Physics 2102 Lecture 6ExampleElectric Potential of a Dipole (on axis)Electric Potential on Perpendicular Bisector of DipoleContinuous Charge DistributionsPotential of Continuous Charge Distribution: ExampleElectric Field & Potential: A Simple Relationship!PowerPoint PresentationSlide 9Equipotentials and ConductorsConductors change the field around them!“Sharp”conductorsSummary:Electric Potential IIElectric Potential IIPhysics 2102Jonathan DowlingPhysics 2102 Physics 2102 Lecture 6Lecture 6ExampleExamplePositive and negative charges of equal magnitude Q are held in a circle of radius R. 1. What is the electric potential at the center of each circle?• VA = • VB =• VC = 2. Draw an arrow representing the approximate direction of the electric field at the center of each circle. 3. Which system has the highest electric potential energy?–Q+QACB€ k +3Q − 2Q( )/r = +kQ /r∑=iiirqkV€ k +2Q − 4Q( )/r = −2kQ/ rk +2Q −2Q( )/r =0UBElectric Potential of a Dipole (on axis)Electric Potential of a Dipole (on axis)( ) ( )2 2Q QV k ka ar r= −− +What is V at a point at an axial distance r away from the midpoint of a dipole (on side of positive charge)?ar-Q+Q( ) ( )2 2( )( )2 2a ar rkQa ar r⎛ ⎞+ − −⎜ ⎟=⎜ ⎟⎜ ⎟− +⎝ ⎠)4(4220arQa−=πε204 rpπε=Far away, when r >> a:VElectric Potential on Perpendicular Electric Potential on Perpendicular Bisector of DipoleBisector of DipoleYou bring a charge of Qo = –3C from infinity to a point P on the perpendicular bisector of a dipole as shown. Is the work that you do:a) Positive?b) Negative?c) Zero?a-Q+Q-3CPU= QoV=Qo(–Q/d+Q/d)=0 dContinuous Charge Continuous Charge DistributionsDistributions•Divide the charge distribution into differential elements•Write down an expression for potential from a typical element — treat as point charge•Integrate!•Simple example: circular rod of radius r, total charge Q; find V at center.dqV kr=∫dqrk Qdq kr r= =∫Potential of Continuous Charge Potential of Continuous Charge Distribution: Example Distribution: Example /Q Lλ =dxdq λ=∫∫−+==LxaLdxkrkdqV0)(λ[ ]LxaLk0)ln( −+−= λ⎥⎦⎤⎢⎣⎡+=aaLkV lnλ•Uniformly charged rod•Total charge Q•Length L•What is V at position P shown?PxLadxElectric Field & Potential: Electric Field & Potential: A Simple Relationship! A Simple Relationship!Notice the following:•Point charge:–E = kQ/r2–V = kQ/r•Dipole (far away):–E ~ kp/r3–V ~ kp/r2•E is given by a DERIVATIVE of V!•Of course! dxdVEx−=Focus only on a simple case:electric field that points along +x axis but whose magnitude varies with x. Note: • MINUS sign!• Units for E -- VOLTS/METER (V/m)fiV E dsΔ =− •∫rrElectric Field & Potential: Example Electric Field & Potential: Example •Hollow metal sphere of radius R has a charge +q•Which of the following is the electric potential V as a function of distance r from center of sphere?+qVr1≈rr=R(a)Vr1≈rr=R(c)Vr1≈rr=R(b)+qOutside the sphere:• Replace by point charge!Inside the sphere:• E =0 (Gauss’ Law) • E = –dV/dr = 0 IFF V=constant2dVEdrd Qkdr rQkr=−⎡ ⎤=−⎢ ⎥⎣ ⎦=Vr1≈Electric Field & Potential: Example Electric Field & Potential: Example E21r≈Equipotentials and ConductorsEquipotentials and Conductors•Conducting surfaces are EQUIPOTENTIALs •At surface of conductor, E is normal to surface•Hence, no work needed to move a charge from one point on a conductor surface to another •Equipotentials are normal to E, so they follow the shape of the conductor near the surface.Conductors change the field Conductors change the field around them!around them!An uncharged conductor:A uniform electric field:An uncharged conductor in the initially uniform electric field:““Sharp”conductorsSharp”conductors•Charge density is higher at conductor surfaces that have small radius of curvature•E = 0 for a conductor, hence STRONGER electric fields at sharply curved surfaces!•Used for attracting or getting rid of charge: –lightning rods–Van de Graaf -- metal brush transfers charge from rubber belt–Mars pathfinder mission -- tungsten points used to get rid of accumulated charge on rover (electric breakdown on Mars occurs at ~100 V/m)(NASA)Summary:Summary:•Electric potential: work needed to bring +1C from infinity; units = V•Electric potential uniquely defined for every point in space -- independent of path!•Electric potential is a scalar -- add contributions from individual point charges•We calculated the electric potential produced by a single charge: V=kq/r, and by continuous charge distributions : V= kdq/r• Electric field and electric potential: E= dV/dx• Electric potential energy: work used to build the system, charge by charge. Use W=qV for each charge.• Conductors: the charges move to make their surface equipotentials. • Charge density and electric field are higher on sharp points of
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