1ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityLecture 12Energy, Force, and Work in Electro- and Magneto-QuasistaticsIn this lecture you will learn:• Relationship between energy, force, and work in electroquasistatic and magnetoquasistatic systemsECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityLorentz Electric ForceThe force on a charge q in an E-field is:EqFrr=qEExample: consider two charges q1and q2:1q2qdThe force that q1exerts upon q2can be obtained by multiplying the E-field that q1produces at the location of q2by the charge q2 :21221212121221ˆ4ˆ4→→→===ndqqndqqEqFooεπεπrr21ˆ→nThe force that q2exerts upon q1is just equal and opposite to what q1exerts upon q2 (Newton’s third law):122212122112ˆ4ˆ4→→→=−= ndqqndqqFooεπεπr2ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityForce Between Charged Plates – Iarea = A++++++++--------dNeed to find the total force on the right plate exerted by the left plate (which is also equal an opposite to the force on the left plate exerted by the right plate)Step (1): Find the E-field at the location of the right plate produced by the left plate ASSUMING THE RIGHT PLATE WAS NOT THERE:()oplaterightnoxdxEεσ2==Step (2): Multiply the E-field calculated above by the total charge on the right plate to get the desired force()()oplaterightnoxRLxAdxEAFεσσ22−==−=→σσ−Consider two charged plates carrying charge per unit areas of +σand –σ, respectively0xECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityPotential Energy and ForceThe force on a charge q in an E-field is:xxqEF =qExThe potential energy of a charge in the E-field is:()xqExqUx−==φSo the force on the charge can also be written as:dxdUFx−=Electric force is the derivative of potential energyThis formula is much more general than it appears (see the next few slides …..)x3ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityEnergy, Force, and Work in ElectromagneticsLets generalize the relationship between energy and forceSuppose one has a CLOSED electromagnetic system with electromagnetic energy UElectromagnetic systemIf the system does some mechanical work “F dx”then its electromagnetic energy must decrease by the same amount: FdxdU −=ordxdUF−=Here “F ” is some force that results in the change of some system length parameter by “dx”ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityForce Between Charged Plates - IINow lets use the energy concepts to calculate the same forceTotal electric energy stored in the field can be calculated as follows:()dAdVEEUooo222.⎟⎟⎠⎞⎜⎜⎝⎛=∫∫∫=εσεεrrThe force between the plates can move the plates thereby doing work that would result in a change of the value of “d”oooAAdUFεσεσε2222−=⎟⎟⎠⎞⎜⎜⎝⎛−=∂∂−=The negative sign indicates the force is in the direction of decreasing “d ” i.e. the force between the plates is attractivearea = A++++++++--------dσσ−0 x4ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityForce Between Charged Metal Plates - IV+-φ = Vφ = 0area = A()dVxEx=++++++++--------dACCVQoε==dNow consider the force between the charged metal plates of a parallel plate capacitor connected to a voltage source221CVU =σσ−dVoεσ=What is the force between the plates in terms of the applied voltage V?Answer should be the same as before:dCVAFo22212−=−=εσdCVVdCdUF222121=∂∂−=∂∂−=But now when you use the previous formula you will get the wrong sign:What went wrong ?The presence of a voltage source means you don’t have a closed system anymoreECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityForce Between Charged Metal Plates - IIV+-φ = Vφ = 0area = A()dVxEx=++++++++--------dACCVQoε==d221CVU =σσ−dVoεσ=• If the system tries to do work by bringing the plates closer orpushing them further apart, the voltage source ensures that the charge Q in the capacitor always satisfies the relation: Q=CV• The voltage source ensures that Q is always equal to CV by bringing in or removing charge from the capacitor while the mechanical work is being performed• This work done by the voltage source in bringing in or removing charges must also be included in the analysis5ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversitySo instead of:FdxdU−=We write:dtVIFdxdU+−=• If the system performs mechanical work Fdx then its electromagnetic energy U must decrease by the same amount • If the voltage source passes a current I in time dtthen it does work and the system electromagnetic energy U must also increase by the same amountxQVxUF∂∂+∂∂−=⇒Lets further generalize the relationship between energy and force by including voltage sourcesSuppose one has an electromagnetic system with electromagnetic energy U that is connected to a FIXED voltage source VElectromagnetic systemV+-Energy, Force, and Work in Electromagnetics – Voltage SourcesAssuming voltage V is held fixedIBut if dQ is the total charge that passed in time dt then: dtIdQ=dQVFdxdU +−=⇒ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityEnergy, Force, and Work in Electromagnetics – Voltage SourcesElectromagnetic systemV+-VdQFdxdU +−=dQdQQUdxxUdUxQ fixed fixed ∂∂+∂∂=More generally the total differential of energy U can be written as:But we had:Therefore, it must be that:fixed QxUF∂∂−=andfixedxQUV∂∂=But if the thing that is kept fixed is the voltage V then we already have the result:fixedVxQVxUF⎟⎠⎞⎜⎝⎛∂∂+∂∂−=6ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityForce Between Charged Metal Plates - IIIV+-φ = Vφ = 0area = A()dVxEx=++++++++--------dACCVQoε==d221CVU =σσ−dVoεσ=221CVU =dCVVdCVVdCdQVdUFV22fixed 2121−=∂∂+∂∂−=⎟⎠⎞⎜⎝⎛∂∂+∂∂−=Electric energy is:Since the voltage V is held fixed the force between the plates can be calculated as:This time we have got the sign right as wellECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityApplication: An Electrostatic ActuatorV+-φ = Vφ = 0dVEx=++++++++--------CVQ=d221CVU =εyLArea of plates = LtDepth of plates = t()dyttyLCoεε+−=221CVU =()22fixed 2121VdtVyCVVyCyQVyUFoVεε−=∂∂+∂∂−=⎟⎠⎞⎜⎝⎛∂∂+∂∂−=Electric energy is:Force on the slab is:The applied voltage can be used to pull in the dielectric slab and when the voltage is removed the slab will come down by gravity+ve sign of the force means