Potentials and Fields
Review: Definition of Potential Potential is defined as potential energy per unit charge. Since change in potential energy is work done, this means V E x dx and E x dv dx etc. The potential difference between any two points is the work required to carry a unit positive test charge between those two points.
Review: Coulomb s Law So we now have a third form of Coulomb s Law: 1. F kqq / r. E kq / r 3. V kq / r
Review: Basics about Potentials U qv V kq/r W AB q ( V B V A ) V E x dv dx E x dx etc.
Field of a Uniformly Charged Sphere Good example for main ideas of this week. Given a non-conducting sphere with a uniform volume charge density. Use Gauss s Law to find the electric field inside the sphere. (Ch. 3) Side effect: again prove the shell theorem! Use that solution to find potential inside and outside this sphere. (Ch. 4)
r>r Uniformly charged sphere ρ volume charge density Charge/Volume Coulomb/Meter Total charge r<r R Q 4 R 3 π 3 ρ Field outside Field inside Q E, r > r 4π 3 Q( r) 4π Q Q( r) r ρ, E ρr, r < 3 ε r 3ε ε R R
Q.3-3 If the uniformly charged sphere has radius R cm and total charge Q 4 nc, how much charge Q in lies within r 1 cm of the center? r<r R 1) 1 nc ) 8 nc 3) 6 nc 4) 4 nc 5) 3 nc
Solution: 4 Vol. πr 3 4 Q ( Vol.) ρ πr 3 Volume of sphere is ( ) 3 So charge inside is 3 ρ Q.3-3 So charge within a sphere is proportional to the radius cubed: Q r 3 So if sphere with r1 has (1/)3 1/8 the volume, and 1/8 the charge. 4 nc Qin ( 1/ 8) Qtot 3 nc 8
Field due to solid charged sphere. So the field inside is: E ρr 3ε And the field outside is: E Q 4 r Exercise for the student: Prove these two formulas agree when rr. Hint: use Q 4 R 3 π 3 ρ
Potential Difference V Continuing with the uniformly charged sphere problem. What is the potential difference between the center of the sphere and the surface? How much work would it require to move an electron from the center to the surface?
By definition V r E r dr R E dr R But we already know E ρr Therefore V R ρ ρ R 3ε r R R dr r dr 3ε ε ε ε 3 3 6 ρ ρ
Suppose we have some given numbers: R 6 3 cm ρ 3 1 C / m 3 ( So that 4 1 3 Q πr ρ 3.4 1 C 3.34 nc ) Then 6 ρr 3 1 (3 1 ) V 5V 6 6 9 1 1 ε
Q.4-3 Using the same numbers, what is the potential at the surface of this sphere? R 3 cm ρ 3 1 6 C / m 3 Q 3.4 1 1 C 1) 5 V ) 1 V 3) 5 V 4) 5 V 5)
Q.4-3 Outside, E is given by Coulomb s Law (shell theorem), so work to bring in test charge from infinity is same as for point charge: E kq / r V kq / r ( 9)( 1 9 1 3.4 1 ) V V kq / R 1 3 1 1) 5 V ) 1 V 3) 5 V 4) 5 V 5)
A new unit of energy One electron-volt (ev) is the energy to move an electron between points with a potential difference of one volt. This is not an SI unit but is universally used in discussing processes involving electrons, atoms, nano-scale structures, etc. U qv So if V1 V and qe then: U e V ( 1V )( e) 1 ev (1V ) (1.6 1 19 C ) 1.6 1 16 J
A new unit of energy 1 ev 1.6 1 19 J So in our previous example, where we have V5 V and qe then the work required to move an electron from the center to the surface is: U qv ( 5 V )( e) 5 ev (1.6 1 19 C ) (5 V ) 8 1 18 J Much simpler to just give 5 ev as the answer.
More about units The SI unit for potential is the volt. The SI unit for electric field is N/C. But E dv/dx so another unit for E could be volts/meter. It is correct to say that the SI unit for E is V/m. This is more commonly used in applications.
Equipotential surfaces Any surface which is everywhere perpendicular to the electric field is an equipotential surface. The potential has the same value everywhere on this surface. This is true because no work is required to move a test charge from place to place on the surface.
For a point charge, equipotentials are spheres: For a dipole, the shape of the equipotentials is more complicated:
Cathode Ray Tube Electron gun: potential V gives electron energy in ev. 1 K mv qv So if V 5 volts, electron energy is K 5 ev
Be careful with SI units Suppose we want to calculate the speed of this electron. We must first convert the energy from ev to SI units. K 1 mv qv 5 ev (5 ev ) (1.6 1 19 J / ev ) 8 1 17 J v 17 K 8 1 J 1.3 1 7 m 9.1 1 31 kg m / s
High energies For X-ray machines, accelerate electrons with potentials of thousands of volts, so we speak of kinetic energies in kev. In nuclear physics, accelerators produce beams of particles with energies in MeV. In elementary-particle physics, high-energy particles beams have energies measured in giga-volts: GeV. 3 1 6 1 9 1 ev ev ev
Capacitance Chapter 5 homework: Questions 1, 1 Problems 1, 9, 5, 39
Capacitors Capacitors store charge and energy. Definition of Capacitance: C Q/V Calculate stored energy: Q U C
Two Conductors Form a Capacitor V + q r E r ds q Given the amount of charge q. Let V be the potential difference V. Define the capacitance C q/v. SI unit is the farad: 1 F 1C/V
Case 1: The Parallel-plate Capacitor The simplest capacitor is made of two large parallel metal plates very close together. Because of its construction, we see that all the charge is on the inner surfaces.
Parallel-Plate Capacitor q σ A E σ V Ed /ε σ A A C q / V d / d σ ε ε
Potential Energy I must do work to charge up a capacitor. This energy is stored in the form of electric potential energy. We will show that this is 1 Q U C Then we will see that this energy is stored in the electric field, with a volume energy density u ε E ( 3 ) J / m
Derivation of Capacitor Energy If capacitor has charge q and I add charge dq, then I must do work dw Vdq (Remember the definition of potential difference!) For each dq I add, I must work harder because there is a stronger field against me. If I start with q and end with qq, then my total work is W Q Vdq q C dq Q C
Energy density of the electric field Special case of parallel-plate capacitor: ε 1 U Q / C CV CE d But C A/ d and Vol. U 1 And so we get u Ad Turns out to be true in general: an electric field always carries this energy density 1 ( ε A/ d ) E d ε E 1 Ad 1 ( U ) /( Vol.) ε E
A Numerical Example 1. Suppose we have metal plates, cm by 3cm, set parallel to each other, separated by a gap of 3 mm. What will be the capacitance of this setup? C ε 1 A 9 1. d.3.3 1 C 1.8 1 F.18 nf 18 pf
. Now suppose we connect a 1-volt battery between these two plates, so that the potential difference across this capacitor is V 1 volts. How much charge will be on the plates?
. If the potential difference is 1 volts how much charge will be on the plates? Use the definition of capacitance: C q / V q CV (.18 nf) (1V ).16 nc Thus the positive plate will have +.16 nc and the negative plate will have.16 nc.
3. What is the strength of the electric field between the plates? Method A: Remember V is work to move 1 C between plates, force times distance, so E V V Ed / d 1V /.3m 4V Method B: From Gauss s Law Eσ/ε, so / m E 9.16 1 ( Q / A) / ε 1..3 9 1 4V / m
4. How much energy is stored? U Q / C U (.16 nc) (18 pf ) 4.67 1 36 1 18 1 SI SI 13 nj
5. What voltage would be required to store 1 joule of energy? U Q / C and Q CV U 1 CV So V U / C 1/1.8 1 1 1.1 1 5 V 11 kv
Case : The Spherical Capacitor Use Gauss to find E(r): Φ out E( r) E( r) Q enc 4πr ε q 4πr ε q ε kq r b V E( r) dr a
Spherical Capacitor Continued b a dr r kq Edr V a b ab a b k ab V q C 4 / / ab a b kq b a kq 1 1
Case 3: The Cylindrical Capacitor Application: capacitance per unit length of a coaxial cable. b V E( r) dr a What is E(r) in this case? Again use Gauss s Law. Cross section of cable length charge per unit λ
Review: Long Line of Charge λ r λ charge/length linear charge density l Φ Q ε Q lλ Gauss law Φ E d A EA E Top view A πrl E λ πrε
Back to Cylindrical Capacitor [ ] a b r r dr dr r dr r E V b a b a b a b a ln ln ) ( λ λ λ λ
Finish Cylindrical Capacitor Capacitance per unit length: C q / V C λ λ so L V λ ln b a ln( b / a)