# 1 Chapter 24--Examples. 2 Problem In the figure to the left, a potential difference of 20 V is applied across points a and b. a) What is charge on each.

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1 Chapter 24--Examples

2 Problem In 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 C 1 = 10  F, C 2 =20  F, and C 3 =30  F. b) What is potential difference across points a and d? c) D and b?

3 Step 1: Find Equivalent Resistance C1&C2: 10+20=30 C1&C2+C3: (1/30)+(1/30)= 2/30 i.e. 30/2=15  F

4 Finding 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 & d

5 Parallel 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  C

6 Problem In the figure, each capacitor C 1 =6.9  F and C 2 =4.6  F. a) Compute equivalent capacitance between a and b b) Compute the charge on each capacitor if V ab =420 V c) Compute V cd when V ab =420 V

7 Analysis 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=6.9

8 Repeat 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  F

9 So 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.

10 In 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  C

11 Problem Two parallel plates have equal and opposite charges. When the space between them is evacuated, the electric field is 3.2 x 10 5 V/m. When the space is filled with a dielectric, the electric field is 2.5 x 10 5 V/m. a) What is the charge density on each surface of the dielectric? b) What is the dielectric constant?

12 ++++++++++++++++++++++ - - - - - - - - - - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - +++++++++++++++++++ E0E0 EiEi E Total =E 0 -E i  i =8.85e-12*(3.2-2.5)*10 5 =4.32x10 6 C/m 2

13 Dielectric Constant K=E 0 /E=3.2e5/2.5e5=1.28

14 Problem A 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?

15 RC Value RC=3  F*7.25k  RC=3e-6*7.25e3 RC=21.75 ms

16 Time 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 ms

17 Since 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 ms

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