Wednesday, Sept. 28, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu1PHYS 1444 – Section 003Lecture #9Wednesday, Sept. 28, 2005Dr. Jaehoon Yu• Quiz Results and Solution• Electric Energy Density• Dielectrics• Molecular Description of Dielectrics• The Electric Battery• Electric CurrentWednesday, Sept. 28, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu2Announcements• Reading Assignment– CH24 – 6 – Early part of 25 – 1 • Quiz results– Do you want to know what your average is?• 37.5/70 Î equivalent to 54/100• Do you want to know what it was last time?– 42.8/60 Î equivalent to 71/100• Hmm…. What do you think???– Do you want to know the top score?• 68/70 Î 97/100Wednesday, Sept. 28, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu3Electric Energy Density• The energy stored in a capacitor can be considered as being stored in the electric field between the two plates• For a uniform field E between two plates, V=Ed and C=ε0A/d• Thus the stored energy is•Since Ad is the gap volume V, we can obtain the energy density, stored energy per unit volume, as U =212CV=()2012AEddε⎛⎞=⎜⎟⎝⎠2012EAdε2012uEε=Electric energy stored per unit volume in any region of space is proportional to the square of E in that region.Valid for any space that is vacuumWednesday, Sept. 28, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu4Dielectrics• Capacitors have an insulating sheet of material, called dielectric, between the plates to– Increase breakdown voltage than that in the air– Higher voltage can be applied without the charge passing across the gap– Allow the plates get closer together without touching• Increases capacitance ( recall C=ε0A/d)– Also increases the capacitance by the dielectric constant– Where C0is the intrinsic capacitance when the gap is vacuum0CKC=Wednesday, Sept. 28, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu5Dielectrics• The value of dielectric constant varies depending on material (Table 24 – 1) – K for vacuum is 1.0000– K for air is 1.0006 (this is why permittivity of air and vacuum are used interchangeably.)• Maximum electric field before breakdown occurs is the dielectric strength. What is its unit?–V/m• The capacitance of a parallel plate capacitor with a dielectric (K) filling the gap is 00ACKC Kdε==Wednesday, Sept. 28, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu6• A new quantity of the permittivity of dielectric is defined as ε=Kε0• The capacitance of a parallel plate with a dielectric medium filling the gap is• The energy density stored in an electric field E in a dielectric is DielectricsACdε=2201122uKE Eεε==Valid for any space w/ dielectric w/ permittivity ε.Wednesday, Sept. 28, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu7• Let’s consider the two cases below: Effect of a Dielectric Material • Constant voltage: Experimentally observed that the total charge on the each plates of the capacitor increases by K as the dielectric material is inserted between the gap Î Q=KQ0– The capacitance increased to C=Q/V0=KQ0/V0=KC0• Constant charge: Voltage found to drop by a factor K Î V=V0/K– The capacitance increased to C=Q0/V=KQ0/V0=KC0Case #1 : constant VCase #2 : constant QWednesday, Sept. 28, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu8• What happens to the electric field within a dielectric?• Without a dielectric, the field is– What are V0and d?•V0: Potential difference between the two plates• d: separation between the two plates• For the constant voltage, the electric field remains the same• For the constant charge: the voltage drops to V=V0/K, thus the field in the dielectric is– The field in the dielectric is reduced. Effect of a Dielectric Material on Field 00VEd=00DVEVEEddKK=== =0DEEK=Wednesday, Sept. 28, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu9Example 24 – 8Dielectric Removal: A parallel-plate capacitor, filled with a dielectric with K=3.4, is connected to a 100-V battery. After the capacitor is fully charged, the battery is disconnected. The plates have area A=4.0m2, and are separated by d=4.0mm. (a) Find the capacitance, the charge on the capacitor, the electric field strength, and the energy stored in the capacitor. (b) The dielectric is carefully removed, without changing the plate separation nor does any charge leave the capacitor. Find the new value of capacitance, electric field strength, voltage between the plates and the energy stored in the capacitor.(a) C =Q =Adε=0KAdε=()212 2 2 834.03.4 8.85 1 0 3 .0 10 3 04.0 10mCNm F nFm−−−×× ⋅ =× =×E=U =CV =()863.0 10 100 3.0 10 3.0FVCCµ−−××=×=Vd=431002.5 104.0 10VVmm−=××212CV =()()28413.0 10 100 1.5 102FVJ−−×=×Wednesday, Sept. 28, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu10Since charge is the same ( ) before and after the removal of the dielectric, we obtainExample 24 – 8 cont’d(b) 0C =0QQ=Since the dielectric has been removed, the effect of dielectric constant must be removed as well. 0E=0U =0V =Where did the extra energy come from?.The energy conservation law is violated in electricity???External force has done the work of 3.6x10-4J on the system to remove dielectric!!Wrong! Wrong! Wrong!CK=()212 2 2 934.08.85 10 8.8 10 8.84.0 10mCNm F nFm−−−×⋅ =×=×0QC = KQ C=KV=3.4 100 340VV×=0Vd=433408.5 10 844.0 10VVm kVmm−=× =×20012CV =()212CKVK=212KCV=KU =443.4 1.5 10 5.1 10JJ−−×× =×Wednesday, Sept. 28, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu11Molecular Description of Dielectric• So what in the world makes dielectrics behave the way they do?• We need to examine this in a microscopic scale.• Let’s consider a parallel plate capacitor that is charged up +Q(=C0V0) and –Q with air in between.– Assume there is no way any charge can flow in or out• Now insert a dielectric– Dielectric can be polar Îcould have permanent dipole moment. What will happen?• Due to electric field molecules may be aligned.Wednesday, Sept. 28, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu12Molecular Description of Dielectric• OK. Then what happens?• Then effectively, there will be some negative charges close to the surface of the positive plate and positive charge on the negative plate– Some electric field do not pass through the whole dielectric butstops at the negative charge– So the field inside dielectric is smaller than the air• Since electric field is smaller, the force is smaller– The work need to move a test charge
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