Ptential and apacitance
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
V due t a Pint harge Electric ptential energy (U e ) due t a pint charge : U e 1 4 1 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
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
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 ()
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 10-19 ) (1 J/) = 1.6 x 10-19 J
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 (9.010 5 J / 9 Nm.810 5 / V ) 3.1410 0.10m 6
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
Euiptential lines G t harges and Fields : http://phet.clrad.edu/new/simulatins /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
Fun with Electric Fields G t electric field hckey : http://phet.clrad.edu/new/simulati ns/sims.php?sim=electric_field_hckey an yu get t level 3?
E and V Relate electric field and electric ptential using wrk. Integral frm: V E dr
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.
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
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
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 )
Dielectric Given a dielectric K, replace Є by KЄ in all electrstatic euatins. A d d A Gauss Law: E da Q
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 8.110 (8.8510 1 N m 1 / N m 1.410 8.110 1 )(11510 m F 4 m 8.1pF ) N m J V F
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
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?
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.
apacitr Prblem Use Gauss euatin t slve fr the (cnstant) electric field in the gap: E da EA E enc enc enc A - + (1)(8.8510 6900N / + - 1 7.010 / N m 6900V + + + + + + + + + + + + + + + + + + + + + + + + + + + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - + + + + + + + + + + + + + + + + + + + + + + + + + + + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - / m 10 )(11510 4 m )
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 10 7.010 1 (.61)(8.8510 / N m )(11510 640N / 640V / m 4 m )
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 = 1 + + 3 +
apacitrs in Series In Series 1 1 e i i V ttal = V 1 + V + V 3 + 1 = = 3 =
Stred Energy Wrk is dne t charge a capacitr apacitrs stre ptential energy U 1 QV 1 V
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? 6 3 6 3 3
apacitrs in ircuits Find the euivalent capacitances f each f the fllwing circuits. All capacitrs have eual capacitance. V V V V
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
apacitr ircuit Prblem Answers: (μf) V (V) (μ) 1 4 5 0 10 0 3 3 10 30 Ttal 3 0 60 1 3
apacitr ircuit Prblem Same circuit. Find the energy stred in the capacitrs 1,, and 3 : U 1 1 1 QV V Q (μf) V (V) (μ) U (μj) 1 4 5 0 50 10 0 100 3 3 10 30 150 Ttal 3 0 60 600
Time Dependence When capacitrs are first cnnected: Initially n charge n capacitr harge rapidly Act like a wire V V R t t
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
R ircuits: harging harge: 1 e t / urrent: i i e t / Ptential: 1 e t / Time cnstant: R
Time Dependence When capacitrs are first discnnected frm the battery: Initially full charge n capacitr Discharge rapidly Act like a wire V t
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
R ircuits: Discharging harge: e t / urrent: i i e t / Ptential: e t / Time cnstant: R
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
apacitrs charging In reality, capacitrs take sme time t charge and t discharge: http://www.splung.cm/cntent/sid/3/page/capacitrs