Conuctors & Capacitance PICK UP YOUR EXAM;; Average of the three classes is approximately 51. Stanar eviation is 18. It may go up (or own) by a point or two once all graing is finishe. Exam KEY is poste on the home page REGRADE requests are DUE Friay at 4 PM in Alexis Olsho s office Put your name on paper explaining your request reason. Take it to Alexis Olsho s office yourself (1 st floor of PAB, near h bar) Same office as Susan Miller Phys 122 Lecture 10 Gray Rybka
Your thoughts I am really confuse by capacitance, especially with how/why it relates to potential. what are some practical applications of capacitors? What is the benefit of having a capacitor? This is a little confusing to me mostly because of resiual confusion from Electric potential, which I still on't unerstan that well conceptually. If you coul talk thoroughly about the link between voltage an capacitance that woul be nice.
We left off with Potential from a ring of charge Consier a uniformly charge ring with total charge of Q V = = = kq r k a 2 z 2 kq a 2 z 2 q This can be efine anywhere with V = 0 assume at infinity
Potential from a isk of charge Uniformly charge isk with area charge ensity σ Can we solve it as a series of charge rings? V = R 0 kσ 2πaa = kσ 2π a 2 z 2 R 0 aa a 2 z 2 = 2πkσ[ a 2 z 2 ] R 0 [ ] = 2πkσ R 2 z 2 z
A short asie: Energy Units MKS: U = QV Þ 1 coulvolt = 1 joule for particles (e, p,...) 1 ev = 1.6x10 19 joules Accelerators Electrostatic: VaneGraaff electrons 100 kev ( 10 5 ev) Electromagnetic: Large Haron Collier protons 7 TeV ( 7 x 10 12 ev)
Conuctors: Main Points Charges free to move E = 0 in a conuctor Surface = Equipotential E at surface perpenicular to surface A great image
A few minutes on Conuctors Q V A < V B Q Same total Q Different raii V at ege ifferent V Small V Big Q 4πε 0 r R small R Big r Checkpoints 1, 3, 5 pretty goo job here overall
A few minutes on Conuctors Q A Q B?? Increases Now the entire thing is a single conuctor E = 0 in a conuctor (implies ΔV = 0 between any 2 points in a conuctor If there was a potential ifference, charges woul move What happens to the charge on conuctor A after it is connecte to conuctor B by the wire? The charges o move initially enough to make V A = V B Potentials: kq A R A = kq B R B Q A = R A R B Q B > 1
Towar Capacitance Capacitance is efine for any pair of spatially separate conuctors. C Q V Two conuctors: one with excess charge = Q the other with excess charge = Q Q E V Q Charges create an electric fiel in the space between We can integrate the electric fiel between them to fin the potential ifference between the conuctor This potential ifference WILL be proportional to Q! The ratio of Q to the potential ifference is the capacitance an it only epens on the geometry of the conuctors
y First etermine E fiel prouce by charge conuctors: Q σ E E = ε o Q x Example from Prelecture What is σ? σ = A = area of plate Secon, integrate E to fin the potential ifference V Q A!! V = E y V 0 = ( Ey) = E y = As promise, V is proportional to Q! 0 0 Q ε A o C Q V = Q Q /ε o A ε0 C = A C etermine by geometry! Not Q
y How I was getting you reay E = σ/2ε 0 for x >0 V = E l = 2πkσ x x > 0 V(x) = 2πkσ x V = x > 0 x σ Conucting slab x x = 0 a b
Clicker: 2 planes σ σ Two infinite oppositely charge parallel plates are locate at an on the x axis. Which graphs best represent the Electric Fiel an the Potential Difference vs x? A B C D E
Clicker σ σ E=0 E=0 V(x) = 2πkσ x V 0 Two infinite oppositely charge parallel plates are locate at an on the x axis. Which graphs best represent the Electric Fiel an the Potential Difference vs x? E constant B
Going green Please explain the green thing an why it oes the things it oes. :/ The charge plate questions were the most ifficult, especially with the green conuctor plates in the mile, I on't know how to account for that. Can we iscuss the secon part of the secon checkpoint question? C=Q/eltaV but C= epsilon_0*area/istance. So what actually happens if you increase Q in such a way that keeps eltav the same?? These formulas seem to contraict. Warning: Lots of Green Thing Questions are next (these Checkpoint issues were more of a struggle)
Clicker: (not the Checkpoint) Q 0 Initial charge on capacitor = Q 0 Q 0 Insert an uncharge conuctor Q 1 t Charge on capacitor now calle Q 1 How is Q 1 relate to Q 0? A) Q 1 < Q 0 B) Q 1 = Q 0 C) Q 1 > Q 0 Q 1 Plates not connecte to anything CHARGE CANNOT CHANGE!
Clicker to reinforce Initial charge on capacitor = Q 0 Q 1 t Q 1 What is the total charge inuce on the bottom surface of the conuctor? A) Q 0 B) Q 0 C) 0 D) Positive but the magnitue unknown E) Negative but the magnitue unknown
Why? Q 0 Q 0 Q 0 E = 0 E E Q 0 E must be = 0 in conuctor! Charges insie conuctor move to cancel E fiel from top & bottom plates.
Now calculate V as a function of istance from the bottom conuctor (efine V=0 there). Q 0 E y E = 0 Calculate V t E 0 V ( y) t y! =! E y 0 y Q 0 What is ΔV = V()? A) ΔV = E 0 B) ΔV = E 0 ( t) C) ΔV = E 0 ( t) V The integral = area uner the curve y
Now, Click on CheckPoint 8 again Two parallel plates with charge Q 0 on each plate. Potential ifference is V 0. Put in uncharge conuctor (green) ADJUST the potential ifference to be the SAME as before: i.e, V 0. How is charge Q 1 relate to the original Q 0? A) Q 1 < Q o B) Q 1 = Q o C) Q 1 > Q o We just foun ΔV = E 2 ( t) In the 1 st case ΔV = E 1 () For same ΔV we nee E 2 > E 1 implies Q 1 > Q o This charge came from some connecte circuit that mae V 0 the same
Let s click this one too Same setup. Potential ifference is V 0. Insert conuctor (green) CHARGE Q 1 is ADJUSTED to make Potential Difference equal to V 0. (see previous Clicker question where Q 1 ha to increase) How oes the Capacitance of the object change? A) C 1 > C o B) C 1 = C o C) C 1 < C o Same V 0 à C 0 = Q 0 / V 0 an C 1 = Q 1 / V 0 Therefore, since Q 1 > Q 0 it means C 1 > C 0 Alternate explanation using Previous situation Same Q 0 V 0 = E 0 an V 1 = E 0 ( t) C 0 = Q 0 /E 0 an C 1 = Q 0 /(E 0 ( t)) C 0 = ε 0 A / an C 1 = ε 0 A /( t) PROPERTY OF GEOMETRY ONLY