clicker 1/25/2011 All C s are 8.00 nf. The battery is 12 V. What is the equivalent capacitance? summary o

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1 /5/0 summary C = ε / d = πε / ln( b / a ) ab C = 4πε 4πε a b a b >> a Capacitance Parallel plates, caxial cables, Earth Series and parallel cmbinatins Energy in a capacitr Dielectrics Dielectric strength Q=C = C = C + C +... C C C C κc U = C = Q C Mre cmplex capacitr systems In general a capacitr system may cnsist f smaller capacitr grups that can be identified as cnnected "in parallel" r "in series" In the example f the figure C and C in fig.a are cnnected in parallel. They can be substituted by the euivalent capacitr C = C + C as 3 shwn in fig.b. Capacitrs C and C in fig.b are cnnected in series. They can be substituted by a single capacitr C C 3 is given by the euatin: 3 = + C C C 3 3 as shwn in fig.c (5 -) Series cmbinatins reduce the capacitance. Eual C reduce by the number invlved. In parallel the capacitance increases. basket f 4 capacitrs, each f C = 6 nf. Hw can yu arrange them t get.5 nf nf 3 nf 4 nf nf 4 nf clicker ll C s are 8.00 nf. The battery is. What is the euivalent capacitance? a. 4 nf b. 6 nf c. 8 nf d. 0 nf e. nf

2 /5/0 ll C s are 8.00 nf. The battery is. What is the euivalent capacitance? C = 4 nf C 3 = nf Q 3 = C 3 x = 44 nc Q 3 = C 3 x = 96 nc Q = C x = 48 nc U 3 = ½ C 3 = ½ x x0-9 x = 864 nj U = ½ C = ½ x 8x0-9 x6 = 44 nj = U U 3 = ½ x 8x0-9 x = 576 nf C 3 stres mst energy, als the highest electric field and mst charge, the stressed part f the circuit.

3 /5/0 Circuits ll capacitrs being the same, rank the euivalent capacitances f the fur circuits. C = C 3 = 8.00 µf, C = C 4 = 6.00 µf, = When the switch S is clsed, hw much charge flws thrugh pint P C = ε / d C 3.4 µf, = 8.8 µc C C 4 = µf C 34 = 3 µf =36 µc = 7. µc C d = ε ( κ + κ ) If the areas are and -. ε C = d [ κ + κ ] ( κ C 3.4 µf, = 8.8 µc C C 4 = µf C 34 = 3 µf =36 µc - ' -' - - C = κc air ' Capacitr with a dielectric In 837 Michael Faraday investigated what happens t the capacitance C f a capacitr when the gap between the plates is cmpletely filled with an insulatr (a.k.a. dielectric) Faraday discvered that the new capacitance is given by : C = κc air Here C air is the capacitance befre the insertin f the dielectric between the plates. The factr κ is knwn as the dielectric cnstant f the material. Faraday's experiment can be carried ut in tw ways:. With the vltage acrss the plates remaining cnstant In this case a battery remains cnnected t the plates. This is shwn in fig.a. With the charge f the plates remaining cnstant. In this case the plates are islated frm the battery This is shwn in fig.b (5-5) 3

4 /5/0 - ' -' - - C = κc air ' Fig.a : Capacitr vltage remains cnstant This is bacause the battery remains cnnected t the plates fter the dielectric is inserted between the capacitr plates the plate charge changes frm t = κ κ The new capacitance C = = = κ = κcair Fig.b : Capacitr charge remains cnstant This is bacause the plates are islated fter the dielectric is inserted between the capacitr plates the plate vltage changes frm t = κ The new capacitance C = = = κ = κcair / κ (5-6) The frce n a filling dielectric as it is inserted between the parallel plates f a capacitr. ε C = C + C = d + dc ( κ ) = C dx With the battery cnnected, U = ½C du F = = U dx ( κ ) x With the battery discnnected, U = Q /C ( κ ) du ( κ ) F = = U dx With the battery cnnected, since x is increasing dwnwards, a negative frce is upwards, pushing the dielectric away. With the battery discnnected, the frce is psitive and pinted dwnwards, pulling in the dielectric. The frce is prprtinal t (κ-) and inversely t. x uestin Questin What is the euivalent capacitance between the pints and B?. μf B. μf C. 4 μf D. 0μF E. Nne f these What wuld a 0 battery d, i.e. hw much charge will it prvide, when it is cnnected acrss and B? 40 μc B parallel-plate capacitr has a plate area f 0.3m and a plate separatin f 0.mm. If the charge n each plate has a magnitude f 5x0-6 C then the frce exerted by ne plate n the ther has a magnitude f abut:. 0 B. 5N C. 0. 9N D. x0 4 N E. 9 x 0 5 N F = E = ε 6 ( 5 0 ) = 4.7N = The electric field = σ/ε why? 4

5 /5/0 Questin parallel-plate capacitr has a plate area f 0.3m and a plate separatin f 0.mm. If the charge n each plate has a magnitude f 5x0-6 C then the frce exerted by ne plate n the ther has a magnitude f abut:. 0 B. 5N C. 9N D. x0 4 N E. 9 x 0 5 N uestin Each f the fur capacitrs shwn is 500 µf. The vltmeter reads 000. The magnitude f the charge, in culmbs, n each capacitr plate is:. 0. B. 0.5 C. 0D. 50E. nne f these HITT parallel-plate capacitr has a plate area f 0.m and a plate separatin f 0. mm. T btain an electric field f.0 x 0 6 /m between the plates, the magnitude f the charge n each plate shuld be:. 8.9 x 0-7 C B..8 x 0-6 C C. 3.5 x 0-6 C D. 7. x 0-6 C E..4 x 0-5 C 5