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Homework assignment: 9,18,28,34,44,51 (Six problems!). Chapter 16: Electric Energy and Capacitance. Work and electric potential energy. Consider a small positive charge placed in a uniform electric field E. A. +. +. +. +. +. +. +. +. +. +. d. conservative force.
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Homework assignment: 9,18,28,34,44,51 (Six problems!) Chapter 16: Electric Energy and Capacitance • Work and electric potential energy • Consider a small positive charge placed in a uniform electric field E. A + + + + + + + + + + d conservative force Potential Difference and Electric Potential - - - - - - - - - - B the force is in the same direction as the net displacement of the test charge The work done by a conservative force can be reinterpreted as the negative of the change in a potential energy associated with that force. SI unit : joule (J)
Work and electric potential energy (cont’d) • In analogy to the gravitational force, a potential can be defined as: • When the test charge moves from height yA to height yB , the work done • on the charge by the field is given by: Electric Potential Energy • U increases (decreases) if the test charge moves in the direction • opposite to (the same direction as) the electric force DPE<0 DPE<0 DPE>0 DPE>0 A B A B + + - - + B + A - B - A
Work and electric potential energy (cont’d) • In analogy to the gravitational force, a potential can be defined as: • When the test charge moves from height yA to height yB , the work done • on the charge by the field is given by: Electric Potential Energy • Define an electric potential difference as: SI unit : joule per coulomb or volt (J/C or V) The change in electric potential energy as a charge q moves from A to B divided by charge q. Then for a special case of a uniform electric field: SI unit : J/C=V=N/C
Example 16.1 : Potential energy differences in an electric field • A proton is released from rest at x=-2.00 cm in a constant electric • field with magnitude 1.50x103 V/C pointing the positive x-direction. • Calculate the change in the electric potential energy associated • with the proton when it reaches x=5.00 cm. Electric Potential Energy (b) Find the change in electric potential energy associated with an electron fired from x=-2.00 cm and reaching x=12.0 m. (b) Find the change in electric potential energy associated with an electron fired from x=3.00 cm to x=7.00 cm if the direction of the electric field is reversed.
Example 16.2 : Dynamics of charged particles • Continuation of Example 16.1. • Find the speed of the proton at x=0.0500 m in part (a) of Example • 16.1. Electric Potential Energy (b) Find the electron’s initial speed, given that its speed has fallen by half at x=0.120 m.
Example 16.3 : TV tubes and atom smasher • Charged particles are accelerated through potential difference. • Suppose a proton is injected at a speed of 1.00x106 m/s between • two plates 5.00 cm apart. The proton subsequently accelerates • across the gap and exits through the opening. • What must the electric potential difference be if the exist speed • is to be 3.00x106 m/s? Electric Potential Energy (b) What is the magnitude of the electric field between the plates?
Electric potential by a point charge • The electric field of a point charge extends throughout space, so does • its electric potential. • The zero point can be anywhere but it is convenient to choose the • point at infinity. Then it can be shown that the electric potential due • to a point charge q at a distance r is given by : Electric Potential and Potential Energy • The superposition principle : the total electric potential at some point • P due to several point charges is the algebraic sum of the electric • potentials due to the individual charges.
Electric potential energy of a pair of point charges • If V1 is the electric potential due to charge q1 at a point P, then • the work required to bring charge q2 from infinity to P without • acceleration is q2V1. • This work is, by definition, equal to the potential energy PE of the • two-particle system when the particles are separated by a distance r. If q1q2>0, PE>0 Electric Potential and Potential Energy If q1q2<0, PE<0 • Electric potential energy with several point charges From the superposition principle
Example 16.4 : Finding the electric potential y(m) P • A 5.00-mC point charge is at • origin and a point charge q2=-2.00mC • is on the x-axis at (3.00,0) m. (0,4.00) r2 r1 • Find the electric potential at point • P due to these charges. q2 q1 x(m) + - 0 Electric Potential and Potential Energy (3.00,0) (b) Find the work needed to bring the 4.00-mC charge from infinity to P.
Surface electric potential of a conductor in electrostatic equilibrium • Work done to move a charge from point A to point B No net work is needed to move a charge between two points that are at the same electric potential. Potentials and Charged Conductors • We will learn that when a charged conductor is in electrostatic • equilibrium: • A net charge placed on it resides entirely on its surface. • The electric field just outside its surface is perpendicular to • the surface and that the field inside the conductor is zero. • -All points on its surface are at the same potential.
Potentials and Charged Conductors • Equipotential surface • E = 0 everywhere inside a conductor - At any point just inside the conductor the component of E tangent to the surface is zero - The tangential component of E is also zero just outside the surface If it were not, a charge could move around a rectangular path partly inside and partly outside and return to its starting point with a net amount of work done on it. vacuum conductor • When all charges are at rest, the electric field just outside a conductor must • be perpendicular to the surface at every point • When all charges are at rest, the surface of a conductor is always an • equipotential surface
Potentials and Charged Conductors • Examples of equipotential surface The equipotentials are perpendicular to the electric field lines at every point.
Potentials and Charged Conductors • Electron volt The electron volt is defined as the kinetic energy that an electron gains when accelerated through a potential difference of 1V. 1 V = 1 J/C 1 eV = 1.60 x 10-19 C.V = 1.60 x 10-19 J In atomic, nuclear and particle physics, the electron volt is used commonly to express energies. • - Electrons in normal atoms have energies of tens of eV’s. • - Excited electrons in atoms that emit x-rays have energies • of thousands of eV’s ( keV = 103 eV). • High energy gamma rays emitted by the nucleus have • energies of millions of eV’s (MeV = 106 eV). • The world most energetic accelerator near Chicago accelerates • protons/anti-proton up to Tera eV’s (TeV = 1012 eV )
Capacitance • Capacitor • Any two conductors separated by an insulator (or a vacuum) form a • capacitor • In practice each conductor initially has zero net charge and electrons • are transferred from one conductor to the other (charging the conductor) • Then two conductors have charge with equal • magnitude and opposite sign, although the net • charge is still zero • When a capacitor has or stores charge Q , the • conductor with the higher potential has charge • +Q and the other -Q if Q>0
Capacitance • One way to charge a capacitor is to connect these conductors to opposite • terminals of a battery, which gives a fixed potential difference Vab between • conductors ( a-side for positive charge and b-side for negative charge). Then • once the charge Q and –Q are established, the battery is disconnected. • If the magnitude of the charge Q is doubled, the electric field becomes • twice stronger and Vab=DV is twice larger. • Then the ratio Q/DV is still constant and it is called the capacitance C. Capacitors and Capacitance Q -Q b a • When a capacitor has or stores charge Q , the conductor • with the higher potential has charge +Q and the other • -Q if Q>0
Parallel-plate capacitor in vacuum • Charge density: • Electric field: • Potential diff.: Parallel-Plate Capacitance • Capacitance: • The capacitance depends only on the • geometry of the capacitor. • It is proportional to the area A. • It is inversely proportional to the separation d • When matter is present between the plates, its • properties affect the capacitance.
Units 1 F = 1 C2/N m (Note [e0]=C2/N m2) 1 mF = 10-6 F, 1 pF = 10-12 F e0 = 8.85 x 10-12 F/m • Example : Size of a 1-F capacitor Parallel-Plate Capacitance
Example : Properties of a parallel capacitor Parallel-Plate Capacitance
Symbols for circuit elements and circuits Combinationsof Capacitors
Capacitors in parallel a b Combinations of Capacitors The parallel combination is equivalent to a single capacitor with the same total charge Q=Q1+Q2 and potential difference.
Capacitors in series a c Combinations of Capacitors b The equivalent capacitance Ceq of the series combination is defined as the capacitance of a single capacitor for which the charge Q is the same as for the combination, when the potential difference V is the same.
Capacitor networks Combinations of Capacitors
Capacitor networks (cont’d) Combinations of Capacitors
Capacitor networks 2 C C C C C A A C C C C C B B C C C C C Combinations of Capacitors C C A A C B B C C
Work done to charge a capacitor • Consider a process to charge a capacitor up to Q with the final potential • difference DV. • Let qi and (Dv)i be the charge and potential difference at an intermediate stage • during the charging process. • At this stage the work (DW)i required to transfer an additional element of • charge Dq is: Energy Stored in a Charged Capacitor DV • The total work needed to increase the capacitor • charge q from zero to Q is: qi (Dv)i • The energy stored: Q Dq
Dielectric materials • Experimentally it is found that when a non-conducting material • (dielectrics) between the conducting plates of a capacitor, the • capacitance increases for the same stored charge Q. • Define the dielectric constant k as: C0 : capacitance w/o dielectric C : capacitance w/ dielectric • When the charge is constant, Capacitor with Dielectrics k k Material Material 3-6 Mica 1 vacuum 3.1 1.00059 Mylar air(1 atm) 2.1 3.40 Teflon Plexiglas 80.4 Water 2.25 Polyethelene
Induced charge and polarization • Consider a two oppositely charged parallel plates with vacuum • between the plates. • Now insert a dielectric material of dielectric constant k. • Source of change in the electric field is redistribution of positive • and negative charge within the dielectric material (net charge 0). • This redistribution is called a polarization and it produces induced • charge and field that partially cancels the original electric field. Dielectrics
Molecular model of induced charge Dielectrics
Molecular model of induced charge (cont’d) Dielectrics