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This document summarizes key topics from a lecture on electric fields: 1) It defines electric field as the force per unit charge and discusses how electric field lines represent the behavior and strength of electric fields graphically. 2) Examples are given for calculating the electric field from point charges and continuous charge distributions using Coulomb's Law and integration. 3) Applications of electric fields including motion of charges in fields and electric dipoles are discussed. Various demonstrations are also listed.

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Summer July 2004 1 Lecture 2 Electric Fields Chp. 23 Cartoon - Analogous to gravitational field Opening Demo - Bending of water stream with charged rod Warm-up problem Physlet Topics Electric field = Force per unit Charge Electric Field Lines and Electric Flux Electric field from more than 1 charge Electric Dipoles Motion of point charges in an electric field Examples of finding electric fields from continuous charges List of Demos Van de Graaff Generator, workings,lightning rod, electroscope, electric wind Smoke remover or electrostatic precipitator Kelvin water drop generator Transparent CRT with visible electron gun Field lines using felt,oil, and 10 KV supply.
Summer July 2004 2 The Electric Field Definition of the electric field. Whenever charges are present and if I bring up another charge, it will feel a net Coulomb force from all the others. It is convenient to say that there is field there equal to the force per unit positive charge. E=F/q0. The direction of the electric field is along r and points in the direction a positive test charge would move. This idea was proposed by Michael Faraday in the 1830s. The idea of the field replaces the charges as defining the situation. Consider two point charges: q0 q1 r
Summer July 2004 3 The force per unit charge is E = F/q0 and then the electric field at r is E = kq1/r2 due to the point charge q1 . The units are Newton/Coulomb. The electric field has direction and is a vector. How do we find the direction.? The direction is the direction a unit positive test charge would move. q1 r E + q0 q1 r The Coulomb force is F= kq1q0/r2 If q1 were positive
Summer July 2004 4 Point negative charge E= kq1/r2 q1 q1 r
Summer July 2004 5 Electric Field Lines Like charges (++) Opposite charges (+ -) This is called an electric dipole.
Summer July 2004 6 Electric Field Lines: a graphic concept used to draw pictures as an aid to develop intuition about its behavior. The text shows a few examples. Here are the drawing rules. E-field lines begin on + charges and end on - charges. (or infinity). They enter or leave charge symmetrically. The number of lines entering or leaving a charge is proportional to the charge The density of lines indicates the strength of E at that point. At large distances from a system of charges, the lines become isotropic and radial as from a single point charge equal to the net charge of the system. No two field lines can cross. Show a physlet 9.1.4, 9.1.7 Show field lines using felt,oil, and 10 KV supply
Summer July 2004 7 Typical Electric Fields (SI Units) 1 cm away from 1 nC of negative charge E = kq /r2 = 1010 *10-9/ 10-4 =105 N /C N.m2/C2 C / m2 = N/C Fair weather atmospheric electricity = 100 N/C downward 100 km high in the ionosphere Field due to a proton at the location of the electron in the H atom. The radius of the electron orbit is 0.5*10-10 m. E = kq /r2 = 1010 *1.6*10-19/ (0.5 *10-10 )2 = 4*1011 N /C q r E + - r Hydrogen atom . Earth - - - - - - - - - E +++++ 1N / C = Volt/meter
Summer July 2004 8 Example of field lines for a uniform distribution of positive charge on one side of a very large nonconducting sheet. How would the electric field change if both sides were charged? How would things change if the sheet were conducting? This is called a uniform electric field.
Summer July 2004 9 Methods of evaluating electric fields Direct evaluation from Coulombs Law or brute force method If we know where the charges are, we can find E from E = kqi/ri 2. This is a vector equation and can be complex and messy to evaluate and we may have to resort to a computer. The principle of superposition guarantees the result. Instead of summing the charge we can imagine a continuous distribution and integrate it. This distribution may be over a volume, a surface or just a line. E = dE = kdqi/r2 r where r is a unit vector directed from charge dq to the field point. dq = dV , or dq = dA, or dq = dl
Summer July 2004 10 Example of finding electric field from two charges Find x and y components of electric field due to both charges and add them up We have q 1 =+10 nC at the origin, q 2 = +15 nC at x=4 m. What is E at y=3 m and x=0? point P x y q1=10 nc q2 =15 nc 4 3 P Use principle of superposition
Summer July 2004 11 Example continued Field due to q1 E = 1010 N.m2/C2 10 X10-9 C/(3m)2 = 11 N/C in the y direction. Recall E =kq/r2 and k=8.99 x 109 N.m2/C2 x y q1=10 nc q2 =15 nc 4 3 Ey= 11 N/C Ex= 0 Field due to q2 5 E = 1010 N.m2/C2 15 X10-9 C/(5m)2 = 6 N/C at some angle f Resolve into x and y components f E Ey=E sin f = 6 * 3/5 =18/5 = 3.6 N/C Ex=E cos f = 6 * (-4)/5 =-24/5 = -4.8 N/C f Now add all components Ey= 11 + 3.6 = 14.6 N/C Ex= -4.8 N/C Magnitude
Summer July 2004 12 Example continued x q1=10 nc q2 =15 nc 4 3 Ey= 11 + 3.6 = 14.6 N/C Ex= -4.8 N/C E Magnitude of electric field f1 f1 = atan Ey/Ex= atan (14.6/-4.8)= 72.8 deg Using unit vector notation we can also write the electric field vector as:
Summer July 2004 13 Example of two identical charges on the x axis. What is the filed on the y axis? Ey=2*E sin f = 2*6 * 3/5 = 7.2 N/C Ex=0 q y 4 x q2 =15 nc 3 q2 =15 nc 4 f 5 E = 1010 N.m2/C2 15 X10-9 C/(5m)2 = 6 N/C q 3 y 4 x q2 = -15 nc q2 =15 nc 4 f 5 Ey=0 Ex=2*E cos f = 2*6 * 4/5 = - 9.6 N/C Ex Ey Example of two opposite charges on the x axis. What is the filed on the y axis?
Summer July 2004 14 4 equal charges symmetrically spaced along a line. What is the field at point P? (y and x = 0) q3 f y x q2 q3 q1 q4 r1 r2 r3 r4 q2 q1 q4 P
Summer July 2004 15 Find electric field due to a line of uniform + charge of length L with linear charge density equal to q y +x dq r dEy= dE cos q -x dE = k dq /r2 dEy dq = dx dEy= k dx cos q /r2 Ey= k q cos q /r2 for a point charge y x -L/2 L/2 0 x= y tanq dx = y sec2 q dq r =y sec q r2 =y2sec2 q dx/r2 = dq/y dEx dE
Summer July 2004 16 What is the electric field from an infinitely long wire with linear charge density of +100 nC/m at a point 10 away from it. What do the lines of flux look like? y =10 cm . Typical field for the electrostatic smoke cleaner Ey +++++++++++++++++++++++++++++++++++++++++
Summer July 2004 17 Electric field gradient When a dipole is an electric field that varies with position, then the magnitude of the electric force will be different for the two charges. The dipole can be permanent like NaCl or water or induced as seen in the hanging pith ball. Induced dipoles are always attracted to the region of higher field. Explains why wood is attracted to the teflon rod and how a smoke remover or microwave oven works. Show smoke remover demo.
Summer July 2004 18 Electrostatic smoke precipitator model Negatively charged central wire has electric field that varies as 1/r (strong electric field gradient). Field induces a dipole moment on the smoke particles. The positive end gets attracted more to the wire In the meantime a corona discharge is created. This just means that induced dipole moments in the air molecules cause them to be attracted towards the wire where they receive an electron and get repelled producing a cloud of ions around the wire. When the smoke particle hits the wire it receives an electron and then is repelled to the side of the can where it sticks. However, it only has to enter the cloud of ions before it is repelled. It would also work if the polarity of the wire is reversed
Summer July 2004 19 Electric Dipoles: A pair of equal and opposite charges q separated by a displacement d is called an electric dipole. It has an electric dipole moment p=qd. P r IEI ~ 2kqd/r3 when r is large compared to d p=qd = the electric dipole moment IEI ~2kp/r3 Note inverse cube law IEI = kq/r2 Note inverse square law for a single charge. + - d p +q -q
Summer July 2004 20 Water (H2O) is a molecule that has a permanent dipole moment. When a dipole is an electric field, the dipole moment wants to rotate to line up with the electric field. It experiences a torque. GIven p = 6.2 x 10 - 30 C m And q = -10 e and q = +10e What is d? d = p / 10e = 6.2 x 10 -30 C m / 10*1.6 x 10 -19 C = 3.9 x 10 -12 m Very small distance but still is responsible for the conductivity of water. Leads to how microwave ovens heat up food
Summer July 2004 21 H2O in a Uniform Electric Field There exist a torque on the water molecule To rotate it so that p lines up with E. x Torque about the com = t F x sin q + F(d-x)sin q = Fdsin q = qEdsin q = pEsin q = p x E Potential Energy = U = -W = -pEcosq = - p. E Is a minimum when p aligns with E t = p x E
Summer July 2004 22 Motion of point charges in electric fields When a point charge such as an electron is placed in an electric field E, it is accelerated according to Newtons Law: a = F/m = qE/m for uniform electric fields a = F/m = mg/m = g for uniform gravitational fields If the field is uniform, we now have a projectile motion problem- constant acceleration in one direction. So we have parabolic motion just as in hitting a baseball, etc except the magnitudes of velocities and accelerations are different. Replace g by qE/m in all equations; For example, In y =1/2at2 we get y =1/2(qE/m)t2
Summer July 2004 23 Example: An electron is projected perpendicularly to a downward electric field of E= 2000 N/C with a horizontal velocity v=106 m/s. How much is the electron deflected after traveling 1 cm. Since velocity in x direction does not change, t=d/v =10-2/106 = 10-6 sec, so the distance the electron falls upward is y =1/2at2 = 0.5*eE/m*t2 = 0.5*1.6*10-19*2*103/10 - 30*(10-8)2 = 0.016m V E d e E Demo Transparent CRT with electron gun
Summer July 2004 24 Back to computing Electric Fields Electric field due to a line of uniform charge Electric field due to an arc of a circle of uniform charge. Electric field due to a ring of uniform charge Electric field of a uniform charged disk Next we will go on to another simpler method to calculate electric fields that works for highly symmetric situations using Gausss Law.
Summer July 2004 25 dEx= k dq cos q /r2 s=r q ds=r dq dEx= k ds cos q /r2 Field due to arc of charge What is the field at the center of a circle of charge? Ans. 0
Summer July 2004 26 Find the electric field on the axis of a uniformly charged ring with linear charge density = Q/2pR. dq = ds r2 =z2+R2 cos q =z/r =0 at z=0 =0 at z=infinity =max at z=0.7R
Summer July 2004 27 Warm-up set 2 1. [153709] Can there be an electric field at a point where there is no charge? Can a charge experience a force due to its own field? Please write a one or two sentence answer for each question. 2. [153707] An insulator is a material that... three are correct is not penetrated by electric fields none of these cannot carry an electric charge cannot feel an electrical force 3. [153708] Which of the following is true of a perfect conductor ? There can be no electric charge on the surface. There cannot be an electric field inside. There cannot be any excess electric charge inside. There cannot be any electric charges inside. Two of the choices are correct
Summer July 2004 28 Kelvin Water Drop Generator Am. J. Phys. 68,1084(2000)

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PHYS632_C2_23_Electric.pptttttttttttttttttt

  • 1. Summer July 2004 1 Lecture 2 Electric Fields Chp. 23 Cartoon - Analogous to gravitational field Opening Demo - Bending of water stream with charged rod Warm-up problem Physlet Topics Electric field = Force per unit Charge Electric Field Lines and Electric Flux Electric field from more than 1 charge Electric Dipoles Motion of point charges in an electric field Examples of finding electric fields from continuous charges List of Demos Van de Graaff Generator, workings,lightning rod, electroscope, electric wind Smoke remover or electrostatic precipitator Kelvin water drop generator Transparent CRT with visible electron gun Field lines using felt,oil, and 10 KV supply.
  • 2. Summer July 2004 2 The Electric Field Definition of the electric field. Whenever charges are present and if I bring up another charge, it will feel a net Coulomb force from all the others. It is convenient to say that there is field there equal to the force per unit positive charge. E=F/q0. The direction of the electric field is along r and points in the direction a positive test charge would move. This idea was proposed by Michael Faraday in the 1830s. The idea of the field replaces the charges as defining the situation. Consider two point charges: q0 q1 r
  • 3. Summer July 2004 3 The force per unit charge is E = F/q0 and then the electric field at r is E = kq1/r2 due to the point charge q1 . The units are Newton/Coulomb. The electric field has direction and is a vector. How do we find the direction.? The direction is the direction a unit positive test charge would move. q1 r E + q0 q1 r The Coulomb force is F= kq1q0/r2 If q1 were positive
  • 4. Summer July 2004 4 Point negative charge E= kq1/r2 q1 q1 r
  • 5. Summer July 2004 5 Electric Field Lines Like charges (++) Opposite charges (+ -) This is called an electric dipole.
  • 6. Summer July 2004 6 Electric Field Lines: a graphic concept used to draw pictures as an aid to develop intuition about its behavior. The text shows a few examples. Here are the drawing rules. E-field lines begin on + charges and end on - charges. (or infinity). They enter or leave charge symmetrically. The number of lines entering or leaving a charge is proportional to the charge The density of lines indicates the strength of E at that point. At large distances from a system of charges, the lines become isotropic and radial as from a single point charge equal to the net charge of the system. No two field lines can cross. Show a physlet 9.1.4, 9.1.7 Show field lines using felt,oil, and 10 KV supply
  • 7. Summer July 2004 7 Typical Electric Fields (SI Units) 1 cm away from 1 nC of negative charge E = kq /r2 = 1010 *10-9/ 10-4 =105 N /C N.m2/C2 C / m2 = N/C Fair weather atmospheric electricity = 100 N/C downward 100 km high in the ionosphere Field due to a proton at the location of the electron in the H atom. The radius of the electron orbit is 0.5*10-10 m. E = kq /r2 = 1010 *1.6*10-19/ (0.5 *10-10 )2 = 4*1011 N /C q r E + - r Hydrogen atom . Earth - - - - - - - - - E +++++ 1N / C = Volt/meter
  • 8. Summer July 2004 8 Example of field lines for a uniform distribution of positive charge on one side of a very large nonconducting sheet. How would the electric field change if both sides were charged? How would things change if the sheet were conducting? This is called a uniform electric field.
  • 9. Summer July 2004 9 Methods of evaluating electric fields Direct evaluation from Coulombs Law or brute force method If we know where the charges are, we can find E from E = kqi/ri 2. This is a vector equation and can be complex and messy to evaluate and we may have to resort to a computer. The principle of superposition guarantees the result. Instead of summing the charge we can imagine a continuous distribution and integrate it. This distribution may be over a volume, a surface or just a line. E = dE = kdqi/r2 r where r is a unit vector directed from charge dq to the field point. dq = dV , or dq = dA, or dq = dl
  • 10. Summer July 2004 10 Example of finding electric field from two charges Find x and y components of electric field due to both charges and add them up We have q 1 =+10 nC at the origin, q 2 = +15 nC at x=4 m. What is E at y=3 m and x=0? point P x y q1=10 nc q2 =15 nc 4 3 P Use principle of superposition
  • 11. Summer July 2004 11 Example continued Field due to q1 E = 1010 N.m2/C2 10 X10-9 C/(3m)2 = 11 N/C in the y direction. Recall E =kq/r2 and k=8.99 x 109 N.m2/C2 x y q1=10 nc q2 =15 nc 4 3 Ey= 11 N/C Ex= 0 Field due to q2 5 E = 1010 N.m2/C2 15 X10-9 C/(5m)2 = 6 N/C at some angle f Resolve into x and y components f E Ey=E sin f = 6 * 3/5 =18/5 = 3.6 N/C Ex=E cos f = 6 * (-4)/5 =-24/5 = -4.8 N/C f Now add all components Ey= 11 + 3.6 = 14.6 N/C Ex= -4.8 N/C Magnitude
  • 12. Summer July 2004 12 Example continued x q1=10 nc q2 =15 nc 4 3 Ey= 11 + 3.6 = 14.6 N/C Ex= -4.8 N/C E Magnitude of electric field f1 f1 = atan Ey/Ex= atan (14.6/-4.8)= 72.8 deg Using unit vector notation we can also write the electric field vector as:
  • 13. Summer July 2004 13 Example of two identical charges on the x axis. What is the filed on the y axis? Ey=2*E sin f = 2*6 * 3/5 = 7.2 N/C Ex=0 q y 4 x q2 =15 nc 3 q2 =15 nc 4 f 5 E = 1010 N.m2/C2 15 X10-9 C/(5m)2 = 6 N/C q 3 y 4 x q2 = -15 nc q2 =15 nc 4 f 5 Ey=0 Ex=2*E cos f = 2*6 * 4/5 = - 9.6 N/C Ex Ey Example of two opposite charges on the x axis. What is the filed on the y axis?
  • 14. Summer July 2004 14 4 equal charges symmetrically spaced along a line. What is the field at point P? (y and x = 0) q3 f y x q2 q3 q1 q4 r1 r2 r3 r4 q2 q1 q4 P
  • 15. Summer July 2004 15 Find electric field due to a line of uniform + charge of length L with linear charge density equal to q y +x dq r dEy= dE cos q -x dE = k dq /r2 dEy dq = dx dEy= k dx cos q /r2 Ey= k q cos q /r2 for a point charge y x -L/2 L/2 0 x= y tanq dx = y sec2 q dq r =y sec q r2 =y2sec2 q dx/r2 = dq/y dEx dE
  • 16. Summer July 2004 16 What is the electric field from an infinitely long wire with linear charge density of +100 nC/m at a point 10 away from it. What do the lines of flux look like? y =10 cm . Typical field for the electrostatic smoke cleaner Ey +++++++++++++++++++++++++++++++++++++++++
  • 17. Summer July 2004 17 Electric field gradient When a dipole is an electric field that varies with position, then the magnitude of the electric force will be different for the two charges. The dipole can be permanent like NaCl or water or induced as seen in the hanging pith ball. Induced dipoles are always attracted to the region of higher field. Explains why wood is attracted to the teflon rod and how a smoke remover or microwave oven works. Show smoke remover demo.
  • 18. Summer July 2004 18 Electrostatic smoke precipitator model Negatively charged central wire has electric field that varies as 1/r (strong electric field gradient). Field induces a dipole moment on the smoke particles. The positive end gets attracted more to the wire In the meantime a corona discharge is created. This just means that induced dipole moments in the air molecules cause them to be attracted towards the wire where they receive an electron and get repelled producing a cloud of ions around the wire. When the smoke particle hits the wire it receives an electron and then is repelled to the side of the can where it sticks. However, it only has to enter the cloud of ions before it is repelled. It would also work if the polarity of the wire is reversed
  • 19. Summer July 2004 19 Electric Dipoles: A pair of equal and opposite charges q separated by a displacement d is called an electric dipole. It has an electric dipole moment p=qd. P r IEI ~ 2kqd/r3 when r is large compared to d p=qd = the electric dipole moment IEI ~2kp/r3 Note inverse cube law IEI = kq/r2 Note inverse square law for a single charge. + - d p +q -q
  • 20. Summer July 2004 20 Water (H2O) is a molecule that has a permanent dipole moment. When a dipole is an electric field, the dipole moment wants to rotate to line up with the electric field. It experiences a torque. GIven p = 6.2 x 10 - 30 C m And q = -10 e and q = +10e What is d? d = p / 10e = 6.2 x 10 -30 C m / 10*1.6 x 10 -19 C = 3.9 x 10 -12 m Very small distance but still is responsible for the conductivity of water. Leads to how microwave ovens heat up food
  • 21. Summer July 2004 21 H2O in a Uniform Electric Field There exist a torque on the water molecule To rotate it so that p lines up with E. x Torque about the com = t F x sin q + F(d-x)sin q = Fdsin q = qEdsin q = pEsin q = p x E Potential Energy = U = -W = -pEcosq = - p. E Is a minimum when p aligns with E t = p x E
  • 22. Summer July 2004 22 Motion of point charges in electric fields When a point charge such as an electron is placed in an electric field E, it is accelerated according to Newtons Law: a = F/m = qE/m for uniform electric fields a = F/m = mg/m = g for uniform gravitational fields If the field is uniform, we now have a projectile motion problem- constant acceleration in one direction. So we have parabolic motion just as in hitting a baseball, etc except the magnitudes of velocities and accelerations are different. Replace g by qE/m in all equations; For example, In y =1/2at2 we get y =1/2(qE/m)t2
  • 23. Summer July 2004 23 Example: An electron is projected perpendicularly to a downward electric field of E= 2000 N/C with a horizontal velocity v=106 m/s. How much is the electron deflected after traveling 1 cm. Since velocity in x direction does not change, t=d/v =10-2/106 = 10-6 sec, so the distance the electron falls upward is y =1/2at2 = 0.5*eE/m*t2 = 0.5*1.6*10-19*2*103/10 - 30*(10-8)2 = 0.016m V E d e E Demo Transparent CRT with electron gun
  • 24. Summer July 2004 24 Back to computing Electric Fields Electric field due to a line of uniform charge Electric field due to an arc of a circle of uniform charge. Electric field due to a ring of uniform charge Electric field of a uniform charged disk Next we will go on to another simpler method to calculate electric fields that works for highly symmetric situations using Gausss Law.
  • 25. Summer July 2004 25 dEx= k dq cos q /r2 s=r q ds=r dq dEx= k ds cos q /r2 Field due to arc of charge What is the field at the center of a circle of charge? Ans. 0
  • 26. Summer July 2004 26 Find the electric field on the axis of a uniformly charged ring with linear charge density = Q/2pR. dq = ds r2 =z2+R2 cos q =z/r =0 at z=0 =0 at z=infinity =max at z=0.7R
  • 27. Summer July 2004 27 Warm-up set 2 1. [153709] Can there be an electric field at a point where there is no charge? Can a charge experience a force due to its own field? Please write a one or two sentence answer for each question. 2. [153707] An insulator is a material that... three are correct is not penetrated by electric fields none of these cannot carry an electric charge cannot feel an electrical force 3. [153708] Which of the following is true of a perfect conductor ? There can be no electric charge on the surface. There cannot be an electric field inside. There cannot be any excess electric charge inside. There cannot be any electric charges inside. Two of the choices are correct
  • 28. Summer July 2004 28 Kelvin Water Drop Generator Am. J. Phys. 68,1084(2000)