rec4 solutions

8/4/2019 Rec4 Solutions

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Recitation Week 4Chapter 24

Problem 14. For the system of capacitors shown in Fig. 24.24, find the equivalent capacitance (a) between b and c, and (b)between a and c.

a

|

=== 15pF

|

b

|

+--+--+

|9.0pF|

=== === 11pF

| |

+--+--+

|

c

(a) The 9 and 11 pF capacitors are in parallel, so the equivalent capacitance between b and c is

C bc = C L + C R = (9.0 + 11) pF = 20 pF (1)

(b) The 15 pF capacitor is in series with the equivalent capacitance C bc, so the equivalent capacitance between a and c is

C ac =

1

C T +

1

C bc

−1

=

1

15+

1

20

−1

pF =

4

12+

3

12

−1

· 5 pF =60

7pF = 8.6 pF (2)

Problem 18. In Fig. 24.26, C 1 = 6.00 µF, C 2 = 3.00 µF, and C 3 = 5.00 µF. The capacitor network is connected to anapplied potential V ab. After the charges on the capacitors have reached their final values, the charge on C 2 is 40.0 µC. (a)What are the charges on capacitors C 1 and C 3? (b) What is the applied voltage V ab?

1+-||-+

| |

a--+ +-+

| 2 | |

+-||-+ |

d

3 |

b----||---+

(a) Because C 1 and C 2 are in parallel, they must have the same voltage differences. Using this and the definition of capacitanceQ = CV , we have

Q1

C 1= V 1 = V 2 = Q2

C 2(3)

Q1 =C 1

C 2Q2 =

6.00

3.0040.0 µC = 80.0 µC . (4)

Since the wire d was initially uncharged, all the charge on C 1 and C 2 had to come from C 3. Assuming V a > V b (if V b > V a just flip all the charge signs in the following solution), the left hand sides of C 1 and C 2 will be at a higher potential than theright, so 0 < Q1L = Q1 = −Q1R and 0 < Q2L = Q2 = −Q2R. Balancing the charge on the wire d

0 = Q1R + Q2R + Q3R (5)

Q3R = −Q1R − Q2R = Q1 + Q2 = 120 µC > 0 (6)

Q3L = −Q3R = −120 µC < 0 . (7)



(b)V ab = V ad + V db (8)

We can find V db easily enough from Q3 = C 3V 3. Because C 1 and C 2 are in parallel, V 1 = V 2 = V ad (as we pointed out in(a)), so

V ab =Q3

C 3+

Q2

C 2=

120 µC

5.00 µF+

40.0 µC

3.00 µF= 24.0 V + 13.3 V = 37.3 V (9)

Problem 32. For the capacitor network shown in Fig. 24.28, the potential difference across ab is 36 V. Find (a) the tota

charge stored in this network; (b) the charge on each capacitor; (c) the total energy stored in the network; (d) the energystored in each capacitor; (e) the potential differences across each capacitor.

a--||--||--b

150 120

nF nF

(a) The equivalent capacitance is

C ab =

1

150+

1

120

−1

nF = 66.7 nF . (10)

The total charge on the capacitors is then

Q = CV = 66.7 nF·

36 V = 2.40 µC . (11)

(b) Each capacitor must also have Q = 2.40 µC, since the wire connecting the capacitors must remain uncharged as a whole

(c) The total capacitative energy is

E C =1

2CV 2 =

1

2· 66.7 nF · (36 V)2 = 43.2 µJ (12)

(d) We don’t know the voltage of each capacitor seperately, but we could find it using Q = CV , since we know Q and C . Wecould also just plug the formula in directly for V in our energy formula

E C =1

2CV 2 =

1

2C

Q

C

2

=1

2

Q2

C (13)

The energy stored in each capacitor is then

E CL =1

2

Q2

L

C L=

1

2

(2.40 µC)2

150 nF= 19.2 µJ (14)

E RL =1

2

Q2

R

C R=

1

2

(2.40 µC)2

120 nF= 24.0 µJ . (15)

Note that E CL + E CR = 43.2 µJ = E C , as it should be, since the equivalent capacitor situation is equivalent (i.e. interchangable) with the two-capacitor situation.

(e) Here we find the potential differences the way we mentioned in (d).

V L =

QL

C L =2.40 µC

150 nF = 16.0 V (16)

V R =QR

C R=

2.40 µC

120 nF= 20.0 V . (17)

Note that V L + V R = 36.0 V = V ab.

Problem 52. Cell membranes (the walled enclosure around a cell) are typically about 7.5 nm thick. They are partiallpermeable to allow charged material to pass in and out, as needed. Equal but opposite charge densities build up on the insidand outside faces of such a membrane,and these charges prevent additional charges from passing throughthe cell wall. Wcan model a cell membrane as a parallel-plate capacitor, with the membrane itself containing proteins embedded in an organicmaterial to give the membrane a dielectric constant of about 10. (See Fig. 24.30.) (a) What is the capacitance per squarecentimeter of such a cell wall? (b) In its normal resting state, a cell has a potential difference of 85 mV across its membraneWhat is the electric field inside this membrane?



Outside axon

++++++++++++++++++++++++

________________________

Axon membrane |- 7.5nm

________________________

------------------------

Inside Axon

(a) For parallel-plate capacitors (from Gauss’ law)

C =Q

V = ε

A

d= kε0

A

d. (18)

So the capacitance per square centimeter will be

C

A=

kε

d=

10 · 8.85 · 10−12 F/m

7.5 · 10−9 m= 1.2 · 10−2 F/m2

·

1 m

100 cm

2

= 1.2 µF/cm2 (19)

(b) Since the electric field generated by parallel plates is constant and perpendicular to the plates, we can easily integrateacross the membrane from the bottom to the top

∆V = t

b

E · dx = E · t

b

dx = E · d . (20)

Notice that for a perpendicular path (cutting straight across the membrane), E and d are in the same direction, so ∆V =E · d = Ed. Flipping this around to give E , we have

E =V

d=

85 mV

7.5 nm= 11 MV/m (21)

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