Today in Physics 1: getting V fom E When it s best to get V fom E, athe than vice vesa V within continuous chage distibutions Potential enegy of continuous chage distibutions Capacitance Potential enegy in capacitos Paallel-plate and cylindical capacitos 8400-faad supecapacito module (ICD) 7 Septembe 01 Physics 1, Fall 01 1
When it s easie to get V fom E Even if a distibution does not stetch to infinity, thee ae cases in which one cannot use the naïve Coulomb s law fom of the potential, V = kdq -. Tivial example: a chaged conducting sphee. Chage unifom on suface, so V = kq/ thee. But E = 0 inside, so the suface and inteio ae an equipotential: so V = kq/ eveywhee inside. Yet b 0 b V = Ed = a kdq - = 0. 0 b 7 Septembe 01 Physics 1, Fall 01 a Q
Example 1: V within a unifom chaged sphee So emembe to include potential fom oute chage, when computing V within a distibution of chage. Usually this is easiest to do by using the foolpoof expession fo potential: ( ) V Example: the unifom sphee of chage, fo which we calculated E using Gauss s law (18 Sept 01): 4πkρˆ 3, < E = 3ˆ 4πkρ 3, = Ed 7 Septembe 01 Physics 1, Fall 01 3 ρ
V within a unifom chaged sphee (continued) Because E changes fom at =, we have to split the integal into a sum of two: 3 4πkρ d 4πkρ V ( ) = E( ) d = dd 3 3 Compae 3 4πkρ 1 4πkρ = 3 3 4πkρ 4πkρ = + 3 6 ( ) 7 Septembe 01 Physics 1, Fall 01 4 ( ) 4π ρ 4π ρ kdq kq k k = = d = ( V!). - 3 0.
Example : V within a non-unifom chaged sphee Conside a sphee with chage density ρ given by ρ0 1, ρ = 0, > Calculate the total chage Q, and ( ) ( ) Fist find E. Since ρ is spheically symmetic we get to use Gauss s Law with a spheical Gaussian suface: E da ( ) ( ) = E 4π = 4πkQ E = kq encl V( ),. encl ρ ( ) 7 Septembe 01 Physics 1, Fall 01 5
V within a non-unifom chaged sphee (cont d) Qencl = ρdv = ρ0 1 4π d 3 5 = 4πρ0 3 5 While we e at it, get the total chage: 3 5 8πρ0 Q = 4πρ 0 = 3 5 15 0 1 4 = 4πρ 0 d d 0 0 ρ ( ) 7 Septembe 01 Physics 1, Fall 01 6 3
V within a non-unifom chaged sphee (cont d) So outside the sphee ( > ), it s ( ) 3 3 15kQ πkρ 0 ˆ 3 E = 4 = ˆ 3 5 3 5 Now we e eady to calculate V. Again the integal beaks into two: ( ) V = Ed ( ) = kq 3 d 15kQ = kq d ˆ, while inside ρ ( ) 7 Septembe 01 Physics 1, Fall 01 7 E 3 3 5
V within a non-unifom chaged sphee (cont d) 4 1 15kQ V ( ) = kq 0 3 6 4 4 kq 15kQ = + 6 6 0 0 kq kq ( 4 4 = + 3 10 + 7 ) 5 8 Check: kq kq ( ) ( 4 4 4 V = + ) 5 8 3 10 + 7 kq = 3 ρ ( ) 7 Septembe 01 Physics 1, Fall 01 8
Electostatic potential enegy of a continuous distibution of chage Similaly, one must be caeful about using U = qv to calculate electostatic potential enegy of continuous distibutions of chage. Hee s an example (poblem 3-64 in the book). A sphee with adius contains a total chage Q, unifomly distibuted though its volume. Calculate its electostatic potential enegy. ρ Imagine building the sphee by binging d infinitesimal shells of the ight chage density in fom infinity one by one. 7 Septembe 01 Physics 1, Fall 01 9
Electostatic potential enegy of a continuous distibution of chage (continued) Since the chage is unifomly distibuted, the density is Q 3Q ρ = = v 3 4π and if the infinitesimal chage bought fom infinity is spead d thick at this density, onto a sphee of adius composed of the shells which aived peviously, its value can be witten as dq = 4πρ d dq ρ d 7 Septembe 01 Physics 1, Fall 01 10
Electostatic potential enegy of a continuous distibution of chage (continued) Binging the shell in fom infinity involves wok, and this wok is the potential enegy of the shell-peviouslyexisting sphee combination: du = dw = V dq Since thee s no chage outside adius befoe the new shell aives, the potential at is calculated fom d V ( ) = Ed = kq ( ) = k = ρ( ) 4π d = 4πkρ 3 0 7 Septembe 01 Physics 1, Fall 01 11 ( ) kq by Gauss s Law dq ( ) q( ) d
Electostatic potential enegy of a continuous distibution of chage (continued) Combine the last two esults and integate: 4πρ du = V ( ) dq = k 4πρ d 3 ( 4πρ ) ( 4πρ ) 5 4 U = k d = k 3 3 5 0 3 4πρ 3 3 = k = 3 5 5 kq Compae to the gavitational potential enegy of a unifom mass 3 GM M (PHY 11): U ==. 5 7 Septembe 01 Physics 1, Fall 01 1 dq ρ d
Capacitance Conside two conductos, chaged up to Q and Q. They ae equipotentials, and the voltage between them is V + = Ed, and nea each suface the electic field magnitude is E= 4πkσ (18 Sept 01). Suppose we double the value of σ. What happens to the othe quantities? E Q -Q 7 Septembe 01 Physics 1, Fall 01 13
Capacitance (continued) Doubling σ doubles the total chage. It also doubles the magnitude of the electic field, but not the patten of field lines (just daw moe of them). And since it doubles the field, it doubles the voltage between the conductos. Appaently, dq is popotional to dv, so Q and V ae popotional. We call the popotionality facto the capacitance, C: Q = C V. E Q -Q 7 Septembe 01 Physics 1, Fall 01 14
What good is C? Enegy stoage in capacitos Capacitos ae impotant as electic cicuit elements. Cicuits can stoe enegy in, and eclaim enegy fom, capacitos. Conside, fo instance, caying a chage Q fom one conducto to the othe, one infinitesimal chage dq at a time: q dw = Vdq = dq C Q 1 1 Q 1 U = W = qdq = = C V. C C 0 Usually we speak loosely about potential V and potential diffeence V in cicuits, and often wite Q = CV o U = CV. 7 Septembe 01 Physics 1, Fall 01 15
Calculation of the capacitance of aangements of conductos Othe mateials besides conductos have capacitance, but aangements of conductos lend themselves to staightfowad calculation of C. Usually this goes as follows: Pesume electic chage to be pesent; say, Q if thee is only one conducto, o ±Q if thee ae two. Eithe: Calculate the electic field fom the chages, and integate it to find the potential diffeence V between the conductos, o Solve fo the potential diffeence diectly, using Then C = Q/V. V = kdq -. 7 Septembe 01 Physics 1, Fall 01 16
Paallel-plate capacito Conside two paallel conducting plates, sepaated by a distance d that is vey small compaed to thei extent in othe dimensions. Suppose each plate has aea A. It doesn t matte what the shape of the flat plates ae, as long as they ae paallel and vey close togethe. With chages ±Q on the plates, the chage densities ae unifom and have values σ = ±Q A. w w w A d A d d A 7 Septembe 01 Physics 1, Fall 01 17
Paallel-plate capacito (continued) At points well inside the gap, the plates can be egaded as infinite, to good appoximation. As we found on 11 Septembe 01, the electic field between two oppositely-chaged infinite paallel plates is unifom, with magnitude E= 4 πkσ. σ = Q A. d E = 4πkσ σ = Q A. 7 Septembe 01 Physics 1, Fall 01 18
Paallel-plate capacito (continued) So 4π kd V = Ed = 4πkσd = Q A C + A = = 4π kd ε A 0 d Q C σ = Q A. d E = 4πkσ σ = Q A. 7 Septembe 01 Physics 1, Fall 01 19
Cylindical capacito Conside two, coaxial, conducting cylindes with adii 1 and > 1. Thei length is L and they cay opposite chages ±Q (chage pe unit length λ = ±Q L). At points well inside the gap, the cylindes can be egaded as infinite, to good appoximation. λ λ 1 L 7 Septembe 01 Physics 1, Fall 01 0
Cylindical capacito (continued) We have shown by use of Gauss s law (18 Septembe 01) that kλ ˆ, 1 < < E = 0,, 1 fo infinite, oppositely-chaged coaxial cylindes. λ λ 1 L 7 Septembe 01 Physics 1, Fall 01 1
Cylindical capacito (continued) Thus d Q ln Q V = Ed = kλ = k ; L 1 C C + L πε0l = = kln ln ( ) ( ) 1 1 1. λ λ 1 L 7 Septembe 01 Physics 1, Fall 01
Units of capacitance In MKS: the unit of capacitance, named in hono of Michael 1 coul Faaday, is the faad : 1F = 1 volt Note that ε 0 1-1 - = 8.85 10 coul Nt m 1-1 -1 = 8.85 10 F m = 8.85 pf m One faad is a Huge capacitance. Those found aound the lab and in cicuits ae usually in the pf- µf ange (10-1 -10-6 F). A 1F paallel-plate capacito with d = 5 µm (0.001 in) has A =.8 km : 1.7 km on a side, if squae..8 cm 7 Septembe 01 Physics 1, Fall 01 3