When a potential difference of 150 V is applied to the plates of a
parallel-plate capacitor, the plates carry a surface charge density of
30.0 nC/cm2. What is the spacing between the plates?
0 A
Q 
 V 
d
8.85  10
12
0  V 
d


30.0  109 C cm

C 2 N m
2
2
 150 V 
1.00  10
4
cm
2
m
2

 4.42 m
Four capacitors are connected as shown in
Figure
(a) Find the equivalent capacitance
between points a and b.
(b) Calculate the charge on each capacitor
if ΔVab = 15.0 V.
Q  C V   5.96 F 15.0 V   89.5 C
Q 89.5 C

 4.47 V
C 20.0 F
15.0  4.47  10.53 V
1
1
1


C s 15.0 3.00
V 
C s  2.50 F
C p  2.50  6.00  8.50 F

1
1 
C eq  

 8.50 F 20.0 F 
Q  C V   6.00 F 10.53 V   63.2 C on 6.00 F
1
 5.96 F
89.5 63.2  26.3 C
Find the equivalent capacitance
between points a and b for the group of
capacitors connected as shown in Figure
P26.27. Take C1 = 5.00 μF, C2 = 10.0 μF,
and C3 = 2.00 μF.
1
1 
 1
Cs  

 3.33  F

 5.00 10.0
C p1  2 3.33  2.00  8.66  F
C p2  2 10.0  20.0  F
C eq
1 
 1


 8.66 20.0
1
 6.04  F
A parallel-plate capacitor is charged and then disconnected from a
battery. By what fraction does the stored energy change (increase
or decrease) when the plate separation is doubled?
C2
d2  2d1
. Therefore, the
1
 C1
2
stored energy doubles
Determine (a) the capacitance and (b) the maximum
potential difference that can be applied to a Teflon-filled
parallel-plate capacitor having a plate area of 1.75 cm2
and plate separation of 0.040 0 mm.
 A
C 0 
d


2.10 8.85  1012 F m 1.75  104 m 2
5
4.00  10 m


  8.13 10
11

F  81.3 pF
Vm ax  Em axd  60.0  106 V m 4.00  105 m  2.40 kV
A parallel-plate capacitor is constructed using a dielectric material
whose dielectric constant is 3.00 and whose dielectric strength is
2.00 × 108 V/m. The desired capacitance is 0.250 μF, and the
capacitor must withstand a maximum potential difference of 4 000
V. Find the minimum area of the capacitor plates.
  3.00
 0 A
C
 0.250  106 F
d
A
C V m ax
Cd

 0  0 Em ax
Vm ax
Em ax  2.00  10 V m 
d
8
0.250  10   4 000



3.00 8.85  10  2.00  10 
6
12
8
0.188 m
2
- A parallel plate capacitor made from 2 squares of metal, 2mm thick and 20cm on a side
separated by 1mm with 1000V between them Find:
a) capacitance
b)charge per plate
c) charge density d)electric field
e) energy stored
f) energy density
4
A
20

20

10
11
12
C   0  8.851012

35
.
4

10
F

3
.
54

10
F
3
d
110
C  3.54 pF
b)
c)
Q  CV  3.54103  3.54nC
Q
3.54109
7
2
 

0
.
885

10
C
/
m
A 20 20104
 
3 2
12
1
10

3
.
54

10
e) U  CV 2 
 1.77106 J
2
2
U
1.77106
3
3
f) u


44
.
25

10
J
/
m
volume 20 20104 1103
Consider the circuit as shown, where C1 = 6.00F and C2= 3.00
F and V =20.0V. Capacitor C1 is first charged by closing of
switch S1. Switch S1 is then opened and the charged capacitor is
connected to the uncharged capacitor by the closing of S2.
Calculate the initial charge acquired by C1 and the final charge
on each.
S1 close, S2 open 
C = Q/V  Q = 120 C
After S1 open, S2 close 
Q1 + Q2 = 120 C
Same potential  Q1 /C1 = Q2 / C2 
(120-Q2)/C1= Q2/C2
(120 - Q2)/6 = Q2/ 3  Q2 = 40 C
 Q 1= 80 C
• An isolated conducting sphere whose radius R is 6.85 cm has a
charge of q=1.25 nC. How much potential energy is stored in the electric
field of the charged conductor?
Answer:
Key Idea: An isolated sphere has a capacitance of C=40R
The energy U stored in a capacitor depends on the charge and the capacitance
according to
… and substituting C=40R gives
April 13, 2015
University Physics, Chapter 24
11
• An isolated conducting sphere whose radius R is 6.85 cm
has a charge of q = 1.25 nC.
Question 2:
What is the field energy density at the surface of the sphere?
Answer:
Key Idea: The energy density u depends on the magnitude of the electric
field E according to
1
u  0E2
2
so we must first find the E field at the surface of the sphere. Recall:
E
1
q
4 0 R2
2
1
q
5
3
3
u  0E 2 

2.54
10
J/m

25.4

J/m
2
32 2 0 R 4
a)
b)
A
2
12
10
C   0  8.8510 

35
.
4

10
F
3
d
5 10
Q  CV  104  35.4 1010  35.4 106 C
4
c)
10
6
E  V/d 

2

10
V /m
3
5 10
A parallel-plate capacitor is charged and then disconnected from a
battery. By what fraction does the stored energy change (increase
or decrease) when the plate separation is doubled?
5- Determine (a) the capacitance and (b) the maximum potential
difference that can be applied to a Teflon-filled parallel-plate
capacitor having a plate area of 1.75 cm2 and plate separation of
0.040 0 mm.
• An air-filled parallel plate capacitor has a capacitance of 1.3 pF.
The separation of the plates is doubled, and wax is inserted
between them. The new capacitance is 2.6pF.
Question:
Find the dielectric constant of the wax.
Answer:
• Key Ideas: The original capacitance is given by
Then the new capacitance is
Thus
rearrange the equation:
April 13, 2015
University Physics, Chapter 24
16
Question 1:
Consider a parallel plate capacitor with capacitance C =
2.00 F connected to a battery with voltage V = 12.0 V as
shown. What is the charge stored in the capacitor?


q  CV  2.00 106 F 12.0 V  2.40 105 C
Question 2:
Now insert a dielectric with dielectric constant  =
2.5 between the plates of the capacitor. What is the
charge on the capacitor?
C   Cair The capacitance of the capacitor is increased:
q  CV  2.50  2.0 106 F  12.0V  6.0 105 C
The additional charge is provided by the battery.
April 13, 2015
University Physics, Chapter 24
17
Given a 7.4 pF air-filled capacitor. You are asked to
convert it to a capacitor that can store up to 7.4 J with
a maximum voltage of 652 V. What dielectric constant
should the material have that you insert to achieve
these requirements?
One common kind of computer keyboard is based on the idea of capacitance. Each key is mounted on one end of a plunger, the other end being attached to a movable
metal plate. The movable plate and the fixed plate form a capacitor. When the key is
pressed, the capacitance increases. The change in capacitance is detected, thereby
recognizing the key which has been pressed.
The separation between the plates is 5.00 mm, but is reduced to 0.150 mm when a key
is pressed. The plate area is 9.50x10-5m2 and the capacitor is filled with a material
whose dielectric constant is 3.50.
Determine the change in capacitance detected by the computer.
If each capacitor has a capacitance of 5 nF, what is the capacitance of
this system of capacitors?
Find the equivalent capacitance
•
•
•
•
We can see that C1 and C2 are in parallel,
and that C3 is also in parallel with C1 and C2
We find C123 = C1 + C2 + C3 = 15 nF
… and make a new drawing
April 13, 2015
University Physics, Chapter 24
20
• We can see that C4 and C123 are in series
• We find for the equivalent capacitance:
1
C1234

1
1
C C

 C1234  123 4
C123 C4
C123  C4
= 3.75 nF
• … and make a new drawing
April 13, 2015
University Physics, Chapter 24
21
• We can see that C5 and C1234 are in parallel
• We find for the equivalent capacitance
C12345  C1234
C123C4
(C1  C2  C3 )C4
 C5 
 C5 
 C5
C123  C4
C1  C2  C3  C4
= 8.75 nF
• … and make a new drawing
April 13, 2015
University Physics, Chapter 24
22
• We have a parallel plate capacitor
constructed of two parallel plates, each
with area 625 cm2 separated by a
distance of 1.00 mm.
Question: What is the capacitance of this
parallel plate capacitor?
 0 A 8.85 10-12 F/m  0.0625 m 2
C

d
0.001 m
 5.53 1010 F
= 0.553nF
Answer: A parallel plate capacitor constructed out of
square conducting plates 25 cm x 25 cm separated by 1 mm
has a capacitance of about 0.5 nF.
April 13, 2015
University Physics, Chapter 24
23
• We have a parallel plate capacitor
constructed of two parallel plates
separated by a distance of 1.00 mm.
Question: What area is required to
produce a capacitance of 1.00 F?
1 F  0.001 m
A

 0 8.8510-12 F/m
Cd
 1.13108 m 2
Answer: Square conducting plates with dimensions 10.6 km
x 10.6 km (6 miles x 6 miles) separated by 1 mm are
required to produce a capacitor with a capacitance of 1 F.
April 13, 2015
University Physics, Chapter 24
24
: A storage capacitor on a random access memory (RAM) chip
has a capacitance of 55 nF. If the capacitor is charged to 5.3
V, how many excess electrons are on the negative plate?
Answer:
Idea: We can find the number of excess electrons on the
negative plate if we know the total charge q on the plate.
Then, the number of electrons n=q/e, where e is the electron
charge in coulomb.
Second idea: The charge q of the plate is related to the
voltage V to which the capacitor is charged: q=CV.
April 13, 2015
University Physics, Chapter 24
25
Capacitor C1 is connected across a
battery of 5 V. An identical
capacitor C2 is connected across a
battery of 10 V. Which one has
more charge?
1) C1
2) C2
3) both have the same charge
4) it depends on other factors
Since Q = CV and the two capacitors are
identical, the one that is connected to
the greater voltage has more charge,
which is C2 in this case.
What must be done to a
capacitor in order to
increase the amount of
charge it can hold (for a
constant voltage)?
1) increase the area of the plates
2) decrease separation between the plates
3) decrease the area of the plates
4) either (1) or (2)
5) either (2) or (3)
+Q –Q
Since Q = CV, in order to increase the charge that a
capacitor can hold at constant voltage, one has to
increase its capacitance. Since the capacitance is
A
given byC   0 , that can be done by either
d
increasing A or decreasing d.
A parallel-plate capacitor
initially has a voltage of 400 V
and stays connected to the
battery. If the plate spacing is
now doubled, what happens?
1) the voltage decreases
2) the voltage increases
3) the charge decreases
4) the charge increases
5) both voltage and charge change
Since the battery stays connected, the voltage
must remain constant! Since
C  0 A ,
d
when the spacing d is doubled, the capacitance
C is halved. And since Q = CV, that means the
charge must decrease.
+Q
–Q
A parallel-plate capacitor initially has a
potential difference of 400 V and is
then disconnected from the charging
battery. If the plate spacing is now
doubled (without changing Q), what is
the new value of the voltage?
1) 100 V
2) 200 V
3) 400 V
4) 800 V
5) 1600 V
+Q –Q
Once the battery is disconnected, Q has to remain
constant, since no charge can flow either to or from
the battery. Since
C  0 A
d
, when the
spacing d is doubled, the capacitance C is halved.
And since Q = CV, that means the voltage must
double.
1) Ceq = 3/2C
What is the equivalent
capacitance, Ceq , of the
combination below?
2) Ceq = 2/3C
3) Ceq = 3C
4) Ceq = 1/3C
5) Ceq = 1/2C
The 2 equal capacitors in series add
up as inverses, giving 1/2C. These
are parallel to the first one, which
add up directly. Thus, the total
equivalent capacitance is 3/2C.
o
Ceq
C
o
C
C
How does the voltage V1 across
the first capacitor (C1) compare to
the voltage V2 across the second
capacitor (C2)?
The voltage across C1 is 10 V. The
combined capacitors C2 + C3 are
parallel to C1. The voltage across C2
+ C3 is also 10 V. Since C2 and C3 are
in series, their voltages add. Thus
the voltage across C2 and C3 each
has to be 5 V, which is less than V1.
1) V1 = V2
2) V1 > V2
3) V1 < V2
4) all voltages are zero
C2 = 1.0 F
C1 = 1.0 F
10 V
C3 = 1.0 F
1) Q1 = Q2
How does the charge Q1 on the
first capacitor (C1) compare to
the charge Q2 on the second
capacitor (C2)?
2) Q1 > Q2
3) Q1 < Q2
4) all charges are zero
We already know that the voltage
C2 = 1.0 F
across C1 is 10 V and the voltage
across both C2 and C3 is 5 V each.
Since Q = CV and C is the same for
all the capacitors, we have V1 > V2
and therefore Q1 > Q2.
10 V
C1 = 1.0 F
C3 = 1.0 F