 2) 20-5 Potential Energy and Potential Difference. A computer monitor accelerates electrons and directs them to create image. If the accelerating plates are 1.05 cm apart, and have a potential difference of 25,500 V, find the change in electric potential energy for an electron that moves from one accelerating plate to the other in the computer monitor. Solution: ΔU = q0 ΔV = ( − e )( 25,500 V ) = −25,500 eV = −25,500 eV × 1.60 × 10−19 J/eV = 4.08 × 10−15 J Electric Field and the Rate of Change of Electric Potential There is a relationship between electric field and electric potential. In Figure 20-2, the work done by the electric field is W = F x ΔS = q0E x ΔS Therefore, the change in electric potential is (q0 E Δs ) −W ΔV = =− = − E Δs q0 q0 Figure 20-2 Electric Field and Electric Potential Electric field and electric potential ΔV E=− Δs (20 − 4) SI unit: volts/meter, V/m Note: 1) 1 N/C = 1 V/m since E has the unit of N/C 2) The electric field E depends the rate of the change of the electric potential with position; 3) The electric potential decreases as one moves in the direction of the electric field. 4) Δs is along the electric field direction. 20-1 Electric Field and Potential Energy A uniform electric field of magnitude 4.1x10-5 N/C points in the positive x direction. Find the change in electric potential energy of a 4.5 μC charge as it moves from the origin to the points: (a) (0, 6.0m); (b) (6.0, 0); and (c) (6.0m, 6.0m). 20-2 Energy Conservation If there is no other resistant force (such as air resistance, friction), for a charged object in an electric field, the total energy must be conservative: The sum of its kinetic and electric potential energies must be the same at any two points (A and B): KA + UA = KB + UB Expressed the kinetic energy as mv2/2, we have 1 2 1 2 mv A + U A = mvB + U B 2 2 The express applies to any conservation force: • For Gravitational field: U=mgy • For electric potential energy: U= qV Energy conversation can also be finally expressed as 1 2 1 2 mv A + q (VA − VB ) = mvB 2 2 Hint: electron has less mass than the proton. 20-19 Energy Conservation A particle with a mass of 3.5 g and a charge of +0.045 μC is located in an electric field of 1200N/C. If it is release from rest at point A in the following figure. (a) In which direction will this charge move? (b) What speed will it have after moving through a distance of 5.0m? Summary 20-1 Electric potential energy and the electric potential 1) The Change of electric potential energy, ΔU ΔU= - W = - (-Fd) = (q0E)d 2) Electric potential, V ΔV=ΔU/q0 3) Electric field and electric potential E=− ΔV Δs 20-2 Energy conversation At two points, kinetic and electric potential energy must be conservative at two points: 1 2 1 2 mv A + qVA = mvB + qVB 2 2 Exercise 20-1 Find the change in electric potential energy, ΔU, as a charge of (a) 2.20 x10-6 C or (b) -1.10x10-6 C moves from point A to point B, given the change in electric potential is ΔV = VB - VA = 24.0 V Solution (a) Solving ΔV= ΔU/q0, we have ΔU=q0 ΔV= (2.20x10-6 C) (24.0 V) = 5.28x10-5 J (b) For q0= -1.10x10-6 C, we obtain ΔU=q0 ΔV= (-1.10x10-6 C) (24.0 V) = - 2.64x10-5 J Example 20-1 Plates at different potentials A uniform electric field is established by connected the plates of a parallel-plate capacitor to a 12-V battery. (a) If the plates are separated by 0.75 cm, what is the magnitude of the electric field in the capacitor? (b) A a charge of +6.24x10-6 C moves from the positive plate to the negative plate. Find the change in electric potential energy for this charge. Picture the problem: Example 20-1 Plates at Different Potentials Solution a) Electric field (ΔV=-12V) E=− ΔV (−12V ) =− = 1600V / m Δs 0.0075m b) The change in electric potential energy ΔU= q ΔV =(6.24x10-6 C)(-12 V)= - 7.5x10-5 J Active Example 20-1 Electric field and potential difference The electric potential at point B in the parallel-plate capacitor shown below is less than that at point A by 4.5 V. The separation between A and B is 0.120 cm, and the separation between the plates is 2.55 cm Find (a) the electric field within the capacitor and (b) the potential difference between the plates. Active Example 20-1 Find the Electric Field and the Potential Difference Solution a) Electric field since ΔV= VB-VA= - 4.5 V E = - ΔV/ΔS = - ( -4.5 V)/0.00120m= 3750 V/m b) The potential different between the plates ΔV = - EΔS = 3750 V/m x 0.0255m = -95.6 V Active Example 20-2 Find the speed with a running start A particle of mass m=1.75x10-5 kg and charge q=5.20x10-5 C has a speed of 5.00 m/s at point A. As the particle moves to position B, the electric potential decreases by 60.0 V; What is the speed at point B? Solution 1) According to the energy conservation: 1 2 1 2 mvB = mv A + q (VA − VB ) 2 2 2) Solve for vB: 1 2 1 2 mvB = mv A + q (VA − VB ) 2 2 1 1 (1.75 × 105 kg ) × vB2 = (1.75 × 105 kg ) × (5.00m / s ) 2 + 5.20 × 10−5 C × (60.0V ) 2 2 vB = 19.5 m / s = 1.95 m / s Example 20-2 From Plate to Plate A particle of charge q=6.24x10-6 C is released from rest at the positive plate and it reaches the negative plate with a speed of 3.4 m/s. The plates is connected to a 12-V battery. What is (a) the mass of the charge and (b) its final kinetic energy? Example 20-2 From Plate to Plate

Electric Potential Energy and Electric Field

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