Chapter 24--ExamplesProblemStep 1: Find Equivalent ResistanceFinding PotentialsParallel NetworkSlide 6AnalysisRepeat AgainSo 420 V *2.3 mF= 966 mCIn the middle network,ProblemSlide 12Dielectric ConstantSlide 14RC ValueTime to 50% of max chargeSince R & C have not changed, RC=21.75 ms1Chapter 24--Examples2ProblemIn the figure to the left, a potential difference of 20 V is applied across points a and b.a) What is charge on each capacitor if C1= 10 F, C2=20 F, and C3=30 F.b) What is potential difference across points a and d?c) D and b?3Step 1: Find Equivalent ResistanceC1&C2: 10+20=30C1&C2+C3: (1/30)+(1/30)= 2/30 i.e. 30/2=15  F4Finding PotentialsQ=CV=15*20=300 CSo charge on C3 is 300 CThe voltage across C3 is  C/30 F= 10 VSo 10 V across points b & d and therefore, 20-10=10 V across a & d5Parallel NetworkIf there is 10 V across this network thenQ1=10 F*10V= 100 CIf there are only 300 C total and 100 C is on this capacitor, then 200 C must be charge of C2Check: Q2=20 F*10V=200 C6ProblemIn the figure, each capacitor C1=6.9 F and C2=4.6 F. a)Compute equivalent capacitance between a and bb)Compute the charge on each capacitor if Vab=420 Vc)Compute Vcd when Vab=420 V7AnalysisLook at the right most connection: it is parallel connection between the C2 capacitor and the 3 C1 capacitors 3 C1’s: 6.9/3=2.3 FC2+3C1’s=2.3+4.6=6.98Repeat AgainNow the 6.9 F capacitor is in series with the other C1 capacitorsSo it is the same circuit as again, so the equivalent is 6.9 FFinally, the total equivalent is 2.3  F9So 420 V *2.3 F= 966 CFor each C1 capacitor on the leftmost network, the voltage across each is 140 V (966/6.9 or 420/3)If there is 140 V across the C2, then 140*4.6=644 C There must be 966-644 = 322 C in the other branch.10In the middle network,Each capacitor has 46.67 V (140/3 or 322 C/6.9) So voltage across c & d is 46.67 VThen C2 capacitor has 214C and the other C1 capacitors have 322-214 =107 C11ProblemTwo parallel plates have equal and opposite charges. When the space between them is evacuated, the electric field is 3.2 x 105 V/m. When the space is filled with a dielectric, the electric field is 2.5 x 105 V/m.a) What is the charge density on each surface of the dielectric?b) What is the dielectric constant?12++++++++++++++++++++++- - - - - - - - - - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - -+++++++++++++++++++E0EiETotal=E0-Ei)(*00TotaliiiTotalEEEEEi=8.85e-12*(3.2-2.5)*105=4.32x106 C/m213Dielectric ConstantK=E0/E=3.2e5/2.5e5=1.2814ProblemA 3.4 F capacitor is initially uncharged and then connected in series with a 7.25 k resistor and an emf source of 180 V which has negligible resistance.a) What is the RC time constant?b) How much time does it take (after connection) for the capacitor to reach 50% of its maximum charge?c) After a long time the EMF source is disconnected from the circuit, how long does it take the current to reach 1% of its maximum value?15RC ValueRC=3F*7.25kRC=3e-6*7.25e3RC=21.75 ms16Time to 50% of max chargeQ(t)=C*V*(1-e-t/RC)Q(t)/CV is the fraction of the maximum charge so let Q(t)/CV =50%.5=1-e-t/RCe-t/RC=.5=1/2 or 2-1-t/RC=-ln(2)t=RC*ln(2)=21.75*.693t=15.07 ms17Since R & C have not changed, RC=21.75 msI(t)=(V/R)*e-t/RCI(t)/(V/R) is the fraction of maximum currentLet I(t)/(V/R) = 1% or 0.010.01=e-t/RCln(0.01)=-4.605=-t/RCt=21.75*4.605=100