Physics 2220 – Module 04 Homework
01.
What is the potential difference between yi = - 5 cm and yf = 5 cm in the uniform electric
field:
⃗
E = (20,000 ̂i − 50,000 ̂j) V/m
Start with the definition of potential difference:
B
Δ V = −∫ ⃗
E ⋅d ⃗s
A
The change in position is all along the y-direction. So in the dot product, only the y
component is going to survive.
⃗
⃗ ⋅dy ̂j = (20,000 ̂i − 50,000 ̂j)⋅dy ̂i = −50,000 dy
E ⋅d ⃗s = E
Now solve the integral:
+ 5 cm
∫
Δ V = 50,000
dy = (50,000 V/m) (0.10 m) = 5000 V
−5 cm
02.
Light from the Sun allows a solar cell to move electrons from the positive to the negative
terminal, doing 2.4 × 10-19 J of work per electron. What is the potential difference of this
solar cell?
Work in this case will be the difference in potential energy:
W =Δ U =q Δ V
Solve for the potential difference
W
2.4 × 10−19 J
Δ V =
=
= −1.5 Volts
−19
q
−1.6 × 10 C
This translates to the EMF of the solar cell being 1.5 Volts.
03.
The electric potential along the x-axis is
x = 0 m and (b) x = 1 m?
V = 100x2, where x is in meters. What is Ex at (a)
First determine the electric field in the x-direction by using the potential:
Ex = −
∂V
= − ∂ (100x2) = −200x
∂ x
∂ x
Now apply it to the problem:
04.
(a)
x=0m
E x = −200x = −200 (0 m) = 0 V/m
(b)
x=1m
E x = −200x = −200 (1.0 m) = −200 V/m
A switch that connects a battery to a 10 μF capacitor is closed. Several seconds later, you
find that the capacitor plates are charged to ±30 μC. What is the potential difference (EMF)
of the battery?
Use the definition of capacitance:
∣Q∣
∣ ∣
∣Δ V∣ = Q
→
∣Δ V∣
C
∣±30 μ C∣
∣Δ V∣ =
= 3 Volts
10 μ F
C=
05.
You need a capacitance of 50 μF, but you don't happen to have a 50 μF capacitor. You do
have a 30 μF capacitor. What additional capacitor do you need to produce a total
capacitance of 50 μF? Should you join the two capacitors in parallel or in series?
The capacitance needed is more than 30 μF.
Capacitors added in parallel will always be individually less than the equivalent
capacitance, so two capacitors in parallel are needed.
C eq = C 1 + C 2
C 2 = C eq − C 1
C 2 = 50 μ F − 30 μ F = 20 μ F
06.
You need a capacitance of 50 μF, but you don't happen to have a 50 μF capacitor. You do
have a 75 μF capacitor. What additional capacitor do you need to produce a total
capacitance of 50 μF? Should you join the two capacitors in parallel or in series?
The capacitance needed is less than 75 μF.
Capacitors added in series will always be individually more than the equivalent
capacitance, so two capacitors in series are needed.
1
1
1
=
+
C eq C 1 C 2
C − C eq
1
1
1
=
−
= 1
C 2 C eq C 1
C eq C 1
C C
(50 μ F) (75 μ F )
C 2 = eq 1 =
= 150 μ F
C 1 − C eq
75 μ F − 50 μ F
07.
A 2.0-cm diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V.
For a parallel-plate capacitor made of circular plates, the capacitance is:
A ϵ0 π r 2 ϵ0
C=
=
d
d
What is:
(a)
The total energy stored in the electric field?
Use the equation for potential energy stored in the electric field and substitute in
that the capacitance for a parallel-plate capacitor:
U=
U=
(b)
(
2
(
)
1
1 π r ϵ0
2
2
C (Δ V ) =
(Δ V )
2
2
d
2
−12
1 π (0.01 m) (8.85 × 10
2
0.00050 m
2
2
C /Nm )
) (200 V) = 1.11 × 10
2
−7
J
The energy density?
Use the equation for energy density for a parallel-plate capacitor, and substitute in
what the electric field will be:
1
1
Δ V 2
2
ϵ0 E = ϵ0
2
2
d
1
200 V
U density = (8.85 × 10−12 C2 / N m 2 )
2
0.00050 m
( )
U density =
(
2
) = 0.71 J / m
3
08.
Two 4.0 cm × 4.0 cm metal plates are separated by a 0.20-mm-thick piece of Teflon.
Note the dielectric constant and dielectric strength of teflon:
κ = 2.1
E max = 60 × 106 V/m
(a)
What is the capacitance?
Capacitance is increased when a dielectric material between the plates:
C = κ C0
For a parallel-plate capacitor:
A ϵ0
x y ϵ0
=κ
d
d
−12
2
2
(0.04 m ) (0.04 m) (8.85 × 10 C / N m )
C = (2.1)
= 1.5 × 10−10 F
0.00020 m
C =κ
(b)
What is the maximum potential difference between the plates?
The maximum potential difference corresponds to the maximum electric field that
can be between the plates before there is a break down. For a parallel-plate
capacitor:
Δ V max = E max d = (60 × 106 V/m) (0.00020 m) = 12000 Volts
09.
What is the equivalent capacitance of the three capacitors in the figure?
Each capacitor has been labeled for reference. Capacitor 2 and 3 are in parallel with each
other, so they may be combined.
C 4 = C 2 + C 3 = 20 μ F + 10 μ F = 30 μ F
Redraw the circuit:
Capacitor 1 and 4 are in series, and they may be combined:
1
1
1
1
1
=
+
=
+
C eq C 1 C 4 10 μ F 30 μ F
C eq = 7.5 μ F