Electric Potential. 1/22/14 Physics for Scientists & Engineers 2, Chapter 23 1

Transcription

1 Electric Potential 1/22/14 Physics for Scientists & Engineers 2, Chapter 23 1

2 Announcements! First exam is next Tuesday, January minute exam during lecture time You can bring a 5 by 8 size cheat sheet note card (filled on both sides) (i.e. ½ letter size piece of paper) Hand-written No computer-printed equation sheets! January 22, 2014 Physics for Scientists & Engineers 2, Chapter 21 2

3 Special Symmetries! We have applied Gauss s Law to a point charge and showed that we get Coulomb s Law! Now let s look at more complicated distributions of charge and calculate the resulting electric field! We will use a charge density to describe the distribution of charge! This charge density will be different depending on the geometry January 22, 2014 Physics for Scientists & Engineers 2, Chapter 22 5

4 Spherical Symmetry! Calculate the electric field from charge distributed as a spherical shell! Assume that we have a spherical shell of charge q with radius r S (gray)! We will assume two different spherical Gaussian surfaces r 2 > r S (purple) i.e. outside r 1 < r S (red) i.e. inside January 22, 2014 Physics for Scientists & Engineers 2, Chapter 22 6

5 Spherical Symmetry! Let s start with the Gaussian surface outside the sphere of charge, r 2 > r S (purple)! We know from symmetry arguments that the electric field will be radial outside the charged sphere! If we rotate the sphere, the electric field cannot change Spherical symmetry! Thus we can apply Gauss Law and get E da = E( 4πr 2 ) 2 = q E = 1 ε 0 4πε 0 q r 2 2 January 22, 2014 Physics for Scientists & Engineers 2, Chapter 22 7

6 Spherical Symmetry! Let s take the Gaussian surface inside the sphere of charge, r 1 < r S (red)! The enclosed charge is zero so E da = E( 4πr 2 ) 1 = 0 E = 0! We find that the electric field is zero everywhere inside spherical shell of charge! Thus we obtain two results The electric field outside a spherical shell of charge is the same as that of a point charge The electric field inside a spherical shell of charge is zero January 22, 2014 Physics for Scientists & Engineers 2, Chapter 22 8

7 Spherical Symmetry: Uniform Distribution! Next, let s calculate the electric field from charge distributed throughout a spherical volume with a uniform charge density ρ > 0 and radius r! We will assume two different spherical Gaussian surfaces r 2 > r r 1 < r (outside, red) (inside, blue)! Let s start with the surface with r 1 < r! From the symmetry of the charge distribution, the electric field is perpendicular to the Gaussian surface everywhere January 22, 2014 Physics for Scientists & Engineers 2, Chapter 22 9

8 Spherical Symmetry: Uniform Distribution! Gauss s Law gives us E da = E( 2 4πr ) 1 = q = ε 0! Solving for E we find ρ 4 3 πr 1 ε 0 3 E inside = ρr 1 3ε 0! The total charge on the sphere is q t = ρv = ρ 4 3 πr 3! The enclosed charge is q = volume inside r 1 volume q t = 4 3 πr 3 1 q 4 t = 3 πr 3 r 1 3 r 3 q t January 22, 2014 Physics for Scientists & Engineers 2, Chapter 22 10

9 Spherical Symmetry: Uniform Distribution! Gauss s Law gives us E da = E( 2 4πr ) 1 = q t! Solving for E we find E inside = q t r 1 4πε 0 r = kq t r 1 3 r 3! Consider r 2 > r ε 0 r 1 E da = E( 2 4πr ) 2 = q t ε 0! Solving for E we find 3 r 3 E outside = q t 4πε 0 r = kq t r 2 same as point charge January 22, 2014 Physics for Scientists & Engineers 2, Chapter 22 11

10 Electric Potential! An electric field has many similarities to a gravitational field! For example, the magnitude of the gravitational force and the electric force are given by F g = G m m 1 2 F r 2 e = k q q 1 2 r 2! Both gravitational and electrostatic forces depend on the inverse square of the distance! Both gravitational and electrostatic forces are conservative! We can define the electric potential energy in analogy with the gravitational potential energy 1/22/14 Physics for Scientists & Engineers 2, Chapter 23 12

11 Electric Potential Energy! For a conservative force, the work is path-independent! When an electrostatic force acts between two or more charges within a system, we can define an electric potential energy, U, in terms of the work done by the electric field, W e, when the system changes its configuration from some initial configuration to some final configuration The change in the electric potential energy is the negative of the work done by the electric field ΔU = U f U i = W e U i is the initial electric potential energy U f is the final electric potential energy 1/22/14 Physics for Scientists & Engineers 2, Chapter 23 13

12 Electric Potential Energy! Like gravitational or mechanical potential energy, we must define a reference point from which to define the electric potential energy! We define the electric potential energy to be zero when all charges are infinitely far apart! We can then write a simpler definition of the electric potential taking the initial potential energy to be zero, ΔU = U f 0 = U = W e,! The negative sign on the work means If E does positive work then U < 0 If E does negative work then U > 0 1/22/14 Physics for Scientists & Engineers 2, Chapter 23 14

13 Constant Electric Field! Let s look at the electric potential energy when we move a charge in a constant electric field! The work done by a constant force is W = F d! For a constant electric field the force is F = q E! The work done by the electric field on the charge is W = q E d = qed cosθ 1/22/14 Physics for Scientists & Engineers 2, Chapter 23 15

14 Constant Electric Field - Special Cases! If the displacement is in the same direction as the electric field W = qed ΔU = qed A positive charge loses potential energy when it moves in the direction of the electric field q d E! If the displacement is in the direction opposite to the electric field W = qed so ΔU = qed A positive charge gains potential energy when it moves in the direction opposite to the electric field d q E 1/22/14 Physics for Scientists & Engineers 2, Chapter 23 16

15 Electric Potential Difference ΔV! The electric potential difference between an initial point i and final point f can be expressed in terms of the electric potential energy of q at each point ΔV = V f V i = U f q U i q = ΔU q! Hence we can relate the change in electric potential to the work done by the electric field on the charge ΔV = W e q! Taking the electric potential energy to be zero at infinity we have V = W e, q 1/22/14 Physics for Scientists & Engineers 2, Chapter 23 19

16 The Volt! The commonly encountered unit joules/coulomb is called the volt, abbreviated V, after the Italian physicist Alessandro Volta ( ) 1 V = 1 J 1 C! With this definition of the volt, we can express the units of the electric field as [E]= [F] [q] = N C = J/m C = V m! For the remainder of our studies, we will use the unit V/m for the electric field 1/22/14 Physics for Scientists & Engineers 2, Chapter 23 20

17 Energy Gain of a Proton! A proton is placed between two parallel conducting plates in a vacuum! The potential difference between the two plates is 450 V! The proton is released from rest close to the positive plate PROBLEM! What is the kinetic energy of the proton when it reaches the negative plate? SOLUTION! The electric potential difference between the plates is 450 V 1/22/14 Physics for Scientists & Engineers 2, Chapter 23 22

18 Energy Gain of a Proton! We can relate the electric potential difference across the plates to the change in electric potential energy ΔV = ΔU q! All of the electric potential energy lost by the proton in crossing between the two plates is converted into kinetic energy! Conservation of energy gives us ΔK + ΔU = 0 ΔK = ΔU = qδv = K f 0 = K f K f = qδv = ( C) ( 450 V) = J 1/22/14 Physics for Scientists & Engineers 2, Chapter 23 23

19 Batteries! A common method of creating an electric potential difference is a battery.! A battery uses chemical processes to maintain a constant potential difference between its anode ( ) and cathode (+).! Lithium ion batteries are the focus of modern research. 1/22/14 Chapter 23 24

20 Battery-Powered Cars! The Tesla sports car uses a 53 kw h lithium ion battery that is charged to 80% of capacity and discharged to 20% capacity.! A typical gasoline powered car carries 50 L of gasoline, which has an energy content of 34.8 MJ/L. PROBLEM:! How does the available energy in the Tesla sports car compare with the energy carried by a gasoline-powered car? SOLUTION:! The energy available in the Tesla battery is: E electric = 60% ( )( 53 kw h) 1000 W 1 kw 3600 s = J =114 MJ 1 h 1/22/14 Chapter 23 25

21 Battery-Powered Cars! The energy available in the gasoline-powered car is: 34.8 MJ E gasoline = ( 50 L) 1 L =1740 MJ! So the gasoline-powered car carried 15 times more energy than the battery-powered car.! However, the typical efficiency of a gasoline-powered car 20% while it is 90% for a battery-powered car, so the comparison of usable energy is: E electric,usable = ( 90% )( 114 MJ)=103 MJ E gasoline,usable = ( 20% )( 1740 MJ)= 348 MJ 1/22/14 Chapter 23 26

22 Van de Graaff Generator! A Van de Graaff generator is a device that creates high electric potential.! The Van de Graaff generator was invented by Robert J. Van de Graaff, an American physicist ( ).! Van de Graaff generators can produce electric potentials up to many 10s of millions of volts. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. W. Bauer and G.D. Westfall 1/22/14 Chapter 23 28

23 The Van de Graaff Generator! The Van de Graaff generator works by applying a positive charge to a non-conducting moving belt using a corona discharge.! The moving belt driven by an electric motor carries the charge up into a hollow metal sphere where the charge is taken from the belt by a pointed contact connected to the metal sphere.! The charge that builds up on the metal sphere distributes itself uniformly around the outside of the sphere. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. W. Bauer and G.D. Westfall 1/22/14 Chapter 23 29

24 Tandem Van de Graaff Accelerator! One use of a Van de Graaff generator is to accelerate particles for condensed matter and nuclear physics studies.! A clever design is the tandem Van de Graaff accelerator:! A large positive electric potential is created by a huge Van de Graaff generator.! Negatively charged C ions get accelerated towards the +10 MV terminal (they gain kinetic energy).! Electrons are stripped from the C and the now positively charged C ions are repelled by the positively charged terminal and gain more kinetic energy. 1/22/14 Chapter 23 30

25 Tandem Van de Graaff Accelerator! Suppose we have a tandem Van de Graaff accelerator that has a terminal voltage of 10 MV (10 million volts).! We want to accelerate 12 C nuclei using this accelerator. PROBLEM 1:! What is the highest energy we can attain for carbon nuclei? SOLUTION 1:! There are two stages to the acceleration: The carbon ions with a -1e charge gain energy accelerating toward the terminal. The stripped carbon ions with a +6e charge gain energy accelerating away from the terminal. 1/22/14 Chapter 23 31

26 Tandem Van de Graaff Accelerator! The kinetic energy gained by each ion is: ΔK = ΔU = q 1 ΔV + q 2 ΔV = K K = eδv + 6eΔV = 7eΔV! Putting in our numerical values we get: ( ) V K = C ( )= J! Physicists often use electron-volts instead of joules to express the kinetic energy of accelerated particles: K = 7e( V)= ev = 70 MeV PROBLEM 2:! What is the highest speed we can attain for carbon nuclei? 1/22/14 Chapter 23 32

27 Tandem Van de Graaff Accelerator SOLUTION 2:! We can relate speed and kinetic energy: K = 1 2 mv2 v = 2K m! The mass of a 12 C nucleus is kg so: v = J ( ) kg = m/s! This is 11% of the speed of light. 1/22/14 Chapter 23 33