Physics 217, Fall 2002 30 September 2002(c) University of Rochester 130 September 2002 Physics 217, Fall 2002 1Today in Physics 217: conductors Conductors: E = 0, V = constant inside. Examples Forces on conductors from surface charge30 September 2002 Physics 217, Fall 2002 2ConductorsConductors are materials that contain charges that can move about freely (within the material) in response to an applied electric field.  Metals are good examples; for most metals, the mobile carriers are electrons.  Better conductors have higher densities of free (mobile) charges. 0EA metal in an externally-applied electric field. Electrons have moved to the “upstream” surface, leaving fixed positive charges behind. -----------+++++++++++30 September 2002 Physics 217, Fall 2002 3Conductors (continued)Implications of the free, mobile charges in conductors in electrostatics: E= 0 inside a conductor. If it weren’t, charges would move; the free charges move whenever the field is nonzero, and stop only when E = 0 inside.ρ= 0 inside a conductor. This is because, for any surface completely within the conductor,because E = 0 there.  Any net charge lies on the surface. That’s as far as it can travel under the influence of the external electric field; also, there’d be nonvanishing E and ρif this weren’t true.enclosed104Qdπ=⋅=∫EavPhysics 217, Fall 2002 30 September 2002(c) University of Rochester 230 September 2002 Physics 217, Fall 2002 4Conductors (continued) V is constant throughout a conductor: because, for any two points a and b inside,Thus the surfaces of conductors are equipotentials.  E is perpendicular to the surface of conductors.Otherwise, free charges would flow along the surface until any parallel component were zero. • Or think of the boundary conditions on E derived on Friday: here, E = 0 inside, and• This is why sharp-pointed conductors make good lightning rods. ()()0VV d−=−⋅=∫baba El, inside , outside0.EE−=&&30 September 2002 Physics 217, Fall 2002 5What can we do with conductors?Or, rather, what kinds of homework problems can I assign about conductors?  Given conductors held at specified potentials, find the fields in between. Given conductors held at specified potentials, find the surface charge density. Given conductors immersed in specified electric fields, find the density of charge induced on the surface.  Or vice-versa, for any of these.30 September 2002 Physics 217, Fall 2002 6Example: concentric conducting spheresGriffiths, problem 2.35:A metal sphere of radius R, carrying charge q, is surrounded by a thick concentric metal shell (inner radius a, outer radius b). The shell carries no net charge. (a) Find the surface charge density, σ, on each surface. (b) Find the potential at the center, using infinity as the reference point.(c) Now the outer surface is grounded; that is, it is connected by a conducting wire to zero potential (same potential as at infinity). How do the answers to (a) and (b) change?Rabq++++++++--------++++++++Physics 217, Fall 2002 30 September 2002(c) University of Rochester 330 September 2002 Physics 217, Fall 2002 7Example: concentric conducting spheres (continued)(a) First, the inner sphere. A conductor carries its charge on its surface, so the density there is the total charge divided bytotal area:Next, the inner surface of the shell. A charge is induced on this surface so as to generate an electric field equal and opposite to that from the sphere’s charge, within the shell, so that E = 0 there. Spherical charged surfaces generate fields outside themselves that appear as if their charge is concentrated at the center, so the charge on the inner surface is –q:24RqRσπ=24aqaσπ=−30 September 2002 Physics 217, Fall 2002 8Example: concentric conducting spheres (continued)Finally, the outer surface of the shell. The shell has no net charge, and –q was induced on its inner surface, so there must be +q there:(b) The field outside the shell is so the potential of its outer surface is This is also the potential at the inner surface, since conductors are equipotentials. The field between the inner surface of the shell and the sphere is soSince conductors are equipotentials, this applies at the center of the sphere, too.24bqbσπ=2ˆ,rqr=E.Vqb=2ˆ,rqr=E()2()RRaaqqqqqqVR Va drbr bRar=− =+=+−∫30 September 2002 Physics 217, Fall 2002 9Example: concentric conducting spheres (continued)(c) Now the shell is grounded. The field from the sphere’s charge still has to be cancelled within the shell, so the surface charge densities on the sphere and on the inner surface of the shell are the same as before. But now the potential on the outer surface of the shell is zero, as at infinity, so the fieldmust be zero outside the shell:and thus there’s no charge on this surface:And since the potential is zero on the inner surface too,() ( )00bVb V d∞−∞=−⋅= ⇒ =∫El E0.bσ=()2()RRaaqqqqVR Va drrRar=− ==−∫Physics 217, Fall 2002 30 September 2002(c) University of Rochester 430 September 2002 Physics 217, Fall 2002 10Example: field in an empty cavity within a conductorBy empty, we mean there’s no charge there. What is the field there?Consider a loop lying partly within the cavity, and partly within the conductor. E = 0 in the conductor, andso Since the path is finite, the field within the cavity must be zero. This is why you can shield external electrostatic fields with a conducting box (Faraday cage), and why you won’t get electrocuted if lightning strikes your car while you’re inside. (Unless you drive a Saturn or a Corvette, that is.)0,d⋅=∫Elvpath thru cavity0d⋅=∫El30 September 2002 Physics 217, Fall 2002 11Forces on conductorsIf there’s a field difference across a surface charge, as there is on the surface of a charged conductor, then there may be a force, too. What will it be? Consider a small patch with area A on a charged sheet.The total field near the centerof the patch is the sumof the patch’s fieldand the field fromthe rest of the charge distribution (“other”).Only the “other” E exerts a force onthe patch. What is otherEσAˆ2πσnˆ2πσ− nother?E30 September 2002 Physics 217, Fall 2002 12Forces on conductors (continued)Just above and just below the center of the patch, we haveWe can also recall the boundary condition derived on Friday,So, for instance, the field just above a charged conductor isthe field