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Question 300 V/m 300 V/m 0 V/m A B 0.02m What is VB-VA? 0.03m A) B) C) D) E) 270 V -270 V -18 V 6V -6 V 0.04m Question EA = 300 x̂ V / m EB = -300 x̂ V / m 0 V/m A B 0.02m 0.04m 0.03m x 0 B .02 DV = VB - VA = - ò E × dl = - ò E × dl A 0 .09 ò E × dl dl = x̂dx .05 VB-VA = -300*(0.02-0) - (-300)*(0.09-0.05)=-6+12 V = +6 V Potential in Metal Not in static equilibrium Metal is not in static equilibrium: • When it is in the process of being polarized • When there is an external source of mobile charges (battery) f DV = - ò E · dl i For each step, the potential difference is: V = -EL Nonzero electric field of uniform magnitude E throughout the interior of a wire of length L. Direction of the field follows the direction of the wire. Electric Field and Potential If we know electric field ( ) everywhere we can compute potential (V) in every point in space. Can we compute ( ) if we know V? Exercise Suppose in some area of space V(x,y,z)=x2+yz. What is E(x,y,z)? ¶V E = - ¶V ¶V Ex = Ez = y ¶y ¶x ¶z ( ¶V Ex = =¶x ¶ x 2 + yz ¶V Ey = =¶y ¶ x 2 + yz ¶V Ez = =¶z ¶ x 2 + yz ( ( ¶x ¶y ¶z ) = - ¶ ( x ) - ¶ ( yz ) = -2x 2 ¶x ¶x ) = - ( ) - ¶ ( yz ) = -z ¶ x2 ¶y ¶y ) = - ( ) - ¶ ( yz ) = -y E ( x, y, z ) = -2x,-z,-y ¶ x2 ¶z ¶z Wire in a Circuit and Electric Field f DV = - ò E · dl i In a current-carrying wire in a circuit, there can be a nonzero electric field, so there is a difference in potential between two locations in the wire. DV = EL Electric field is not confined to the wire in a circuit – there must be an electric field E around the wire in the air! Potential of a Uniformly Charged Ring Q z 2 + R2 Method 1: Divide into point charges and add up contributions due to each charge 1 DQ DV = 4pe 0 r V = å DV = å V= V= 1 DQ 4pe 0 z 2 + R2 1 1 4pe 0 z +R 1 Q 4pe 0 2 2 z2 + R2 å DQ Potential of a Uniformly Charged Ring Q z 2 + R2 Method 2: Integrate electric field along a path z V = - ò E · dl ¥ z V = -ò ¥ Ez = 1 ( Qz 4pe 0 R + z dl = ẑdz 2 V =- ) 2 3/2 1 Qz ( 4pe 0 R 2 + z 1 4pe 0 z Qò ¥ (R ) 2 3/2 z 2 +z ) dz 2 3/2 z é -1 ù V =Qê 4pe 0 ë z 2 + R 2 úû ¥ 1 Q V= 4pe 0 z 2 + R 2 1 dz Potential of a Uniformly Charged Ring Q z 2 + R2 V= 1 4pe 0 Q z2 + R2 What is V for z>>R ? 1 Q V» 4pe 0 z The same as for a point charge! The difference between metals and insulators -Q metal +Q E inside metal = 0 -Q insulator +Q E inside insulator is non zero Dielectric Constant +Q -q - +q -Q + + + + + + + + + + Electric field in capacitor filled with insulator: Enet=Eplates-Edipoles qdipoles = b*Q Enet Q / A) ( = Ke0 Enet = K – dielectric constant E plates K Dielectric Constant Inside an insulator: Enet = Eapplied K Dielectric constant for various insulators: vacuum air typical plastic NaCl water strontium titanate 1 (by definition) 1.0006 5 6.1 80 310 Potential Difference in a Capacitor with Insulator B DV = - ò E · dl E plates A DV = Es = s E plates s K Q / A) ( DV = Es = s Ke0 DVvacuum DVinsulator = K Q / A) ( = e0 Potential Difference in Partially Filled Capacitor B d +Q -Q DV = - ò E · dl E plates = - A (Q / A) e0 K DV = DVvacuum + DVinsulator DVvacuum = DVinsulator A s B x DV = (Q / A) e0 (s - d) (Q / A) = d Ke0 (Q / A) e0 [s - d(1 - 1 / K )] x̂ Finding Potential Difference 1. Subtract the potential at the initial location A from the potential at final location B DV = VB - VA A B 2. Travel along a path from A to B adding up - E · dl at each step: B DV = - ò E · dl A A dl B E Common Pitfall Assume that the potential V at a location is defined by the electric field E at this location. A Example: E = 0 inside a charged metal sphere, but V is not! A negative test charge Q = -0.6C was moved from point A to point B In a uniform electric field E=5N/C. The test charge is at rest before and after the move. The distance between A and B is 0.5m and the line connecting A and B is perpendicular to the electric field. How much work was done by the net external force while moving the test charge from A to B? A 0.5m B A. B. C. D. E. 1.5J 0J –1.5J 3.0J –3.0J E = 5 N/C After moving the -0.6C test charge from A to B, it was then moved from B to C along the electric field line. The test charge is at rest before and after the move. The distance between B and C also is 0.5m. How much work was done by the net external force while moving the test charge from A to C? A E = 5 N/C 0.5m A. B. C. D. E. 1.5J 0J –1.5J 3.0J -3.0J B C Instead of moving the test charge from A to B then to C, it is moved from A to D and then back to C. The test charge is at rest before and after the move. How much work was done by the net external force while moving the test charge this time? A D 0.5m B A. B. C. D. E. 0.5m C 1.5J 0J –1.5J Infinitely big Do not know at this time. E = 5 N/C Conducting charged sphere and concentric charged conducting shell Q2 Q1 rA rB rC A solid conducting sphere of radius rA has charge Q1 uniformly distributed over its surface. Concentric with the solid sphere is a conducting shell of inner radius rB and outer radius rC with charge Q2 on its outer surface. What is the electric potential at the center of the solid sphere? Take the potential at infinity to be equal to zero and the origin at the center of the solid sphere. Energy Density of Electric Field How much work do I need to do to increase separation of plates? Wby_you=Fby_you*s Note that by doing so you increase area in space where electric field (E = Q/A0) is present. Fby _ you (Q / A) = QE = Q 2e 0 Eone _ plate Q / A) ( = 2e 0 (for small s) Energy Density of Electric Field Energy can be stored in electric fields DU electric = Fby _ you Ds = Q (Q / A) Ds 2e 0 Multiply by A/A and 0/0 2 1 æQ / Aö ÷÷ ADs DU el = e 0 çç 2 è e0 ø E DU el 1 2 = e E Field energy density: (J/m3) 0 D ( volume) 2 Volume in which we created electric field Formula holds for any charge configuration. Energy expended by us was converted into energy stored in the electric field Potential Energy and Field Energy A different way to express all space instead of calculating a work it takes to assemble charges. The idea of energy stored in fields is a general one: Magnetic and gravitational fields can also carry energy. An Electron and a Positron System Surroundings e+ e- Release electron and positron – the electron (system) will gain kinetic energy Conservation of energy surrounding energy must decrease Does the energy of the positron decrease? - No, it increases Where is the decrease of the energy in the surroundings? - Energy stored in the fields must decrease An Electron and a Positron System Surroundings e+ Single charge: 1 E~ 2 r eEnergy: 1 2 e E ò 2 0 dV Dipole: 1 (far) E~ 3 r Energy stored in the E fields decreases as e+ and e- get closer!