23 September 2002 Physics 217, Fall 2002 1Today in Physics 217: electric potential Finish Friday’s discussion of the field from a uniformly-charged sphere, and the gravitational analogue of Gauss’ Law. Electric potential Example: a field and its potential Poisson’s and Laplace’s equations23 September 2002 Physics 217, Fall 2002 2Electric potentialBecause in electrostatics, we can express E as the gradient of a scalar function:where of course Vis called the electric scalar potential. By the gradient theorem, we can writeSuppose we have agreed upon a standard reference point, O; then0×=E—V=−E —() ()()dVdVbVa−⋅= ⋅= −∫∫bbaaEl l—() ()dVdlVdl−⋅= ⋅+ ⋅∫∫ ∫bbaaEl——OO23 September 2002 Physics 217, Fall 2002 3Electric potential (continued)This leads us to an integral definition of V:Properties of the electric potential:Arbitrariness. An arbitrary constant can be added to the potential without changing the field (which, after all, is the fundamental quantity). To each constant corresponds a potential reference point. Thus there is always a large selection of appropriate reference points in any electrostatics problem. ()Vd=− ⋅∫ElPOP23 September 2002 Physics 217, Fall 2002 4Electric potential (continued)Convention: take O to lie at infinity, unless the charge distribution itself extends to infinity. Just remember that you can actually put the reference point anywhere that doesn’t lead to an infinite result for the potential; sometimes you will find reference points not at infinity that will be more convenient for your calculation. Significance. The magnitude of the electric potential therefore has no physical significance; only differencesin potential do. Superposition. The electric potential superposes. If E =12 12VddVV=− ⋅ − ⋅ − = + +∫∫El El……PPOO12, then++EE…23 September 2002 Physics 217, Fall 2002 5Electric potential (continued) Thus at a given point P, one could compute Vs due to different parts of a charge distribution, and add the results to get the real thing. This may seem trivial until we find, later this week, that the closely-related electrostatic potential energy does not superpose. From point charges and superposition we can induce the potential from a continuous charge distribution:()()1riiNiNiiQVd EdrrdQVρτ→∞∞→∞=′=− ⋅ →− =′′=→∫∫∑∫ElrPOOPOPrr23 September 2002 Physics 217, Fall 2002 6Units. In CGS, it’s the statvolt:In MKS, it’s the volt:Correspondence: Electric potential (continued)dyne cm ergstatvoltesu esu==jouleNt mvoltcoul coul==1 statvolt 299.792458 volts↔23 September 2002 Physics 217, Fall 2002 7Electric potential (continued)What it’s good for. It’s often easier to calculate V, and take its gradient to find E, than to calculate E directly. Reasons:•Vis a scalar; no vector addition to get it. • There are many situations in nature in which Vcan be regarded as constant over a region in space near where one would like to know E. The solution of Vfor space between the constant-V(“equipotential”) locations and the reference point – the process of which is called a boundary-value problem – can be shown to be unique. Finding Vby boundary-value solution, and then calculating E, is in these cases usually much easier than calculating E directly.23 September 2002 Physics 217, Fall 2002 8ExampleGriffiths problem 2.20One of these is an impossible electrostatic field. Which one?Here, kis a constant with the appropriate units. For the possibleone, find the potential, using the origin as your reference point. Check your answer by computing First take the curl of each function:() ( )()() ( )()22ˆˆˆa. 2 3ˆˆˆb. 2 2kxy yz xzky xyz yz=++=+++ExyzEx yz.V−—23 September 2002 Physics 217, Fall 2002 9Example (continued)The first one can’t, but the second one can.()() ()()() () ( )ˆˆa. 3 2 3ˆˆˆˆ2230k xz yz k xy xzyz zxkyzxykyzxxy∂∂ ∂∂×= − + −∂∂ ∂∂∂∂+−=−−−≠∂∂Exyzxyz—()() ()()()()2222ˆˆb. 2 2 2ˆ20kyz xyz ky yzyz zxkxyz yxy∂∂ ∂∂×= − + + −∂∂ ∂∂∂∂++−=∂∂Exyz—23 September 2002 Physics 217, Fall 2002 10Example (continued)Integrate E to get V: choose path for convenience since the result is path-independent.()()2,0,00,0,000xdkydxd⋅= =⋅=∫ElEl()2(,,0)2(,0,0) 0222xy yxd k xy z dy kxydyd kxydy kxy⋅= + =⋅= =∫∫ElEl()()(,,)2(,,0) 022xyz zxyd k yz dzd k yz dz kyz⋅=⋅= =∫∫ElEl()()()()0,0,0 ,0,0 , ,0 , ,xxyxyz→→→23 September 2002 Physics 217, Fall 2002 11Example (continued)ThusCheck:()()22 22,, 0V x y z kxy kyz k xy yz=− − =− +()()()22ˆˆˆ ˆ ˆ ˆ22VVVVkyxyzyzxyz∂∂∂−=− − − = + + +∂∂∂=xyz x y zE—23 September 2002 Physics 217, Fall 2002 12Differential equations for the electric potentialTo get Vwithout referring first to E:In regions where there are no electric charges,These equations, plus boundary conditions, provide the boundary-value-problem way to calculate V. ()244VVπρπρ⋅=⋅− =∇=−E———Poisson’s equation20V∇=Laplace’s