560 likes | 706 Views
How do most objects interact with each other?. Electric Charge ( q or Q ) An intrinsic property of matter (as is mass) S.I. unit: coulombs (C), (also: μ C) Two types (+) or (-): add algebraically to give a net charge. Like charges repel; unlike charges attract .
E N D
How do most objects interact with each other? Electric Charge (q or Q) An intrinsic property of matter (as is mass) S.I. unit: coulombs (C), (also: μC) Two types (+) or (-): add algebraically to give a net charge. Like charges repel; unlike charges attract. Coulomb’s Law: The force between two charges (q1) and q2) separated by a distance r is: Fe=k q1 q2 / r2. k is Coulomb’s constant: k = 9 × 109 Nm2/C2 Force by q1 on q2is equal and opposite to the force by q2 on q1.
More about Charge. • Quantization of charge (e): charge comes in discrete amounts of e = 1.6×10-19C • e is the “elementary charge”; we always treat e as positive. • Electrons have a charge of –e • Protons have a charge of +e and about 2000 times the mass of electrons. • Conservation of charge: In any reaction, net charge remains the same. (Is this different from conservation of mass?)
Examples • Book sitting on a table • Pushing off • A hard wall • A soft wall • Electric Force is a measure of hardness in repulsion. • Sodium Metal vs. Aluminum Metal. • Electric Force is a measure of hardness in attraction. • What happens if we hold a • Negatively charged rod held near a neutral atom? • Positively charged rod held near a neutral atom?
How can charge be transferred between objects? • Friction • Electrons transferred by rubbing. • Rubbing plastic with fur; fur loses electrons. • Rubbing glass with silk; silk gains electrons. • Induction • A charged object causes electrons to move in a neutral object. • Atoms become polarized; they turn into dipoles. • Charged rod (+) or (-) near a neutral object; what happens?
How can charge be transferred between objects? (cont’d) • Conduction • Electrons can move freely through some materials, called conductors. • Conductors can be charged directly through contact with charged objects (not friction). • Conductors can be charged through induction • Electrons move freely to one side or another; object as a whole becomes polarized.
Van de Graf Generator • Device to concentrate positive charge on a conducting sphere. • Operation: • Lower roller (G) rubs e-off belt. • Belt becomes (+) • (+) belt attracts e- from brush (B), bleeding them off the conducting sphere. • Belt is neutralized as it travels back to lower roller. • Sphere (A) becomes (+) more and more.
Electrical Properties of Materials • Conductors: Outer electrons can move nearly freely through the material. (What is it about metallic bonding that allows this?) • Insulators: Outer electrons are bound tightly to the atoms. Cannot move freely.
Electrical Properties of Materials(cont’d) • Semiconductors: Outer electrons are bound, but not so tightly. (e.g. silicon, germanium) • Can be made to conduct electricity in very controlled ways. • Doping: impurities to increase or decrease electrons in crystal structure (e.g. transistors) • Photoelectric Effect: electrons can be energized by absorbing light ot leave their atoms and travel through the material. (e.g. photocells)
Using Coulomb’s Law • A charge (Q) of +2µC is fixed in the ground. A charge (q) of +3µC having a mass of 0.24 kg is held 0.15m directly above the fixed charge and released. How does the charge q move? Give the magnitude and direction of its acceleration. Use (g = 10 m/s2 and k = 9 × 109 Nm2/C2)
Using Coulomb’s Law (cont’d) • Two protons are separated by 5×10-15m. • Calculate the force on one proton due to the other. (qp= + e = 1.6×10-19 C). Does the force seem small or large as felt by: a person? a fly? a proton? b. Calculate the acceleration of the proton. Is this a large number? (mp= 1.67×10-27 kg)
Using Coulomb’s Law (cont’d) • Three charges (q1 and q3 are fixed): q1 = -86 µC at (52 cm, 0) q2 = +50 µC at (0, 0) q3 = +65 µC at (0,30 cm) What is the net force on q2? (magnitude and direction) What is its acceleration if it has a mass of 0.5 kg?
Electric Field • How does one charge know that another charge is there? (“Action at a distance”). • A charge actually changes the space around it; a force field is caused. • Other charges interact with this “Electric Field”. • The Electric Field due to a “source” charge (qs): • E = k qs /r2 ř (a vector) • The direction of E at a point is in the same direction that a positive point charge would move if placed at that point. (BY DEFINITION)
Electric Field (cont’d) • Suppose we havemany charged objects creating an E-field in space. • At some point (r) (position vector from origin) in space: • Put a small positive test charge (qT) at r. • The magnitude of E at r equals the Force on qT divided by qT. E = Fe/qT (By definition) • The direction of E at r is in the direction of the force on qT. (By definition)
Advantages of the E-Field • The Electric fields caused by several charges add up as vectors. • For a system of charges: • Find the resultant (net) E-Field • Now you can easily find the force on any charge (q) you place in the system: Fe= q E • If q is (+), then Fe and E are in the same direction, otherwise they are opposite to each other. • If you change q, you don’t need to recalculate the E-field. • The E-field is physical; energy is actually stored in space.
Common Types of E-Field Problems • Given acollection of point charges: • Find the resulting E-field at a specified location • Find the location where the E-field has a specified magnitude and direction (such as zero.) • Given the E-field: • find the forces and/or accelerations produced on a charge or charged system.
Common Mistake to Avoid • E = k q /r2 ONLY when it is CAUSED by a POINT CHARGE we call “q”. • If you have are given an E-field due to some other source or sources, the above equation is FALSE. • Example: A uniform E-field has the same value everywhere in space. For such a field, the above equation is FALSE.
Using the E-field • For charges: q1 = -3 μC placed at (0,0) q2 = +6 μC placed at (2 m, 0) a. Find the E-field at (-1m, 0) b. Find the force on an electron placed at (-1 m , 0) c. Where on the x-axis is E = 0?
Using the E-field (cont’d) • A pendulum with: charge (+Q) and mass (m) is in equilibrium in a uniform E-field applied in the +x-direction. In terms of Q, m, and θ, • Find the magnitude of the E-field. • Find the tension in the string.
Using the E-field (cont’d) • A charge of (+q) and mass (m) is at the origin and moving in the +x-direction with a speed v. A uniform E-field (with magnitude: E) acts in the –x-direction. • Find the acceleration of the charge. • Find the maximum value of +x reached by the charge. • Find the time for the charge to return to its start point.
Electric Field Lines • Used to visualize an electric field’s • Direction: Directed path along which a positive test charge would move. • Magnitude: Proportional to the number of field lines coming through an area (line density).
Drawing Electric Field Lines To draw the electric field lines for a single charge • Draw lines directed away from (+) charges. • Draw lines directed towards (-) charges. • The number of lines entering or leaving the charge is proportional to the amount of charge. • Examples to do: Using 4 lines for a (+1 C) charge, draw the field lines for: • +1C charge, +2C charge, -1C charge, -2C charge.
Drawing Electric Field Lines (cont’d) To draw electric field lines for a system of charges: • Draw field lines near each charge in the system. • Connect field lines smoothly between (+) and (-) charges. • Field lines between like charges can go to infinity. • Draw field lines perpendicular (normal) to conducting surfaces (why?).
Examples • Dipole (+ -) • Two (+) charges (+ +) • Quadrupole + - - + 4. Oppositely charged plates that are close together.(*Important*) Check answers at Electric Field Applet
How does a Conductor behave in an E-Field? • In electrostatics, charges don’t move. (Fields due to stationary charges are called electrostatic fields.) • Recall what happens when we put a charged rod near a conductor: • Transient current • Polarization of entire object • Electrons moved so as to cancel the E-field inside the conductor. (Why must this be so?) • Excess charges move to surface and distribute themselves out of mutual repulsion. • And so ....
How does an E-field behave in a Conductor? • No electrostatic field can exist in the material of a conductor. • At the surface of a charged conductor: • E-field is perpendicular to the surface. • Component tangential to the surface is zero. • Higher concentrations of charge occur at surfaces of higher curvature. • In an empty cavity within a conductor • E-field is zero • Region is shielded from any external charges.
Charge densities at curved surfaces • On flat surfaces of low curvature, repulsive forces are directed mostly parallel to surface, keeping charges further apart. • On highly curved surfaces, parallel components of the repulsive forces are smaller, allowing charges to be closer together. • Result: Strong E-fields near highly curved surfaces. (Ex: Lightning Rods)
Conductor in an E-field(summary) Charge concentrates at high curvature. Field is normal to surface Field is NOT zero in a cavity containing charge. Field is ZERO in an empty cavity.
How do we describe the ability of an Electric Field to do Work? • We want to discuss the ability of a field to do work without regard to the material being worked on. • Approach: • Develop an intuitive gravitational analog. • Extend the concept to the electrical case. • Apply the concept to different combinations of positive/negative source/test charges.
What is the Gravitational Equivalent to an E-field? • For uniform fields: • Fg = mg Fe = q E • For fields due to a point mass or charge: • Fg = m GM/r2 ř Fe = q kqs/r2 ř So E-field is to charge as ____________ is to mass.
Gravitational Potential Energy b _______________ a _______________ • Mass falling from b to a loses P.E.: ΔUg = -mgh • Work done by the gravitational field: Wfield = mgh • It takes work to lift mass from a to b: Wexternal = mgh • More mass will lose more P.E. • More mass will require more work to lift.
Gravitational Potential • Define Gravitational Potential: Vg = Ug/m (gravitational potential energy per unit mass). • Gravitational Potential Difference: ΔVg = -gh in going from b to a. • ΔVgis a measure of • How fast and how far material will fall from b to a. • How difficult it is to lift material from a to b. • ΔVgdescribes the terrain through which we can climb or fall without regard to our mass. • For a uniform field (such as near Earth’s Surface) Vg = gh or Vg = -GM/r
Topographical Maps • Topo map: maps contours or lines of equal gravitational potential. • No work is done to move objects along these lines. • Moving across lines requires work • Closely spaced lines indicates steepness. • Steepness indicates the strength of the field in the direction of your movement. (e.g. “field strength” here would be analogous to an an inclined plane: g sinθ; direction of fall would be down the plane.)
Electrical Potential Energy Let’s extend the concept to the E-field: • A (+q) charge “falls” from high (P.E.)b to low (P.E.)a : ΔUe = Uea - Ueb < 0 • Work done by the field in this falling: Wfield = F d = qE d • Change in P.E. : ΔUe = -qEd • More charged material means • More work done by the field in falling • More P.E. lost in falling • More external work to push the charge “uphill” against the field. + - | | | | (b) (a) E
Electric Potential • Define Electric Potential: V= Ue/q (Electric potential energy per unit charge). (Units: joule/coulomb or “volt”; J/C or V) • Potential Difference or (“Voltage”): (going from b to a) ΔV = Va-Vb=ΔUe/q = -qEd/q = -Ed • ΔVis a measure of • How fast and how far charged material will fall from b to a. • How difficult it is to lift charged material from a to b. • ΔUe = q ΔV (the signs of q and ΔV are important) • ΔV describes the electrical terrain through which we can climb or fall without regard to our charge. • BUT: The direction of your fall depends on the sign of your charge.
Electric Potential and Charged Material + - | | | | (b) (a) High V Low V E • E-field direction specifies region of high V to region of low V. • (+) charges fall to: • ___________ potential • _________ potential energy • (-) charges fall to: • ___________ potential • _________ potential energy ΔUe = q ΔV
Some Examples • An electron is accelerated across a voltage of 5000 volts between two plates. • What is the change in potential energy of the electron? • What is the work done on the electron by the E-field between the two plates? • What speed does the electron attain? • Who works with these kinds of problems? • What is an “electron-volt (eV)”?
The Electron Volt • The amount of work needed to push an electron across a potential difference of 1 volt. • Allows MUCH easier calculations of energies. • Allows MUCH easier grading of the calculations. • Use them when charges are given in units of “e”. • 1 eV = (1.60×10-19C)(1 J/C) = 1.60×10-19J • Notice: instead of multiplying voltage by the electronic charge in coulombs, you just put an “e” in front of it. • If you’re asked for speeds; you’re back to S.I. units.
Examples (cont’d) 2. Two plates are charged to a voltage 50V. If their separation is 5.0 cm, what is the E-field between them? The E-field is uniform between the plates. + - | | | | | | | | +50 V 0V | d = 5 cm |
Visualizing Electric Potential • Draw Equipotentials:Lines along which the electric potential is constant. • Similar to contours on a topo map. • Questions • How would you calculate the work done in moving a charge along an equipotential? • How are equipotentials related to electric field lines? They are perpendicular to each other at every point.
Mapping Equipotentials • What do equipotentials look like for • A point charge? • A dipole? • Two like charges? • Applet as before: Electric Field Applet • For a charge distribution, how could you experimentally map • Equipotentials? • Electric field lines?
Finding the Electric Potential due to a Point Charge • Suppose we have a positive charge: Q • How can we calculate (estimate) the work done by Q’s E-field as a (+) charge q falls away from rA to rB? • WTOT = Σ Wi = Σ Fi Δr = Σ q Ei Δr = q Σ kQ/ri2 Δr = q (kQ/ rA - kQ/ rB) (By Calculus) • What is the resulting change in potential energy of q? • ΔUe = - WTOT = q (kQ/rB - kQ/rA) • What is the potential difference (voltage) between rA and rB? • ΔV = ΔUe /q = kQ/rB - kQ/rA
Electric Potential due to a Point Charge • Due to a point charge (Q) • V = kQ/r at a point a distance r from Q • V 0 as r ∞ • V increases as you get closer to a (+) charge. • V decreases as you get closer to a (-) charge. • Be able to plot V vs. r for a point charge. • Potential Difference or Voltage between two points (rA and rB) (due to a point charge): • ΔV = kQ/rB - kQ/rA • Work done in moving a charge from one point to another is independent of the path taken between the two points. • ΔV between two points is independent of any path between the two points.
Electric Potential Energy(of two point charges) • If we have a (+q1) charge fixed in space • If we push another (+q2) charge towards it from ∞, how does the P.E. of the system of two charges change? • If a (-q2) charge falls towards it from ∞, how does the P.E. of the system change? • Change in P.E. of two point charges (q1 and q2) (The q’s include the signs) • ΔUe = q2 ΔV = q2 (V1 - V∞) = q2 (k q1 /r12 - 0) = k q1 q2/r12 = U12 • How is this different from gravitational P.E. ?
Electric Potential Energy(for various charge configurations) • For several charges, the potential energies add (same as in the gravitational case): • For three charges: • Utotal = U12 + U13 + U23 • For four charges • Utotal = U12 + U13 + U23 + U14 + U24 + U34 • The potentials at a point due to a collection of charges (q1, q2, q3, …) also add: • Vtotal = V1 + V2 + V3 + ….
Examples Ex 1: Looking at some charge configurations (Two charges at different spacing as shown) • +q -q • +q -q • +q +q • Which set has a positive P.E.? • Which set has the most negative P.E.? • Which set requires the most work to separate the charges to ∞?
Examples (cont’d) Q1 is at (0, 26 cm) Q2 is at (0, -26 cm) 2. Potential above two charges: B 40 cm 40 cm A 30 cm Find the potential at points A and B due to charges Q1 and Q2. 60 cm Q2=+50μC Q1=-50μC At A: VA = VA1 + VA2 = k(Q1/rA1 + Q2/rA2) = At B: VB = VB1 + VB2 = k(Q1/rB1 + Q2/rB2) =
Examples (cont’d) • For the configuration of charges shown • Calculate the energy necessary to build the configuration. • Calculate the potential at the center of the triangle. q1=+4µC S=.2m q3= -4µC q2=+4µC
How can we store Electric Energy? (How can we store Electric Charge?)