Download Exercise 24.5 Exercise 24.21

Heimadæmi 3
Due: 11:00pm on Thursday, February 4, 2016
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Exercise 24.5
A parallel­plate capacitor with circular plates and a capacitance of 13.8 μF is connected to a battery which provides a
voltage of 10.5 V .
Part A
What is the charge on each plate?
ANSWER:
= 1.45×10−4 C Q
Correct
Part B
How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the
battery?
ANSWER:
= 7.25×10−5 C Q
Correct
Part C
How much charge would be on the plates if the capacitor were connected to the battery after the radius of each plate
was doubled without changing their separation?
ANSWER:
= 5.80×10−4 C Q
Correct
Exercise 24.21
For the system of capacitors shown in the the figure below, a potential difference of 25.0 V is maintained across ab.
Part A
What is the equivalent capacitance of this system between a and b?
ANSWER:
C
= 19.3 nF Correct
Part B
How much charge is stored by this system?
ANSWER:
= 482 nC Q
Correct
Part C
How much charge does the 6.50 nF capacitor store?
ANSWER:
= 162 nC Q
Correct
Part D
What is the potential difference across the 7.50 nF capacitor?
ANSWER:
V
= 25.0 V Correct
Exercise 24.44
A parallel­plate capacitor has plates with area 2.30×10−2 m2 separated by 1.80 mm of Teflon.
Part A
Calculate the charge on the plates when they are charged to a potential difference of 10.0 V .
ANSWER:
= 2.37×10−9 C Q
Correct
Part B
Use Gauss's law to calculate the electric field inside the Teflon.
ANSWER:
E
= 5560 N/C Correct
Part C
Use Gauss's law to calculate the electric field if the voltage source is disconnected and the Teflon is removed.
ANSWER:
E
= 1.17×104 N/C Correct
Problem 24.45
Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the
flash lasts for a time interval of 1.48×10−3 s with an average light power output of 2.20×105 W .
Part A
If the conversion of electrical energy to light has an efficiency of 95.0 % (the rest of the energy goes to thermal
energy), how much energy must be stored in the capacitor for one flash?
ANSWER:
U
= 343 J Correct
Part B
The capacitor has a potential difference between its plates of 140 V when the stored energy equals the value
calculated in part A. What is the capacitance?
ANSWER:
C
= 3.50×10−2 F Correct
Problem 24.66
A parallel­plate capacitor is made from two plates 12.0 cm on each side and 4.50 mm apart. Half of the space between
these plates contains only air, but the other half is filled with Plexiglas of dielectric constant 3.40. (See the figure below.) An
18.0 V battery is connected across the plates.
Part A
What is the capacitance of this combination?
ANSWER:
C
= 6.23×10−11 F Correct
Part B
How much energy is stored in the capacitor?
ANSWER:
U
= 1.01×10−8 J Correct
Part C
If we remove the Plexiglas, but change nothing else, how much energy will be stored in the capacitor?
ANSWER:
U
= 4.59×10−9 J Correct
Exercise 22.17
A very long uniform line of charge has charge per unit length 4.64 μC/m and lies along the x­axis. A second long uniform
line of charge has charge per unit length ­2.24 μC/m and is parallel to the x­axis at y1 = 0.410 m .
Part A
What is the magnitude of the net electric field at point y2 = 0.180 m on the y­axis?
ANSWER:
E
= 6.39×105 N/C Correct
Part B
What is the direction of the net electric field at point y2 = 0.180 m on the y­axis?
ANSWER:
­y ­axis
+ y ­axis
Correct
Part C
What is the magnitude of the net electric field at point y3 = 0.612 m on the y­axis?
ANSWER:
E
= 6.31×104 N/C Correct
Part D
What is the direction of the net electric field at point y3 = 0.612 m on the y­axis?
ANSWER:
­y ­axis
+ y ­axis
Correct
Prelecture Concept Question 23.07
Part A
The electric potential at a certain distance from a point charge can be represented by V. What is the value of the
electric potential at twice the distance from the point charge?
ANSWER:
At twice the distance, the electric potential is V/2.
At twice the distance, the electric potential remains V.
At twice the distance, the electric potential is V/4.
At twice the distance, the electric potential is 2V.
At twice the distance, the electric potential is 4V.
Correct
Problem 23.34
Part A
If the electric potential in a region is given by V (x)
=
6/x
2
, the x component of the electric field in that region is
ANSWER:
−6x.
12x .
12x−3 .
6x.
−12x−3 .
Correct
Bouncing Electrons
Two electrons, each with mass m and charge q, are released from positions very far from each other. With respect to a
certain reference frame, electron A has initial nonzero speed v toward electron B in the positive x direction, and electron B
has initial speed 3v toward electron A in the negative x direction. The electrons move directly toward each other along the x
axis (very hard to do with real electrons). As the electrons approach each other, they slow due to their electric repulsion.
This repulsion eventually pushes them away from each other.
Part A
Which of the following statements about the motion of the electrons in the given reference frame will be true at the
instant the two electrons reach their minimum separation?
ANSWER:
Electron A is moving faster than electron B.
Electron B is moving faster than electron A.
Both electrons are moving at the same (nonzero) speed in opposite directions.
Both electrons are moving at the same (nonzero) speed in the same direction.
Both electrons are momentarily stationary.
Correct
If at a given moment the electrons are still moving toward each other, then they will be closer in the next instant. If
at a given moment the electrons are moving away from each other, then they were closer in the previous instant.
The electrons will be traveling in the same direction at the same speed at the moment they reach their minimum
separation. Only in a reference frame in which the total momentum is zero (the center of momentum frame) would
the electrons be stationary at their minimum separation.
Part B
What is the minimum separation rmin that the electrons reach?
Express your answer in term of q, m, v , and k (where k =
1
4πϵ0
).
Hint 1. How to approach the problem
Since no external or nonconservative forces act on the system of the two electrons, both momentum and total
energy (kinetic plus potential) are conserved. Find one expression for the energy when the electrons are far
apart, and another when they reach their minimum separation rmin . This will give you an equation in which the
only unknown is the speed of the electrons at the moment of their minimum separation. Apply conservation of
momentum, using the same initial and final states, to obtain a second equation involving the speed of the
electrons. Solve the simultaneous energy and momentum equations to obtain rmin .
Hint 2. Find the initial energy
What is the total energy Einitial of the two electrons when they are initially released? Assume that the electrons
are so far apart that their potential energy is zero.
Express your answer in terms of m and v .
ANSWER:
Einitial
= 5mv
2
Hint 3. Find the final energy
What is the total energy Ef inal of the electrons when they reach their minimum separation rmin ? Assume that
the (identical) speed of the two electrons is u.
Express your answer in terms of m, u, q, rmin , and k (where k =
Hint 1. Find the final kinetic energy
What is the final kinetic energy (both electrons)?
Express your answer in terms of u, and m.
ANSWER:
1
4πϵ0
).
= Kf inal
mu
2
Hint 2. Find the final potential energy
What is the final potential energy of this 2­electron system?
Express your answer in terms of k, q, and rmin .
ANSWER:
Uf inal
= kq
2
rmin
ANSWER:
Ef inal
= mu
2
+
kq
2
rmin
Hint 4. Find the initial momentum
⃗ What is the total momentum p initial
of the two electrons when they are initially released?
Express your answer as a vector in terms of m, v , and ^i .
ANSWER:
p ⃗ initial
= ^
−2mv i
Hint 5. Find the final momentum
What is the total momentum p f⃗ inal of the two electrons when they reach their minimum separation rmin ?
Assume that the (identical) velocity of the two electrons is u⃗ .
Express your answer as a vector in terms of m and u⃗ .
ANSWER:
p f⃗ inal
= 2mu⃗ Hint 6. Some math help
From the momentum equations, ∣u⃗ ∣
find rmin .
ANSWER:
rmin
= kq
2
2
4mv
; that is, u
= v
= v
. Substitute for u in the energy conservation equation to
Correct
An experienced physicist might approach this problem by considering the system of electrons in a reference frame
in which the initial momentum is zero. In this frame the initial speed of each electron is 2v. Try solving the
problem this way. Make sure that you obtain the same result for rmin , and decide for yourself which approach is
easier.
Score Summary:
Your score on this assignment is 97.8%.
You received 8.8 out of a possible total of 9 points.