Example: stock market

Electrostatic Force and Electric Charge

R. D. FieldPHY 2049 Chapter Force and ElectricElectrostatic Force and ElectricChargeChargeElectrostatic Force ( charges at rest): Electrostatic Force can be attractive Electrostatic Force can be repulsive Electrostatic Force acts through emptyspace Electrostatic Force much stronger thangravity Electrostatic forces are inverse square law forces (proportional to1/r2) Electrostatic Force is proportional to the product of the amount of chargeon each interacting objectMagnitude of the Electrostatic Force is given by Coulomb'sLaw:F = K q1q2/r2 (Coulomb's Law)where K depends on the system of unitsK = Nm2/C2 (in MKS system)K = 1/(4 0) where 0 = C2/(Nm2) Electric Charge :electron Charge = -ee = Cproton Charge = eC = CoulombElectric Charge is a conserved quantity (net Electric Charge is nevercreated or destroyed!)q1q2rR. D. FieldPHY 2049 Chapter System (meters-kilograms-seconds):also Amperes, Volts, Ohms, WattsForce:F = maNewtons = kg m / s2 = 1 NWork:W = FdJoule = Nm = kg m2 / s2 = 1 JElectric Charge :QCoulomb = 1 CF = K q1q2/r2 K = Nm2/C2 (in MKS system)CGS System (centimeter-grams-seconds): Force :F = ma1 dyne = g cm / s2 Work:W = Fd1 erg = dyne-cm = g cm2 / s2 Electric Charge :Qesu ( Electrostatic unit)F = q1q2/r2 K = 1 (in CGS system)Conversions (MKS - CGS): Force :1 N = 105 dynesWork:1 J = 107 ergsElectric Charge :1 C = esuFine Structure Const

A dipole with charge Q and separation d is located on the x-axis with its midpoint at the origin. A charge q is on the x-axis a distance x from the midpoint of the dipole. What is the electric force on q due to the dipole and how does this force behave in the limit x >>d (dipole approximation )? R. D. Field PHY 2049 ...

Tags:

  Charges, Separation

Information

Domain:

Source:

Link to this page:

Please notify us if you found a problem with this document:

Other abuse

Transcription of Electrostatic Force and Electric Charge

1 R. D. FieldPHY 2049 Chapter Force and ElectricElectrostatic Force and ElectricChargeChargeElectrostatic Force ( charges at rest): Electrostatic Force can be attractive Electrostatic Force can be repulsive Electrostatic Force acts through emptyspace Electrostatic Force much stronger thangravity Electrostatic forces are inverse square law forces (proportional to1/r2) Electrostatic Force is proportional to the product of the amount of chargeon each interacting objectMagnitude of the Electrostatic Force is given by Coulomb'sLaw:F = K q1q2/r2 (Coulomb's Law)where K depends on the system of unitsK = Nm2/C2 (in MKS system)K = 1/(4 0) where 0 = C2/(Nm2) Electric Charge :electron Charge = -ee = Cproton Charge = eC = CoulombElectric Charge is a conserved quantity (net Electric Charge is nevercreated or destroyed!)q1q2rR. D. FieldPHY 2049 Chapter System (meters-kilograms-seconds):also Amperes, Volts, Ohms, WattsForce:F = maNewtons = kg m / s2 = 1 NWork:W = FdJoule = Nm = kg m2 / s2 = 1 JElectric Charge :QCoulomb = 1 CF = K q1q2/r2 K = Nm2/C2 (in MKS system)CGS System (centimeter-grams-seconds): Force :F = ma1 dyne = g cm / s2 Work:W = Fd1 erg = dyne-cm = g cm2 / s2 Electric Charge :Qesu ( Electrostatic unit)F = q1q2/r2 K = 1 (in CGS system)Conversions (MKS - CGS): Force :1 N = 105 dynesWork:1 J = 107 ergsElectric Charge :1 C = esuFine Structure Constant (dimensionless): = = K 2 e2/hc (same in all systems of units)h = Plank's Constant c = speed of light in vacuumR.

2 D. FieldPHY 2049 Chapter Force versus GravityElectrostatic Force versus GravityElectrostatic Force :Fe = K q1q2/r2 (Coulomb's Law)K = Nm2/C2 (in MKS system)Gravitational Force :Fg = G m1m2/r2 (Newton's Law)G = Nm2/kg2 (in MKS system)Ratio of forces for two electrons : e = C m = kge, me, mrFe / Fg = K e2 / G m2 = (Huge number !!!)R. D. FieldPHY 2049 Chapter ForcesVector ForcesqQr^The Electrostatic Force is a vector:The Force on q due to Q points along the direction r and is givenbyrFKqQrr=2$q1QF1q2q3F2F3 Vector Superposition of Electric Forces:If several point charges q1, q2, q3, .. simultaneously exert electricforces on a Charge Q thenF = F1 + F2 +F3 + ..R. D. FieldPHY 2049 Chapter & Vector AdditionVectors & Vector AdditionThe Components of a vector:x-axisy-axis AAx =A cos Ay =A sin Vector Addition:x-axisy-axisACBTo add vectors you add the components of the vectors as follows:rrrrrAAxAyAzBBxByBzCABABxAByABzx yzxyzxxyyzz=++=++=+=+++++$$$$$$()$()$()$ R.

3 D. FieldPHY 2049 Chapter Electric DipoleThe Electric Dipole+Q-QdAn Electric "dipole" is two equal and opposite point chargesseparated by a distance d. It is an electrically neutral "dipole moment" is defined to be the Charge times theseparation (dipole moment = Qd).Example Problem:+Q-QdqxA dipole with Charge Q and separation d is located on the y-axis with itsmidpoint at the origin. A Charge q is on the x-axis a distance x from themidpoint of the dipole. What is the Electric Force on q due to the dipole andhow does this Force behave in the limit x >>d (dipole approximation)?Example Problem:-Q+QdxA dipole with Charge Q and separation d is located on the x-axis with itsmidpoint at the origin. A Charge q is on the x-axis a distance x from themidpoint of the dipole. What is the Electric Force on q due to the dipole andhow does this Force behave in the limit x >>d (dipole approximation)?R. D. FieldPHY 2049 Chapter Electric FieldThe Electric Field+QqEThe Charge Q produces an Electric field which in turnproduces a Force on the Charge q.

4 The Force on q is expressedas two terms:F = K qQ/r2 = q (KQ/r2) = q EThe Electric field at the point q due to Q is simply the Force perunit positive Charge at the point q:E = F/q E = KQ/r2 The units of E are Newtons per Coulomb (units = N/C).The Electric field is a physical object which can carry bothmomentum and energy. It is the mediator (or carrier) of theelectric Force . The Electric field is Electric Field is a Vector Field:rEKQrr=2$R. D. FieldPHY 2049 Chapter Field LinesElectric Field Lines+Q-QElectric field line diverge from ( start) on positive chargeand end on negative Charge . The direction of the line is thedirection of the Electric number of lines penetrating a unit area that isperpendicular to the line represents the strength of the electricfield.+2Q+QR. D. FieldPHY 2049 Chapter Field due to a DistributionElectric Field due to a Distributionof Chargeof ChargedQdE = K dQ/r2 rrThe Electric field from a continuous distribution of Charge isthe superposition ( integral) of all the (infinite)contributions from each infinitesimal dQ as follows:rEKrrdQ= 2$ and QdQ= Charge Distributions: Linear Charge density : ( (x) ) = Charge /unitlengthLdQ = dx For a straight line dQ = (x) dx andQdQxdx== ()If (x) = is constant then dQ = dx and Q = L,where L is the D.

5 FieldPHY 2049 Chapter DistributionsCharge DistributionsCharge Distributions: Linear Charge density : ( ) ( ) = Charge /unit arclengthRdQ = ds = R d For a circular arc dQ = ( ) ds = ( ) Rd andQdQdsRd=== ()()If ( ) = is constant then dQ = ds and Q = s, where s is thearc length. Surface Charge density : ( (x,y) ) = Charge /unit areadQ = dA For a surface dQ = (x,y) dA andQdQxydA== (,)If (x,y) = is constant then dQ = dA and Q = A, where Ais the area. Volume Charge density : ( (x,y,z) ) = Charge /unit volumedQ = dV For a surface dQ = (x,y,z) dV andQdQxyzdV== (,,)If (x,y,z) = is constant then dQ = dV and Q = V, where V isthe D. FieldPHY 2049 Chapter the Electric FieldCalculating the Electric FieldExample:PxLA total amount of Charge Q is uniformily distributed along a thin straight rodof length L. What is the Electric field a a point P on the x-axis a distance xfrom the end of the rod?

6 Answer: rEKQxxLx=+()$Example:A total amount of Charge Q is uniformilydistributed along a thin straight rod oflength L. What is the Electric field a a pointP on the y-axis a distance y from themidpoint of the rod?Answer: rEKQyyLy=+222(/)$Example:A infinitely long straight rod has a uniformcharge density . What is the Electric field apoint P a perpendicular distance r from therod?Answer: rEKrr=2 $PyLPr R. D. FieldPHY 2049 Chapter Useful MathSome Useful MathApproximations:()111+ +<< pp()111 << ppe +<<11tansin <<<<11 Indefinite Integrals:()axadxxxa2223222+=+ /()xxadxxa2232221+= + /R. D. FieldPHY 2049 Chapter the Electric FieldCalculating the Electric FieldExample:A total amount of Charge Q is uniformilydistributed along a thin semicircle ofradius R. What is the Electric field a apoint P at the center of the circle?Answer: rEKQRx=22 $Example:A total amount of Charge Q isuniformily distributed along athin ring of radius R.

7 What isthe Electric field a point P onthe z-axis a distance z from thecenter of the ring?Answer:()rEKQzzRz=+2232/$Example:A total amount of Charge Q isuniformily distributed on thesurface of a disk of radius is the Electric field apoint P on the z-axis a distancez from the center of the disk?Answer:rEKQRzzRz= + 21222$RPx-axisRPz-axiszRPz-axiszR. D. FieldPHY 2049 Chapter the Electric FieldCalculating the Electric FieldExample:What is the Electric field generated by alarge (infinite) sheet carrying a uniformsurface Charge density of coulombs permeter?Answer: rEz= 20$Example:What is the Electric field at a point Pbetween two large (infinite) sheetscarrying an equal but opposite uniformsurface Charge density of ?Answer: rEz= 0$Pz P- R. D. FieldPHY 2049 Chapter of a Vector FieldFlux of a Vector FieldFluid Flow:n^Flux = vAn^Flux = 0n^Flux = vA cos Consider the fluid with a vectorrv which describes the velocity of the fluidat every point in space and a square with area A = L2 and normal$n.

8 Theflux is the volume of fluid passing through the square area per unit to the Electric Field: Electric flux through the infinitesimal areadA is equal todEdA = rrwheredAAnr=$Total Electric Flux through a Closed Surface: ESEdA= rrdAE n^d =E dA cos EnormalSurface SR. D. FieldPHY 2049 Chapter Flux and Gauss' LawElectric Flux and Gauss' LawThe Electric flux through any closed surface isproportional to the net Charge = 0 For the discrete case the total Charge enclosed is the sum over allthe enclosed charges :QqenclosediiN== 1 For the continuous case the total Charge enclosed is the integral ofthe Charge density over the volume enclosed by the surface S:QdVenclosed= Simple Case: If the Electric field is constant over the surface andif it always points in the same direction as the normal to thesurface then ESEdAEA= = rrThe units for the Electric flux are Nm2 SR. D. FieldPHY 2049 Chapter in Static EquilibriumConductors in Static EquilibriumConductor: In a conductor someelectrons are free to move (withoutrestraint) within the volumn of thematerial (Examples: copper, silver, aluminum, gold)Conductor in Static Equilibrium:When the Charge distribution on aconductor reaches static equilibrium( nothing moving), the net electricfield withing the conductingmaterial is exactly zero (and the Electric potential is constant).

9 Excess Charge : For a conductorin static equilibrium all the (extra) Electric Charge reside on thesurface. There is no net electriccharge within the volumn of theconductor ( = 0). Electric Field at the Surface:The Electric field at the surface of aconductor in static equilibrium isnormal to the surface and has a magnitude, E = / 0, where isthe surface Charge density ( Charge per unit area) and the netcharge on the conductor isQdASurface= .ConductorConductor instatic equilibriumE = 0V = constantConductor instatic equilibriumE = 0V = constant = 0 Surface Charge Density ER. D. FieldPHY 2049 Chapter ' Law ExamplesGauss' Law ExamplesProblem: A solid insulating sphere of radius Rhas Charge distributed uniformly throughout itsvolume. The total Charge of the sphere is Q. Whatis the magnitude of the Electric field inside andoutside the sphere?Answer:Problem: A solid conducting sphere of radius Rhas a net Charge of Q.

10 What is the magnitude of theelectric field inside and outside the sphere? Whereare the charges located?Answer: charges are on the surface and rrEKQrrEoutin==20$Problem: A solid conducting sphere of radiusb has a spherical hole in it of radius a and has anet Charge of Q. If there is a point Charge -qlocated at the center of the hole, what is themagnitude of the Electric field inside and outsidethe conductor? Where are the charges on theconductor located?Answer: charges are on the inside and outsidesurface with Qin=q and Qout=Q-q and rrEKQqrrEEKqrrrbarbra> <<<= == ()$$220 Total Charge Q = constantInsulating SphereRrrEKQrrEKQrRroutin==23$$Net Charge QConductingSphereRNet Charge Qon conductorb-qaR. D. FieldPHY 2049 Chapter Potential EnergyGravitational Potential EnergyGravitational Force : F = G m1m2/r2 Gravitational Potential Energy GPE:U = GPE = mgh (near surface of the Earth)Kinetic Energy: KE = 122mvTotal Mechanical Energy: E = KE +UWork Energy Theorem:W = EB-EA = (KEB-KEA) + (UB-UA)(work done on the system)Energy Conservation: EA=EB(if no external work done on system)Example:A ball is dropped from a height h.


Related search queries