Ch26 11 12 13 21 25 34 35 36 37 43 59

Thank God for yellow !!
Chpt. Summary:

A. C = Q/(V+ - V-)

Parallel plate (derive this !):

Cylindrical (derive this !)

Spherical (derive this !)

B. Energy density (energy/volume) in a region of space with E:

C. Energy of a capacitor:

The energy of a capacitor can  heat up objects. See #37 and the Sample Test 3.

D. Dieletric materials with k > 1 increase the capacitance. See # 43.

E.
Capacitors in series: 1/Cs = 1/C1  + 1/ C2 + ...
Capacitors in parallel: Cp = C1 + C2 + ...

Answers
11.
(a) Derive the capacitance C of this system. Read example 26.2. 
(b) C = Q/(V+  - V-) . See your class notes! I derived this. Solve for  V+  - V-  .
12.  See equation 26.6 in example 26.3 . Note that b = 2a  in this problem. Substitute b = 2a in equation 26.6.  Set this expression equal to 20.0 µF. Solve for the two radii  a  and b.  Then find the volume of a shell  of thickness b - a. Here is how you do that: First,  find the volume of a sphere of radius b. Second,  find the volume of a sphere of radius a.  Then subtract the second volume from the first volume.
13. Guess what? This problem is almost identical to problem 2 of sample test 2. Click here and scroll down to review it. In both cases, the electric field is constant and horizontal. In this case, 
V +  -  V- = Ed so that   E = ( V +  -  V-   )/d . Use this expression for E and follow the same method as problem 2 of sample test 2 to find E and then  solve for ( V +  -  V-   ). Note you can also read  example 23.4 in Ch. 23 for some of the "flava" associated with this problem.
21.(a) Let 15µF = C1   and 3.00 µF = C2. These two capacitors are in series, so  Cs = C1*C2/(C1 + C2) . Now,  Cs is in parallel with C3 = 6.00 µF. Thus, we can combine them as so: Cp = Cs + C3  .
 Finally, Cp is in series with  C4 = 20.0µF. Do you know how to combine Cp and C4 in series to get the final expression Cf for the equivalent capacitor? I'm sure you do. 
(b) The charge on Cf  is Qf = 15 Volts/Cf  . This also  is the charge on C4  because that capacitor is in series with Cp. The next step is to find the charge across C3. That's easy. The total voltage of 15 Volts  is the sum of the voltage across C3 and the voltage across C4 : 15 Volts = Q3/C3.+ Qf/C4 . Solve for Q3.  Finally, you can find the voltage across C1 and C2. The voltage across the series combo of  C1 and C2 is Q3/C3  since the series combo is in parallel with C3. Thus, the charge across both C1 and C2 is Cs* Q3/C3 . 
34. You should read  page 815 for the low-down on energy density. The derivation is easy.  Use the fact that the energy of a parallel plate capacitor is  U = ½CV2  .  Then,  use the fact that V =Ed and use the expression for C in equation 26.3. Then divide by the volume Ad of the space between the plates to get that the energy density (in J/m3 ) is

To solve this problem , use the fact that uE*(volume) = 1.00x10-7 J. Solve for the volume in m3 and liters.
35. We should do 35 first, since the result will be applied to 25. Let's write down the potential energy of the parallel plate capacitor:

where,

Thus,
.

We see that as the coordinate x, the separation between the plates, increases, the potential energy increases. Think of the gravitational potential energy near the surface of the earth:

As y increases, so does the potential energy. The force is attractive, towards the earth. The force is directed downward in the negative y direction, as indicated by the negative sign:

(The relationship between the component of the force and the derivative of the potential energy function is discussed in Ch. 8.)

Well, just as in the gravitational case, we get:

The force component negative, so the force points in the negative x - direction. Imagine this as the force on the right plate in the diagram above. That force points left. It is attractive. From Newton's third law, the force on the left plate is of equal magnitude, but points right.

The magnitude of the mutually attractive force (experienced by both plates) is:
25. After thinking about , we can approach 25 with greater clarity.
(a)



(b)

Here, x is the distance that one spring stretches. What is x? Think about it. If the distance between the plates decreases from d to d/2, then what is the distance that one spring stretches ?

Note that Q2 = C2V. Solve for k.
36. Let Q = the charge on plate a. Thus, from Chapter 23:

Fx =QE

The force points in the positive x-direction. E = the magnitude of the electric field caused by the negative charge on plate b. From Chapter 24:

Note that:
,

where A = the area of the plate. Finally,

Solve for x in terms of the other symbols shown above.
37.  Use the fact that:
 
energy absorbed by tree=0.01*energy stored in capacitor 
energy absorbed by tree=0.01* ½CV2

 Note that C = Q/V, so  ½(Q/V)V2 =  ½QV.

Now set  0.01* ½QV  = mCW(100 - 30) + mLv. 

Solve for m.
43. 

k = 3.7 for paper.

Solve for the length. Remember to convert to meters (m) !
59. Recall the derivation for a simple parallel plate capacitor (with plate 
separation  s) given in class and in the text:

Here,

and ,
image744.gif (1285 bytes)
In the expression for V above, we integrated in the space between the plates separated by a distance s.
Thus,

Now, in the current problem, the numerator Q in the definition for C is the same as in the case of the simple parallel plate system. What has changed is the integral for V. E = 0 inside the conductor of thickness d between the plates. Thus, we get:

image745.gif (1354 bytes)
Use this information to compute the capacitance for the system shown in fig. P26.59.