midterm 01 – SAENZ, LORENZO – Due: Feb 14 2008, 11:00 pm
E & M - Basic Physical Concepts
Current and resistance
Current: I = ddtQ = n q vd A
Ohm’s law: V = I R, E = ρJ
I , R = ρℓ
E = Vℓ , J = A
A
Electric force and electric field
Electric force between 2 point charges:
|q | |q |
|F | = k 1r2 2
k = 8.987551787 × 109 N m2 /C2
ǫ0 = 4 π1 k = 8.854187817 × 10−12 C2 /N m2
qp = −qe = 1.60217733 (49) × 10−19 C
mp = 1.672623 (10) × 10−27 kg
me = 9.1093897 (54) × 10−31 kg
~
~ =F
Electric field: E
2
Power: P = I V = VR = I 2 R
Thermal coefficient of ρ: α = ρ ∆ρ
0 ∆T
Motion of free electrons in an ideal conductor:
a τ = vd → qmE τ = nJq → ρ = n qm2 τ
|Q|
~2 + · · ·
~ =E
~1 + E
Point charge: |E| = k r2 , E
Field patterns: point charge, dipole, k plates, rod,
spheres, cylinders,. . .
Charge distributions:
Linear charge density: λ = ∆Q
∆x
Surface charge density: σsurf =
∆Q
Volume charge density: ρ = ∆V
∆Qsurf
∆A
Electric flux and Gauss’ law
~ · n̂∆A
Flux: ∆Φ = E ∆A⊥ = E
Gauss law: Outgoing Flux from S, ΦS = Qenclosed
ǫ0
Steps: to obtain electric field
~ pattern and construct S
–Inspect E
H
~ · dA
~ = Qencl , solve for E
~
–Find Φs = surf ace E
ǫ
0
Spherical: Φs = 4 π r2 E
Cylindrical: Φs = 2 π r ℓ E
Pill box: Φs = E ∆A, 1 side; = 2 E ∆A, 2 sides
σ
k
~ = 0, Esurf
Conductor: E
= 0, E ⊥ = surf
in
surf
Potential
ǫ0
Potential energy: ∆U = q ∆V 1 eV ≈ 1.6 × 10−19 J
Positive charge moves from high V to low V
Point charge: V = krQ V = V1 + V2 = . . .
1 q2
Energy of a charge-pair: U = k rq12
Potential difference: |∆V | = |E ∆sk |,
R
~ · ∆~s, V − V = − B E
~ · d~s
∆V = −E
B
A
A
¯
¯
d
V
∆V
E = − dr , Ex = − ∆x ¯
= − ∂V
∂x , etc.
f ix y,z
Capacitances
Q=CV
Series: V = CQ = CQ + CQ + CQ + · · ·, Q = Qi
eq
1
2
3
Parallel: Q = Ceq V = C1 V + C2 V + · · ·, V = Vi
ǫ A
Q
Parallel plate-capacitor: C = V
= EQd = 0d
2
RQ
Q
Energy: U = 0 V dq = 12 C , u = 12 ǫ0 E 2
2
1
2
Uκ = 21κ Q
C0 , uκ = 2 ǫ0 κ Eκ
Q
Q
Spherical capacitor: V = 4 π ǫ r1 − 4 π ǫ r2
0
0
~
Potential energy: U = −~
p·E
Dielectrics: C = κC0 ,
V =IR
Direct current circuits
q
Area charge density: σA = ∆Q
∆A
1
Series: V = I Req = I R1 + I R2 + I R3 + · · ·, I = Ii
V + V + V + · · ·, V = V
Parallel: I = RV = R
i
R2
R3
eq
1
Steps: in application of Kirchhoff’s Rules
–Label currents: i1 , i2 , i3 , . . .
P
P
i
–Node equations:
i =
P in
Pout
–Loop equations: “ (±E) + (∓iR)=0”
–Natural: “+” for loop-arrow entering − terminal
“−” for loop-arrow-parallel to current flow
RC circuit: if ddty + R1C y = 0, y = y0 exp(− RtC )
Charging: E − Vc − R i = 0, 1c ddtq + R ddti = ci + R ddti = 0
Discharge: 0 = Vc − R i = qc + R ddtq , ci + R ddti = 0
Magnetic field and magnetic force
µ0 = 4 π × 10−7 T m/A
µ a2 i
µ i
0
Wire: B = 2 π0 r
Axis of loop: B =
2 (a2 +x2 )3/2
~ → q ~v × B
~
Magnetic force: F~M = i ~ℓ × B
~ × B,
~
Loop-magnet ID: ~τ = i A
µ
~ = i A n̂
2
r
Circular motion: F = mrv = q v B, T = f1 = 2 π
v
~ + q ~v × B
~
Lorentz force: F~ = q E
~
Hall effect: V = FM d , U = −~
µ·B
H
q
~ and magnetism of matter
Sources of B
µ
~
µ
v ×r̂
0 q~
~ = 0 i ∆ℓ×r̂
Biot-Savart Law: ∆B
4 π r2 , B = 4 π r2
2
µ0 i ∆y
∆B = 4 π
sin θ, sin θ = ar , ∆y = r a∆θ
r2
H
~ · d~s = µ I
B
Ampere’s law: M =
L
0
encircled
Steps: to obtain magnetic field
~ pattern and construct loop L
–Inspect B
~
–Find M and Iencl , and solve for B.
d (E A)
ΦE = ǫ
Displ. current: Id = ǫ0 d dt
0
dt
Magnetism in atom:
Orbital motion: µ = i A = 2 em L
L = m v r = n h̄,
QA
= d dt
h̄ = 2hπ = 1.06 × 10−34 J s
h̄ = 9.27 × 10−24 J/T
µB = 2em
µspin = µB
Magnetism in matter:
0
B = B0 + BM = (1 + χ) B0 = (1 + χ) µ0 B
µ0 = κm H
Ferromagnetic: χ ≫ 1
Diamagnetic: −1 ≪ χ < 0
Paramagnetic: 0 < χ ≪ 1, M = C
TB
µorbit = n µB ,
Spin: S = h̄2 ,
midterm 01 – SAENZ, LORENZO – Due: Feb 14 2008, 11:00 pm
Question 1, chap 24, sect 5.
part 1 of 1
10 points
A point charge q1 is concentric with two
spherical conducting thick shells, as shown
in the figure below. The smaller spherical
conducting shell has a net charge of q2 and
the larger spherical conducting shell has a net
charge of q3 .
r4
q3
r3
r2
r1
q2
q1
R1
R2 R3 R4 R5
10. Qr3 = −q1 − q2 + q3
Explanation:
The net charge inside a Gaussian surface
located at r = R4 must be zero, since the field
in the conductor must be zero. Therefore,
the charge on the inner surface of the large
spherical conducting shell must be −(q1 + q2 ).
Question 2, chap 22, sect 5.
part 1 of 1
10 points
Two uncharged metal balls X and Z stand
on insulating glass rods. A third ball, carrying
a negative charge, is brought near the ball Z
as shown in the figure. A conducting wire is
then run between X and Z and then removed.
Finally the third ball is removed.
conducting wire
−
Under static conditions, the charge on a
conductor resides on the surface of the conductor. What is the charge Qr3 on the inner surface of the larger spherical conducting
shell?
2
Z
X
When all this is finished
1. balls X and Z are both positive, but ball
Z carries more charge than ball X.
2. ball X is neutral and ball Z is positive.
1. Qr3 = +q1
3. ball X is positive and ball Z is neutral.
2. Qr3 = +q1 − q2
4. ball X is positive and ball Z is negative.
3. Qr3 = −q1 + q2
5. balls X and Z are both negative.
4. Qr3 = −q1 − q2 − q3
5. Qr3 = +q1 + q2 + q3
6. Qr3 = −q1 − q2 correct
7. Qr3 = −q1
6. ball X is negative and ball Z is positive.
correct
7. balls X and Z are still uncharged.
8. balls X and Z are both positive, but ball
X carries more charge than ball Z.
8. Qr3 = 0
9. ball X is neutral and ball Z is negative.
9. Qr3 = +q1 + q2
10. ball X is negative and ball Z is neutral.
midterm 01 – SAENZ, LORENZO – Due: Feb 14 2008, 11:00 pm
Explanation:
When the conducting wire is run between
X and Z, some negative charge flows from
Z to X under the influence of the negative
charge of the third ball.
Therefore, after the wire is removed, X
is negatively charged and Z is positively
charged.
Question 3, chap 23, sect 1.
part 1 of 1
10 points
Four point charges are placed at the four
corners of a square. Each side of the square
has a length L.
q4 =
q
q3 =
q
b
b
Pb
b
1.
2.
3.
4.
5.
6.
7.
8.
9.
Explanation:
L
Each charge is at a distance √ from P, so
2
the magnitude of each field is
kq
2kq
2 = 2 .
L
L
√
2
Consider the direction of each field:
E=
E2
E1
E3
b
Find the magnitude of the electric field vec~ at the center P of the square.
tor E
k q2
L2
√
2 2kq
L2
√
2 2 k q2
L2
4 k q2
L
4 k q2
L2
kq
L2
4kq
correct
L2
√
2 2kq
L
kq
L
The electric field is
X
X k qi
~ =
~i =
E
E
r̂ .
2 i
r
i
i
i
L
q1 = −q L q2 = q
3
E4
E2 and E4 cancel, so
~ = kE
~ 1 k + kE
~ 3k
kEk
= 2 |E1|
=
4kq
.
L2
Question 4, chap 25, sect 3.
part 1 of 1
10 points
Consider an equilateral triangle with sides
of lengths 3.3 µm and charge −0.2 µC, 1.2 µC
and 1.5 µC located at the corners of the triangle.
Find the minimum work required to move
the first point charge to infinity.
Correct answer: 1470.69 (choice number 1).
Explanation:
Let : ǫ0 = 8.85419 × 10−12 C2 /N · m2 ,
q1 = −0.2 µC = −2 × 10−7 C ,
q2 = 1.2 µC = 1.2 × 10−6 C ,
q3 = 1.5 µC = 1.5 × 10−6 C , and
a = 3.3 µm = 3.3 × 10−6 m .
To move the charge q1 out of the electric
field due to q2 and q3 , we need to do work
midterm 01 – SAENZ, LORENZO – Due: Feb 14 2008, 11:00 pm
against the electric field. So, the minimum
work required is the electric potential difference.
W = −∆U
= −q1
q2
q3
+
4 π ǫ0 a 4 π ǫ0 a
,
Since
4 π ǫ0 a = 4 π 8.85419 × 10−12 C2 /N · m2
× (3.3 × 10−6 m)
= 3.67175 × 10−16 C2 /N · m ,
Consider a solid conducting sphere with an
radius R1 surrounded by a concentric thick
conducting spherical shell which has an inner
radius R2 and outer radius R3 . There is a
charge Q on the sphere and no net charge on
the shell.
For this problem, we adopt the standard
convention of setting the electric potential at
infinity to zero.
q2 = 0
R1
R2
Ob
R3
−7
W = (−2 × 10 C)
1.2 × 10−6 C
×
3.67175 × 10−16 C2 /N · m
1.5 × 10−6 C
+
3.67175 × 10−16 C2 /N · m
= 1470.69 J .
Question 5, chap 25, sect 2.
part 1 of 1
10 points
The magnitude of a uniform electric
field between the two plates is about
1.3 × 106 N/C.
If the distance between these plates is
0.1 cm, find the potential difference between
the plates.
Correct answer: 1300 (choice number 8).
Explanation:
Let : E = 1.3 × 106 N/C and
∆d = 0.1 cm = 0.001 m .
The magnitude of the potential difference is
∆V = E ∆d
= (1.3 × 106 N/C) (0.001 m)
= 1300 V .
Question 6, chap 25, sect 3.
part 1 of 1
10 points
4
b
B
bA
q1 = Q
Find the potential at O, where RO < R1 .
kQ kQ
+
R2
R1
2kQ
2. VO =
R1 + R2
1. VO =
3. VO = ∞
√
2 2kQ
4. VO =
R1
5. VO = 0
2kQ
R
√ 1
2kQ
7. VO =
R1
kQ
8. VO =
R1
kQ kQ
9. VO =
+
R3
R1
kQ kQ kQ
10. VO =
−
+
correct
R3
R2
R1
Explanation:
We are still inside the spherical shell. The
potential due to the shell requires knowing
something about where charge exists on the
shell. To this end, consider a Gaussian spherical surface inside the shell. This surface can
6. VO =
midterm 01 – SAENZ, LORENZO – Due: Feb 14 2008, 11:00 pm
5
have no flux through it since no electric field
can be upheld inside a conductor. By Gauss’s
Law, there can therefore be no charge enclosed. But we are already enclosing Q on
the sphere, so a charge −Q must reside on the
inner surface of the shell:
This potential is the same at O as it is everywhere inside the sphere. A conductor is an
equipotential object.
qinner = −Q .
Consider the two cases shown below. In
Case One two identical capacitors are connected to a battery with emf V . In Case Two,
a dielectric slab with dielectric constant κ fills
the gap of capacitor C2 .
Since the net charge on the shell is zero, the
charge on the outer surface must be +Q for
the inner and outer charges to add up to zero:
Question 7, chap 26, sect 3.
part 1 of 1
10 points
Case One
C1
C2
qouter = +Q .
Now that we know the exact distribution
of charge, we can utilize the expression for
potential inside of a spherical charge distribution:
q
V =k ,
a
for a thin shell of radius a.
Thus the potential due to the inner shell
(radius R2 ) is
V2 = k
V
Case Two
C1
C′ 2
Q
qinner
= −k
R2
R2
and that due to the outer shell (radius R3 ) is
κ
qouter
Q
V3 = k
=k
,
R3
R3
so the contribution from the two surfaces of
the shell is
V1 = k
Q
Q
−k
.
R3
R2
However, we are now looking at a point
inside the sphere. The charge is all on the
surface of the sphere, so similarly to the situation for the shell we have
Q
V =k
R1
everywhere inside the sphere.
Thus the total potential at O is
VO = k
Q
Q
Q
−k
+k
.
R3
R2
R1
V
V2′
of potential differences across
V2
capacitor C2 for the two cases is
The ratio
V2′
V2
V′
2. 2
V2
V′
3. 2
V2
V′
4. 2
V2
1.
=
1+κ
.
2
= κ.
1+κ
.
2κ
1+κ
=
.
3κ
=
midterm 01 – SAENZ, LORENZO – Due: Feb 14 2008, 11:00 pm
V2′
= 2κ.
V2
2
V′
. correct
6. 2 =
V2
1+κ
V′
1+κ
7. 2 =
.
V2
3
2κ
V′
.
8. 2 =
V2
1+κ
3
V′
.
9. 2 =
V2
1+κ
V2′
2κ
10.
=
.
V2
1 + 3κ
Explanation:
For two capacitors in series,
5.
Q1 = Q2
C1 V1 = C2 V2
The
charge
−19
on an electron is 1.60218 × 10
C and its
mass is 9.10939 × 10−31 kg .
How far does the electron travel before it is
brought to rest?
Correct answer: 7.39292 (choice number 8).
Explanation:
Let :
v = 4.3 × 106 m/s ,
qe = 1.60218 × 10−19 C ,
m = 9.10939 × 10−31 kg ,
E = 711 N/C .
For case One,
For case Two,
1
V.
2
C2′
= κ, so
C1
V2′ =
Therefore
W =
1
V
1+κ
Z
F dx = F x = qe E x
since the force is constant. When the electron
comes to rest, all its kinetic energy has been
converted, so
1
m v 2 = qe E x .
2
C2
= 1, so
C1
V2 =
and
1
The kinetic energy K = m v 2 is depleted
2
by the amount of work done by the electric
force F = qe E on the particle:
In addition,
C2
V1 + V2 =
V2 + V2 = V
C1
1
V
V2 =
C2
1+
C1
6
m v2
2 qe E
(9.10939 × 10−31 kg)(4.3 × 106 m/s)2
=
2 (1.60218 × 10−19 C) (711 N/C)
100 cm
×
1m
= 7.39292 cm .
x=
V2′
2
=
.
V2
1+κ
Question 8, chap 23, sect 4.
part 1 of 1
10 points
An electron moves at 4.3 × 106 m/s into a
uniform electric field of magnitude 711 N/C.
The field is parallel to the electron’s velocity
and acts to decelerate the electron.
Question 9, chap 24, sect 5.
part 1 of 1
10 points
Consider a conducting spherical shell with
inner radius 1 m and outer radius 1.4 m.
There is a net charge 2 µC on the shell. At
its center, within the hollow cavity, there is a
point charge −9 µC.
midterm 01 – SAENZ, LORENZO – Due: Feb 14 2008, 11:00 pm
1 m,
Q′2 inside
−9 µC
S
2 µC
7
surfaces. Hint: Notice that with the electric field assumed, there is no electric charge
throughout the region, including the space
within the pyramid.
Correct answer: 604.478 (choice number 5).
Explanation:
Q′′2 outside,
1.4 m
0.3 m
Determine the flux through the spherical
Gaussian surface S, which has a radius of
0.3 m. The permitivity of free space is
8.8542 × 10−12 C2 /N m2.
Correct answer: −1.01647 × 106 (choice
number 1).
Explanation:
Let : q1 = −9 µC = −9 × 10−6 C and
ǫ0 = 8.8542 × 10−12 C2 /N m2 .
The surface S encloses only the point charge
q1 , so by Gauss’ Law
I
~ · dA
~ = qencl = q1
ΦS =
E
ǫ0
ǫ0
S
−6
−9 × 10 C
=
8.8542 × 10−12 C2 /N m2
= −1.01647 × 106 N m2 /C .
Question 10, chap 24, sect 1.
part 1 of 1
10 points
A (3.22 m by 3.22 m) square base pyramid
with height of 3.34 m is placed in a vertical
electric field of 58.3 N/C, which is uniform
throughout the region.
Let : s = 3.22 m ,
h = 3.34 m , and
E = 58.3 N/C .
~ · A.
~ Since there is no
By Gauss’ law Φ = E
charge contained in the pyramid, the net flux
through the pyramid must be 0 N/C. Since
the field is vertical, the flux through the base
of the pyramid is equal and opposite to the
flux through the four sides. Thus we calculate
the flux through the base of the pyramid,
which is
Φ = E A = E s2
= (58.3 N/C) (3.22 m)2
= 604.478 N m2/C .
Question 11, chap 26, sect 1.
part 1 of 1
10 points
When a potential difference of 123 V is
applied to two plates of a parallel-plate capacitor, the inner surface of the positive plate
carries a surface charge density of 24 nC/cm2 .
The permittivity of a vacuum is 8.85419 ×
10−12 C2 /N · m2 .
What is the spacing between the plates?
Correct answer: 4.53777 (choice number
10).
Explanation:
3.34 m
3.22 m
58.3 N/C
Calculate the total electric flux which
comes out through the pyramid’s four slanted
Let : σ = 24 nC/cm2 = 0.00024 C/m2 ,
V = 123 V , and
ǫ0 = 8.85419 × 10−12 C2 /N · m2 .
The electric field within the gap is given by
σ
V
= , where s stands for the spacing,
E=
s
ǫ0
midterm 01 – SAENZ, LORENZO – Due: Feb 14 2008, 11:00 pm
ǫ0 V
σ
(8.85419 × 10−12 C2 /N · m2 ) (123 V)
=
(0.00024 C/m2 )
= 4.53777 × 10−6 m
s=
= 4.53777 µm .
Question 12, chap 24, sect 3.
part 1 of 1
10 points
Consider a long cylindrical charge distribution of radius R with a uniform charge density
ρ > 0.
Find the electric field at distance r from the
axis where r < R .
~ = − π r ρ r̂
1. E
ǫ0
r
~ = + ρ r̂
2. E
ǫ0
~ = − π R ρ r̂
3. E
ǫ0
π
~ = + R ρ r̂
4. E
2 ǫ0
~ = + π R ρ r̂
5. E
ǫ0
r
~ = − ρ r̂
6. E
2 ǫ0
~ = + r ρ r̂ correct
7. E
2 ǫ0
~ = + π r ρ r̂
8. E
ǫ0
Explanation:
By Gauss’ Law,
I
~ · dA
~ = q .
E
ǫ0
If our Gaussian surface is a cylinder of radius r and height h, then the area of the surface is 2 π r h + 2 π r 2 and the charge enclosed
by the surface is π r 2 h ρ . By symmetry, the
field points radially outward or inward from
the cylinder axis, so the flux through the top
and bottom of the cylinder (the sides with
area π r 2 each) is zero. Then Gauss’ Law
becomes
π r2 h ρ
ǫ0
rρ
.
E=
2 ǫ0
E ·2πrh =
The field E points radially outward for positive ρ, and we know that unit normal from
the Gaussian surface points radially outward,
so
~ = + r ρ r̂ .
E
2 ǫ0
Question 13, chap 25, sect 3.
part 1 of 1
10 points
Four charges are placed at the corners of a
square, where q is positive (q > 0).
Q1 = −q
Q2 = +q
a
or the width of the gap. Solving for s leads to
8
Q4 = −q
Q3 = +q
The magnitude of the total electrostatic
energy of the system is given by
√ ke q 2
1. |U | = 2 2 2 .
a
2
ke q
2. |U | = 2 2 .
a
ke q 2
3. |U | = 4 2 .
a
√ ke q 2
4. |U | = 4 2
.
a
√ ke q 2
5. |U | = 2 2 .
a
ke q 2
6. |U | = 2
.
a
ke q 2
7. |U | = 8 2 .
a
ke q 2
8. |U | = 4
.
a
midterm 01 – SAENZ, LORENZO – Due: Feb 14 2008, 11:00 pm
√ ke q 2
. correct
2
a
√ ke q 2
10. |U | = 4 2 2 .
a
Explanation:
The total electrostatic energy is
9. |U | =
U = U12 + U13 + U14 + U23 + U24 + U34
ke q 2
1
1
=
−1 − √ + 1 + 1 − √ − 1
a
2
2
2
√ ke q
,
=− 2
a
which has a magnitude of
√ ke q 2
|U | = 2
.
a
5.
6.
7.
8.
9.
10.
Ut
Ub
Ut
Ub
Ut
Ub
Ut
Ub
Ut
Ub
Ut
Ub
= 2κ
=
1
κ+1
= 2κ + 1
= κ correct
1
2κ + 1
1
=
κ
=
Explanation:
For a capacitor with dielectric κ,
C=
Question 14, chap 26, sect 3.
part 1 of 1
10 points
A capacitor is constructed from two metal
plates. The bottom portion is filled with air
and the top portion is filled with material of
dielectric constant κ. The plate area in the
top region is the same as that in the bottom
region. Neglect edge effects.
κ
9
κǫ0 A
d
Energy stored in a charged capacitor:
U=
1
1
1 Q2
QV = V2C =
.
2
2
2 C
Ct
Cb
E
Ut
of energy stored in
Ub
the top portion to the bottom portion.
Determine the ratio
Ut
Ub
Ut
2.
Ub
Ut
3.
Ub
Ut
4.
Ub
1.
= κ2
=κ+1
1
2κ
1
= 2
κ
=
The capacitance of a capacitor is determined solely by the dielectric constant of the
material and geometric configuration of the
capacitor.
The bottom portion and the top portion
may be treated as two capacitors connected
in parallel. Here Vb = Vt = V . Therefore,
1 2
V Ct
Ut
= 2
1 2
Ub
V Cb
2
Ct
=
Cb
= κ.
midterm 01 – SAENZ, LORENZO – Due: Feb 14 2008, 11:00 pm
+
+
+
+
−
−
−
−
x
How is the net field aligned at the origin?
12 µF
63 µF
Consider the capacitor circuit
41 µF
Consider symmetrically placed rectangular
insulators with uniformly charged distributions of equal magnitude as shown in the figure below.
y
Question 16, chap 26, sect 2.
part 1 of 1
10 points
34 µF
Question 15, chap 23, sect 3.
part 1 of 1
10 points
10
68 V
What is the effective capacitance of the
circuit?
Correct answer: 37.5 (choice number 9).
Explanation:
1. aligned with the negative y-axis
2. along the 30◦ degree direction
3. along the 225◦ degree direction
4. aligned with the positive y-axis
Let : C1
C2
C3
C4
EB
= 34 µF ,
= 41 µF ,
= 63 µF ,
= 12 µF ,
= 68 V .
C2
C3
and
5. along the 60◦ degree direction
6. aligned with the negative x-axis correct
C1
C4
7. along the 45◦ degree direction
8. aligned with the positive x-axis
9. along the 135◦ degree direction
10. along the 315◦ degree direction
Explanation:
At the origin, the positive slab of charge
produces an electric field pointed toward
quadrant III (away from the positively
charged slab). The negatively charged slab
produces an electric field of equal magnitude
(as the positively charged slab) but pointing
toward quadrant II (toward the negatively
charged slab). The x-components of the two
fields add (producing Ex < 0), while the ycomponents cancel. Thus the electric filed is
along the negative x-axis.
EB
For capacitors in series,
X 1
1
=
Cseries
C
X i
Vseries =
Vi ,
and the individual charges are the same.
For parallel capacitors,
X
Cparallel =
Ci
X
Qparallel =
Qi ,
and the individual voltages are the same.
C1 and C2 are in parallel, so
C12 = C1 + C2
= 34 µF + 41 µF
= 75 µF .
midterm 01 – SAENZ, LORENZO – Due: Feb 14 2008, 11:00 pm
C3 and C4 are in parallel, so
C34 = C3 + C4
= 63 µF + 12 µF
= 75 µF .
C12
EB C34
C12 and C34 (Note: C12 = C34 ) are in series
with the battery, so
1
1
1
+
C12 C34
C12 C34
=
C12 + C34
(75 µF) (75 µF)
=
75 µF + 75 µF
C1234 =
= 37.5 µF .
11