Chapter 16: Electric Fields
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Chapter 16: Electric Fields Electrically charged particles have electric fields around them. In these fields, other electrically charged particles will exhibit certain things which are governed by the principles in this chapter. Electric Fields (E) The electric field strength in an electric field is defined as the force per unit charge that acts on a small positive point charge. As such, it is given by the formula; πΈ= πΉ π In a radial electric field, produced by a large charge Q, the electric field strength E is given by; π π2 , π0 = the electric constant, and π = the distance from the charge. πΈ=π 1 Where π = 4ππ 0 When drawing field lines, the lines and their direction show what path a positive charge would take when placed at that point. Electric Potential (V) The electric potential at a point is the potential energy per unit charge of a small positive charge placed at that point. This is the same as the work done per charge to move a small positive charge from infinity to that point. The potential energy of a point charge q is; πΈπ = ππ Where V is the potential at that point. Electric potential is a scalar quantity and has the unit Volt. Field Lines and Equipotentials An equipotential is a surface of constant potential. No change of potential energy occurs when an object is moved along an equipotential. Lines of force are always perpendicular to the equipotential surface. Electric Field Strength and Potential Gradient The work done on a small positive charge in moving it from infinity to another point is given by; πΈπ = ππ or, in terms of work, βπ = πβπ The work done by the charge; βπ = −πβπ Work = Force x Distance, so; βπ = πΉπ , and πΉ = πΈπ and π = βπ where π = distance so βπ = ππΈβπ ππΈβπ = −πβπ πΈ=− βπ βπ Electric Field Strength = - Potential Gradient For a uniform field between a pair of plates the above simplifies to; πΈ= π£ π Measuring Electric Charge Millikan’s oil drop experiment can be used to determine the fundamental electric charge. A drop of oil is placed between two charged plates with a significant potential across them. The potential difference is adjusted until the oil drop is suspended in between the two plates. In this position, the resultant vertical force on the drop is 0. Therefore; πΈππππ‘πππππ πΉππππ = πΊπππ£ππ‘ππ‘πππππ πΉππππ πΈπ = ππ The electrical force in a uniform field is given πΈ= π π π= πππ π π= πππ ππ Since q is some number of electrons ππ; Since, π, π, π, π are know, e can be found to be 1.6 × 10−19 C Accelerating Charges An electric field accelerates a charge in the same way a gravitational field accelerates a mass. 1 As a charge loses electric potential energy πΈπ = ππ it gains kinetic energy 2 ππ£ 2 . πΊπππ ππ πππππ‘ππ = πππ π ππ πππππ‘πππππ πππ‘πππ‘πππ 1 ππ£ 2 = ππ 2 2ππ π£=√ π for low energies; π£ ∝ √π So as the charge π moves closer to the charge providing the field, it gains speed, and loses speed in the other direction as it gains potential. The amount of energy gained by an electron in moving through a p.d. of 1 volt is 1 electron volt or 1 eV. Particles with the same charge (proton) but with different masses gain different energies. With the above expressions, it would appear that a p.d. of only 256,000 volts would accelerate an electron t o the speed of light. However, this equation has ignored relativistic effects. Relativistic Momentum The speed of a particle π£ must reach a limit at π. However, the particle’s momentum π, can increase indefinitely. Newton says π = ππ£, however π£ cannot be greater than π. Einstein says instead, π = πΈππ When v is a very small fraction of c, πΎ = 1, π π, π = ππ£ Relativistic momentum (blue) against classical momentum (red) as a particle approaches light speed (c). As a particle approaches the speed of light, its momentum approaches infinity. Relativistic Energy The energy of a free particle is composed of two parts, its kinetic energy and its rest energy. πΈπ‘ππ‘ππ = πΈπππππ‘ππ + πΈπππ π‘ πΈπ‘ππ‘ππ = πΎππ 2 When the particle is at rest, π£ = 0 and πΎ = 1, so; πΈπππ π‘ = ππ 2 πΈπ‘ππ‘ππ πΈπππ π‘ + πΈπππππ‘ππ πΎππ 2 = = =πΎ πΈπππ π‘ πΈπππ π‘ ππ 2 Electrons in an Electric Field (E) + Potential (V) Electron y F 0 Potential We know that πΉ = ππΈ, so for a single electron, πΉ = ππΈ, and from Newton’s second law πΉ = ππ ππΈ = ππ π= ππΈ π The electron enters the field with 0 vertical velocity and a horizontal velocity π£. Its vertical displacement π¦ can be given by; 1 π¦ = ππ‘ 2 2 π= ππΈ π 1 ππΈ π¦ = ( ) π‘2 2 π π₯ Horizontally, displacement = speed x time, π₯ = π£π‘, π‘ = π£ 1 ππΈ π₯ 2 π¦ = ( ) ( 2) 2 π π£ therefore, for a particle with charge π; ππΈ π¦=( ) π₯2 2ππ£ 2 where m is the rest mass of the particle Giving a parabolic path, like the graph of π¦ = π₯ 2, The electron gains a vertical velocity (and thus gains kinetic energy) but its horizontal component stays the same, the electric field lines are perpendicular to this line so have no effect on it. Charged Particles in Magnetic Fields (B) The force on a current carrying conductor in a magnetic field is given by; πΉ = π΅πΌπΏ πΌ = πΆβππππ πππ π’πππ‘ π‘πππ, πΌ = π π‘ A particle that moves a distance πΏ in a time π‘ has velocity; πΏ π£ π£ = ,π‘ = π‘ πΏ πΌ= ππ£ πΏ πΉ = ππ£π΅ This can be used to find the force acting on a single charged particle (q) moving through a magnetic field (B) with a velocity (v). Electrons in a Magnetic Field (B) A uniform magnetic field will deflect particles in a curved path. The force on a moving charge in a magnetic field is always perpendicular to its direction of travel. The direction of velocity is the same as the flow of charge (current) so velocity can be used in place of current in Fleming’s left hand rule. B field is directed into the page The force πΉ = π΅ππ£ is always perpendicular to the motion of the particle, and so, like an orbiting planet, always provides a centripetal force. The momentum of a particle moving in such a way is given by; π = π΅ππ, or, for high speeds, π = πΎπ΅ππ Crossed Fields Velocity Selector If a uniform electric field and uniform magnetic field are placed perpendicular to each other in such a way that they produce opposite deflections, they are called crossed fields. In such a set up, each field is the same size; π΅ππ£ = ππΈ π£= πΈ πΈππππ‘πππ πππππ π π‘πππππ‘β = π΅ ππππππ‘ππ πππ’π₯ ππππ ππ‘π¦ These two fields will have no effect (being equal) for particles whose velocity is equal to πΈ π΅ This means physicists will know the velocity of any particles that make it through the crossed field without being deflected. The radius of the path that the ions take depends on their charge/mass ratio. If all the ions have the same charge, π is a function only of the particle’s mass. π= π΅ππ π£ Circular Electric Fields Equipotential lines in a circular field are like contours on a map, equipotential lines always meet field lines at right angles. 1 The electric field strength E in a circular field is proportional to π2 and π. π π Therefore, πΈ = π 2 , which is analogous to π = πΊ π π2 Electrical Potential near a Charged Sphere We already know that πΈ=− ππ ππ Therefore π π = − ∫ πΈ ππ, and πΈ = π π2 π = −ππ ∫ π=π 1 ππ π2 π π Gauss’ Contribution All the field lines leaving a spherical charge must spread out equally, and the field strength at a given surface perpendicular to all these lines is the same. This surface is given by 4ππ 2 . The number of field lines on a surface is given by; πΈ × 4ππ 2 , and by convention, this is equal to π where π0 is the permittivity of free space. π 0 πΈ × 4ππ 2 = πΈ= π π0 π 4ππ0 π 2 Comparing this with our equation for Coulomb’s law; πΈ=π π π2 1 π must equal 4ππ 0 This is called Coulomb’s constant. Coulomb’s Law πΉ= 1 ππ × 2 4ππ0 π The force between two charges Q and q is directly proportional to the product of the charges and inversely proportional to the distance between the charges squared. Forces between like charges are repulsive and between unlike charges are attractive. Similarities between Gravitational Force and Electric Force Line of force/field line Gravitational Force Electric Force The path that a mass would take in that field The path that a positive charge would take in that field Newton’s law of gravitation; Coulomb’s law of force; Inverse square law of force πΉ=πΊ× ππ π2 πΉ= 1 ππ × 2 4ππ0 π Field strength Force per unit mass; πΉ π= π Force per unit charge; πΉ πΈ= π Unit of field strength Nkg −1 or ms −2 NC −1 or Vm−1 In uniform fields π is the same everywhere, Field lines are parallel and equally spaced. πΈ is the same everywhere, Field lines are parallel and equally spaced. Potential Gravitational potential energy per unit mass. Electric potential energy per unit positive charge. Unit of potential Jkg −1 V or JC −1 Potential energy between two masses/charges πππππ£ = −πΊ ππ π From a point mass or uniform spherical mass; πΈπ = From a point charge or a uniform spherical charge; Radial fields π=πΊ× Action at a distance π π2 Between any two masses. 1 ππ × 4ππ0 π πΈ= 1 π × 2 4ππ0 π Between any two charged objects. Force Attractive only. Attractive and repulsive, depending on charges.
