PHYS 222
Worksheet 7 – Capacitance
Supplemental Instruction
Iowa State University
Leader:
Course:
Instructor:
Date:
Alek Jerauld
PHYS 222
Dr. Paula Herrera-Siklódy
2/2/12
Useful Equations
Q
V
A
C  0
d
C
Capacitance is the ratio of charge over voltage (units: Farad [F])
Capacitance is geometrically dependent. A is affective area, d
is the distance between surfaces, ε0 is the dielectric constant
Ceq  C1  C2  C3  ...
Finding equivalent capacitance of capacitors in parallel
1
1 1 1
    ... Finding equivalent capacitance of capacitors in series
Ceq C1 C2 C3
C  4 0
ab
ba
Capacitance of a double spherical-shelled capacitor
Diagrams:
Series:
Parallel:
C1
C1
C2
C3
C2
C3
Related Problems
1) A spherical capacitor contains a charge of 3.30 nC when connected to a potential
difference of 230 V. If its plates are separated by vacuum and the inner radius of the outer
shell is 4.60 cm. (Book 24.13) Calculate:
(a) Capacitance
Q 3.3(109 )
C 
 14.3(1012 ) F  14.3 pF
V
230
(b) Radius of the inner sphere
C  4 0
a
ab
1
,k 
ba
4 0
0.046(14.3)(1012 )(9)(109 )
bCk

 3.39 cm
b  Ck 0.046  (14.3)(1012 )(9)(109 )
(c) The electric field just outside the surface of the inner sphere
E
Q
4 0a
2
 2.58(104 ) N / C
2) The potential across ab is 50.0 V. (Book 24.22)
(a) Find the equivalent capacitance of the
system between a and b
C23  C2  C3  13  F
1
C1234 
 3.47  F
1 1  1
C1 C23 C4
(b) How much charge is stored by this
combination of capacitors?
Ceq 
Qeq
 Qeq  CeqV  174C
V
(c) How much charge is stored in the 10-µF capacitor?
Since all the charge must go through the 10- µF capacitor prior to going through any other
capacitor, Q1 = 174 µF
(d) How much charge is stored in the 9.0-µF capacitor?
All the charge must pass through the last capacitor, thus Q4 = 174 µF
3) In the figure, each capacitor is 4.60 µF and Vab=31.0 V.
(Book 24.18)
(a) Calculate the charge on C1 and C4
Q1  Q2  Q12  C12V12
1
C12 
 2.3  F
1  1
C1 C2
V12  V3  Vab  V4
Q
V4  4
C4
Q4  Qeq  CeqVab
1
Ceq 
1  1
C123 C4
C123  C12  C3  6.9 F
 Ceq  2.76 F
 Q4  85.58C
 V4  18.6043V
 V12  12.3957V
 Q1  28.5 C
(b) Calculate the voltage between points a and d
Vad  Vab V4  12.4V
4) The circuit below has the following specifications: C1 = 10 mF, C2 = 30 µF, C3 = 250
mF, C4 = 80 mF, C5 = 90 µF, and V = 25 V
(a) What is the equivalent capacitance of the circuit?
C123  C1  C2  C3  0.26 F
C45  C4  C5  0.08009 F
1
Ceq 
 61.23 mF
1  1
C123 C45
C1
C2
C3
C4
(b) What is the total charge on each plate of C4?
(c) What is the voltage across each of the
capacitors?
V4  V5  19.1 V
V1  V2  V3  Vs V4  5.9V
V
C5
C4
V4
C
Q5  5
V5
Q4  Q5  Qeq  VsCeq
 V4C4  V5C5  VsCeq
V4  V5
Q4 
 V4C4  V4C5  VsCeq  V4 
Q4  C4V4  C4
VsCeq
C4  C5
VsCeq
C4  C5
 1.529 C