1. Reading Assignment
Serway Sections 26.1–5, 28.4–6. Next week Chapter 29.
2. Serway 26.16
OPTIONAL
A 1-megabit computer memory chip contains many 60-fF capacitors. Each capacitor has a plate area of
21
×
10
−
12 m 2 . Determine the plate separation of such a capacitor (assuming a parallel-plate configura-
A, where 1 ˚
≡
10
−
10 m is roughly the size of an atom.
SOLUTION: Each chip has a capacitance C = 60 femtoFarads = 60
×
10
− 15 capacitor C =
0
A/d , so :
F. For a parrallel plate d =
0
A
C
=
(8 .
85 × 10
− 12
F / m)(21 × 10
− 12
(60 × 10
− 15 F) m
2
)
= 3 .
1
×
10
−
9
3. Serway 26.22
A cylindrical capacitor has outer and inner conductors whose radii are in the ratio of b/a = 4 / 1. The inner conductor is to replaced by a wire whose radius is one-half the radius of the original inner conductor.
By what factor should the capacitor’s length be increased in order to obtain a capacitance equal to that of the original capacitor?
SOLUTION: See Serway Example 26.2. The capacitance of a length l of two such cylinders is l
C =
2 k e ln b/a
=
2 π
0 l ln b/a
Change the radius of the inner conductor to a
0 that the capacitance remain unchanged
= a/ 2 and find the necessary length l
0 from the condition
C =
2 π
0 l ln b/a
=
2 π
0 l
0 ln b/a
0
= C
0 l
0 l
= ln b/a
0 ln b/a
= ln 2 b/a ln b/a
= ln 8 ln 4
=
3 ln 2
2 ln 2
= 1 .
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4. Serway 26.34A
The circuit in Figure P26.34 of your text consists of two identical parallel metal plates connected by identical metal springs to a battery of voltage V . With the switch open, the plates are uncharged, are separated by a distance d , and have a capacitance C . When the switch is closed, the distance between the plates decreases by a factor 0.5. (a) How much charge collects on each plate and (b) what is the spring constant for each spring? (Hint: see problem 26.51 in your text.)
SOLUTION: a)After the switch is closed, the new distance between the plates is d
0
C
0
= 2 C
= d/ 2, giving an increased capacitance
. The definition of capacitance then gives the charge on each plate as :
Q = C
0
V = 2 CV.
b)Problem 26.51 states that the force on each capacitor plate due to the other is, (and using
0
A = Cd ) :
Q
2
F electric
=
2
0
A
=
(2 CV )
2
2( Cd )
=
2 CV
2
.
d
1

This forces pulls the plates together. It must be equal and opposite to the force on each plate due to its attached spring, since we have mechanical equilibrium. Since the plate separation has decreased by d/2, each spring has increased in length by d/4, and :
F spring
= − kd
4 the minus sign reflects the fact that the spring pulls the plates apart. Setting F electric find :
8 CV 2 k = d 2
+ F spring
= 0, we
5. Serway 26.55A+
A parallel-plate capacitor in air has a plate separation d and a plate area A . The plates are charged to potential difference V and disconnected from the power supply. The capacitor is then immersed in a liquid of dielectric constant κ . Determine (a) the charge on the plates before and after immersion,
(b) the capacitance and voltage after immersion and (c) the change in the energy of the capacitor.
Finally, (d) determine the change in the energy of the capacitor if the power supply remains connected when the capacitor is immersed.
SOLUTION: a)Before immersion: Q = CV = place to go.
0
( A/d ) V . The charge remains the same after immersion since it has no b)After immersion the capacitance is κ
0
A/d , and the voltage is V f and voltage still has to be Q .
= V /κ , since the product of capacitance c)The energy stored in a capacitor is U = (1 / 2) CV 2 = (1 / 2) QV . The change in stored energy is
U f
−
U i =
1
2
Q ( V f
−
V i
) =
1
2
0
AV d
V
κ
−
V
After immersion the stored energy is less than before.
d)Use U = (1 / 2) CV
2 and remember that V is constant now, not Q :
=
0
AV 2
2 d
1 − κ
κ
U f
− U i
=
1
2
V
2
( C f
− C i
) =
1
2
V
2
(
κ
0
A d
− 0
A
) = d
So now the stored energy has increased .
0
AV
2 d
2
( κ − 1)
6. Serway 26.72
A capacitor is constructed from two square plates of sides l and separation d . A material of dielectric constant κ is inserted a distance x into the capacitor, as in Figure P26.72 in your text. (a) Find the equivalent capacitance of the device. (b) Calculate the energy stored in the capacitor if the potential difference is V . (c) Find the direction and magnitude of the force exerted on the dielectric, assuming a constant potential difference V . Neglect friction. (d) Obtain a numerical value for the force assuming that l = 5 .
0 cm, V = 2000 V, d = 2 .
0 mm, and the dielectric is glass ( κ = 4 .
5). (Hint: the system can be treated as two capacitors in parallel.)
SOLUTION: a)Add the two parallel capacitances. Their areas are lx and l ( l
− x ), respectively, so
C = κ
0 lx d
+
0 l ( l − x ) d
=
0 l d
(( κ − 1) x + l ) .
b)
U =
1
2
CV
2
=
0 l
2 d
(( κ − 1) x + l ) V
2
.
2

c)The potential energy is a function of x , the position of the dielectric. The force on the dielectric is equal to the negative gradient of the potential energy
~
=
−
∂U
∂x
ˆ =
− 0 lV
2 d
2
( κ
−
1) ˆ
Since κ > 1, the force points to the left, in a direction to push the dielectric out of the capacitor. Qualitatively we can predict this by noting that the capacitance, and for fixed V , the potential energy is less for smaller x , that is if the dielectric is inserted less far.
d)
F =
8 .
85 × 10
− 12 C 2 / Nm 2 × 0 .
05 m × (2000 J / C)
2 × 0 .
002 m
2 (4 .
5 − 1)
= 1 .
55 × 10
− 3
N
7. Serway 28.72
Before the switch is closed in Figure P28.72 of your text, there is no charge stored by the capacitor.
Determine the currents in R
1
, R
2 and C (a) at the instant the switch is closed (that is, at after the switch is closed for a long time (that is, as t → ∞ ).
t = 0) and (b)
SOLUTION: a)At t = 0 there is no charge yet on the capacitor, and therefore no voltage across R
2 which is connected in parallel. The full voltage is across R
1
. At t = 0 i i
1
2
10 V
=
10 kΩ
= 0
= 1 .
0 mA b)As t
→ ∞ the capacitor is fully charged. Current flows through the series combination of R
1 and
10 V i
1
= i
2
= = 0 .
67 mA
10 kΩ + 5 kΩ and R
2 only
One can analyze this circuit more generally by applying Kirchoff’s Laws. Let I
1 be the current through R
1
, let I c
= dq/dt be the current charging the capacitor, and let I
2 be the current through R
2
.
Note that all of these currents will vary with time in this circuit.
Application of Kirchoff’s Laws results in the following three equations:
1.
I
1
= I c
+ I
2
= dq/dt + I
2 or I
2
= I
1
2. V
−
I
1
R
1
− q/C = 0 (inner loop).
3. V
−
I
1
R
1
−
I
2
R
2
= 0 (outer loop)
− dq/dt
By combining (1) and (3), we find
I
1
=
V + R
2 dq dt
R
1
+ R
2 and substitution of this expression into (2) yields
V 1 −
R
1
R
1
+ R
2
−
R
R
1
1
R
+
2
R
2 dq dt
− q c
= 0
Note that this is the same form as Equation 28.9 in Serway for a simple series circuit with a resistance R c
, capacitance C , and a voltage source V c
, with
V
0
= V 1
−
R
1
R
1
+ R
2
R
2
= V
R
1
+ R
2 and
R
1
R
2
R c
=
R
1
+ R
2
As in the development by Serway, we can rearrange the above to obtain
C V dq
0 − q
= dt
CR c integrating yields
− ln( C V
0 − q ) = t
CR c
+ A
3

and finally, exponentiating and evaluating the constant of integration A by using the initial condition q (0) = 0 gives the final solution q ( t ) = C V
0
1 − e
− t
CRc = C V
R
1
R
+
2
R
2
1 − e
− t ( R
1 +
R
2)
CR
1
R
2
We can obtain the current I c
= dq/dt from our solution: dq dt
=
V
R
1 e
− t ( R
1 +
R
2)
CR
1
R
2 and substitution into our previous equations for I
1 and I
2 yields
I
I
1
2
=
=
V
R
1
R
1
+ R
2 e
− t ( R
1+
R
2)
CR
1
R
2
R
1
+ R
2
R
1
V
+ R
2
1 − e
− t ( R
1 +
R
2)
CR
1
R
2
For the two limits t = 0 and t → ∞ , the above formula yield the same results we obtained above.
4