Potential and Capacitance

Transcription

1 Ptential and apacitance

2 Electric Ptential Electric ptential (V) = Electric ptential energy (U e ) per unit charge () Define: ptential energy U e = 0 at infinity (r = ) lim U 0 r e Nte the similarity f U e t U G

3 V due t a Pint harge Electric ptential energy (U e ) due t a pint charge : U e r Electric ptential (V) due t a pint charge 1 : V U 1 r e 1 4 Fr many pint charges: V 1 4 i r i

4 Define Electric Ptential (V) W = wrk dne by an electric field n a charged particle as the particle mves frm infinity ( ) t r. Define electric ptential: V W Ptential difference: V W

5 Units Units fr ptential: Rewrite the units fr electric field using V: W V J V N m V N m V m N J J V N Euivalent units fr electric field Electric ptential (V) Wrk (J) harge ()

6 Units f Energy J = jule = large amunt f energy 1 ev = 1 electrn vlt = the wrk dne t mve 1 electrn thrugh a ptential difference f 1 V 1 ev = e (1V) = (1.6 x ) (1 J/) = 1.6 x J

7 alculate Electric Ptential A rd f unifrm linear charge density λ and ttal charge = 3.14 x 10-6 is bent int an arc f radius R=0.1m frm 10º t 40º frm the +x axis, with its center at the rigin. At the rigin the electric field is.34 x 10 6 N/ t the right (+x). Determine the electric ptential at the rigin. V 1 4 r.810 ( J / 9 Nm / V ) m 6

8 Euiptential Lines Euiptential lines are lines f eual ptential Similar t cntur lines n a tpgraphical map Are always perpendicular t electric field lines N wrk is reuired t mve a charge alng an euiptential line

9 Euiptential lines G t harges and Fields : /sims.php?sim=harges_and_fields Add charges and draw euiptential lines fr: 1) like charges ) unlike charges 3) Oppsitely charged parallel plates 4) Smething different

10 Fun with Electric Fields G t electric field hckey : ns/sims.php?sim=electric_field_hckey an yu get t level 3?

11 E and V Relate electric field and electric ptential using wrk. Integral frm: V E dr

12 apacitrs apacitrs are used t stre ptential energy in an electric field. When a capacitr is charged t a charge f Q, its plates (ppsite sides) have eual but ppsite charges f +Q and Q.

13 apacitrs The charge Q and ptential difference V fr a capacitr are prprtinal: Q = V = capacitance depends nly n gemetry (nt n Q r V) Unit: 1 farad = 1 F = 1/V Practical unit: 1 pf = 1 x 10-1 F

14 alculating frm V Parallel plate capacitr: ylindrical capacitr: d A V d A d A Ed V a b a b a b L L V a b L r dr L Eds V ln ln ln ) E( rl EA

15 Dielectric A dielectric is an insulating material placed between the plates f a capacitr increases the capacitance (farads = culmbs per vlt) Dielectric cnstant = Κ ( Kappa )

16 Dielectric Given a dielectric K, replace Є by KЄ in all electrstatic euatins. A d d A Gauss Law: E da Q

17 apacitr Prblem A parallel-plate capacitr has a plate area A=115cm and separatin d=1.4cm. A ptential difference V =85.5V is applied. a) What is the capacitance f the capacitr? A d ( N m 1 / N m )(11510 m F 4 m 8.1pF ) N m J V F

18 apacitr Prblem Same capacitr: A=115cm d=1.4cm V =85.5V =8.1pF What free charge appears n the plates? V ( 8.1pF)(85.5V ) 70 p

19 apacitr Prblem Same capacitr. The battery is nw discnnected and a dielectric slab f thickness b=0.780cm and dielectric cnstant K=.61 is placed between the plates. What is the electric field in the gaps between the plates and the dielectric slabs?

20 apacitr Prblem harge n the capacitr surface will induce charge n the clsest surface f the dielectric reate a Gaussian surface that enclses nly the charge n the + surface f the capacitr.

21 apacitr Prblem Use Gauss euatin t slve fr the (cnstant) electric field in the gap: E da EA E enc enc enc A - + (1)( N / / N m 6900V / m 10 )( m )

22 apacitr Prblem Same capacitr. What is the electric field inside the dielectric (K=.61)? Same as befre, except electric field pints in ppsite directin. enc E A (.61)( / N m )( N / 640V / m 4 m )

23 apacitrs in Parallel A cmbinatin f capacitrs in a circuit can be replaced by an euivalent capacitr In parallel: e i i V 1 = V = V 3 = ttal =

24 apacitrs in Series In Series 1 1 e i i V ttal = V 1 + V + V = = 3 =

25 Stred Energy Wrk is dne t charge a capacitr apacitrs stre ptential energy U 1 QV 1 V

26 apacitrs in ircuits In the fllwing circuits with capacitance and initial charge as shwn, where will the charge mve (if at all) when the switches are clsed?

27 apacitrs in ircuits Find the euivalent capacitances f each f the fllwing circuits. All capacitrs have eual capacitance. V V V V

28 apacitr circuit prblem Redraw the circuit Find V and fr capacitrs 1,, and 3 = μf 0 V μf 3 μf 4 μf 1 = 4 μf 3 = 3 μf

29 apacitr ircuit Prblem Answers: (μf) V (V) (μ) Ttal

30 apacitr ircuit Prblem Same circuit. Find the energy stred in the capacitrs 1,, and 3 : U QV V Q (μf) V (V) (μ) U (μj) Ttal

31 Time Dependence When capacitrs are first cnnected: Initially n charge n capacitr harge rapidly Act like a wire V V R t t

32 Time Dependence After being cnnected fr a lng time apacitr is fully charged an nt charge any mre Act like an pen circuit (n current) V V R t t

33 R ircuits: harging harge: 1 e t / urrent: i i e t / Ptential: 1 e t / Time cnstant: R

34 Time Dependence When capacitrs are first discnnected frm the battery: Initially full charge n capacitr Discharge rapidly Act like a wire V t

35 Time Dependence When capacitrs have been discnnected fr a lng time: Twards n mre charge n the capacitr Expnentially t n mre current V t

36 R ircuits: Discharging harge: e t / urrent: i i e t / Ptential: e t / Time cnstant: R

37 apacitr prblem A capacitr f capacitance is discharging thrugh a resistr f resistance R. In terms f the time cnstant R, when will the charge n the capacitr be half its initial value? 1 1 e e e t / t / R t / R ln 1 t ln R e t / R ln 1 R ln 0.69R t R

38 apacitrs charging In reality, capacitrs take sme time t charge and t discharge: