Physics Lecture: 15 FRI 20 FEB

Transcription

1 Physics 2113 Jonathan Dowling Physics 2113 Lecture: 15 FRI 20 FEB Electric Potential III

2 Conservative Forces, Work, and Potential Energy W = ( ) F r dr Work Done (W) is Integral of Force (F) U = W F ( r) = du dr Potential Energy (U) is Negative of Work Done Hence Force is Negative Derivative of Potential Energy

3 Coulomb s Law for Point Charge F 12 = kq 1q 2 r 2 Force [N] = Newton! E = kq 2 12 r 2 Electric Field [N/C]=[V/m] U 12 = kq 1q 2 r Potential Energy [J]=Joule V 12 = kq 2 r Potential Voltage [J/C]=[V] =Volt F 12 = du 12 dr E 12 = dv 12 dr q 2 q 1 q 2 P 2 P 1 P 2 P 1 U 12 = F 12 dr V 12 = E 12 dr

4 Potential At Center of Ring of Charge Divide the charge distribution into differential elements Write down an expression for potential voltalge from a typical element treat as point charge Integrate! Simple example: circular rod of radius r, total charge Q; find V at center. Same result holds along z- axis! dq V = = k r r dv = Units : Nm2 C 2 dq = k r Q C = Nm m C kdq r = J C V [ ]

5 Potential Along Axis of Ring of Charge Divide the charge distribution into differential elements Write down an expression for potential voltalge from a typical element treat as point charge Integrate! dq V = R dv = kdq r = k r dq = k r Q dq = k Q R 2 z 2

6 Potential Voltage of Continuous Charge Distribution: Return of the Rod! Uniformly charged rod Total charge Q Length L What is V at position P? λ = V P = Q / L kdq r = rod! dq La a = kλ[ ln(x) ] La a = λ dx kλdx x r = x dq=λdx L a P V P = kλ ln L a ( ) ln( a) dq Units: [Nm 2 /C 2 ][C/m]=[Nm/C]=[J/C]=[V]

7 Potential Voltage of Continuous Charge Distribution: Return of the Rod! Uniformly charged rod Total charge Q Length L What is V at position P? What about a>>l? r = x dq=λdx L a P V P = kλ ln L a = kλ ln L a a ( ) ln( a) kλ L a>>l a = k λl a = kq a = kλ ln( 1 L / a) In this limit we recover V=kq/r for point charge!

8 Potential Voltage of Continuous Charge Uniformly charged rod Total charge Q Length L What is V at position P? r = x 2 d 2 L 0 ( ) V P = dv = kdq = k λdx r r L dx = kλ = kλ ln x x 2 d 2 x 2 d 2 0 ( ) L 0 λ = Q / L dq = λdx ( ) ln d = kλ ln L L 2 d 2 ( ) = kλ ln L L2 d 2 d

9 Potential Voltage of Continuous Charge What about d>>l? V P = kλ ln L L2 d 2 d = kλ ln L d kλ ln 1 L d 1 L2 d 2 kλl d = kq d r = x 2 d 2 λ = Q / L dq = λdx We recover formula V=kq/r for a point charge Again!

10 Electric Field & Potential: A Simple Relationship! Notice the following: Point charge: E = kq/r 2 V = kq/r Dipole (far away): E = kp/r 3 V = kp/r 2 E is given by a DERIVATIVE of V! Of course! Δ V = E!! ds f i Focus only on a simple case: electric field that points along x axis but whose magnitude varies with x. E x = dv dx Note: MINUS sign! Units for E: VOLTS/ METER (V/m)

11 Uniformly charged rod Total charge Q Length L We Found V at P! Find E from V? dq L x E from V: Example λ = Q / L a P V = kλln L a a { [ ] ln[ a] } = kλ ln L a { [ ]} E = dv da = kλ d da ln [ L a ] ln a = kλ 1 a 1 L a Units: Nm 2 C 2 C m m = N C = V m m 2 Electric Field!

12 Electric Field & Potential: ICPP Hollow metal spherical shell of radius R has a charge on the shell q R Which of the following is magnitude of the electric potential voltage V as a function of distance r from center of sphere? V (a) 1 r V (b) 1 r r=r r r=r r Hint: Inside sphere there are no charges so E=0. But E=dV/dr=0. What can V be?

13 E x = dv dx ΔV Δx (exact for constant field) (a) Since Δx is the same, only ΔV matters! ΔV 1 =200, ΔV 2 =220, ΔV 3 =200 E 2 > E 3 = E 1 Δx E!" (b) = 3 E!" The bigger the voltage drop the stronger the field. E!" (c) F = qe = ma accelerate leftward

14 Equipotentials and Conductors Conducting surfaces are EQUIPOTENTIALs At surface of conductor, E is normal (perpendicular) 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. V E

15 Conductors Change the Field Around Them! An Uncharged Conductor: A Uniform Electric Field: An Uncharged Conductor in the Initially Uniform Electric Field:

16 Sharp Conductors Charge density is higher at conductor surfaces that have small radius of curvature E = σ/ε 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)

17 Ben Franklin Invents the Lightning Rod!

18 LIGHTNING SAFE CROUCH If caught out of doors during an approaching storm and your skin tingles or hair tries to stand on end, immediately do the "LIGHTNING SAFE CROUCH. Squat low to the ground on the balls of your feet, with your feet close together. Place your hands on your knees, with your head between them. Be the smallest target possible, and minimize your contact with the ground.

19 Summary: Electric potential: work needed to bring 1C from infinity; units = V = Volt 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 : dv = kdq/r Electric field and electric potential: E= -dv/dx Electric potential energy: work used to build the system, charge by charge. Use W = U = qv for each charge. Conductors: the charges move to make their surface equipotentials. Charge density and electric field are higher on sharp points of conductors.

20