Physics 2212 Quiz #3 Solutions Spring 2009 ϵ0 Permittivity constant ...
Physics 2212 Spring 2009
Quiz #3
Solutions
²0 Permittivity constant mp Mass of a proton µ0 Permeability constant me Mass of an electron e Fundamental charge c Speed of light K Coulomb constant g Acceleration due to gravity Unless otherwise directed, friction, drag, and gravity should be neglected, and all batteries and wires are ideal. Any integrals in free-response problems must be evaluated. I. (16 points) A cylindrical wire with radius R carries a total current I0 . The current density magnitude in the wire is given by 2 ¯ ¯ ¯J~¯ = J0 r R where J0 is a constant and r is the distance from the center of the wire. What is J0 , in terms of R, I0 , and physical or mathematical constants? . . . . . . . . . . . . . . . . . . . . . . . Remember that current density is an area density. Z I=
~ J~ · dA
~ that are small in the radial direction, Since the current density varies radially, choose area elements dA namely, rings of area 2πr dr. These rings have an area vector (perpendicular to their surface) that is parallel to the current density. Z
R
I0 = 0
r2 J0 J0 cos 0◦ 2πr dr = 2π R R
Z
R 0
· ¸R J0 r4 J0 4 r dr = 2π R =π R 4 0 2R 3
So J0 =
2I0 πR3
Quiz #3 Solutions Page 1 of 5
II. (16 points) Two conductors carrying equal and opposite charges ±Q constitute a capacitor (not a parallel-plate capacitor!). The conductors are separated by a distance d along the electric field line marked i. The electric field magnitude along this line depends on the charge Q and the distance s from the negatively-charged conductor according to ¯ ¯ ~ ¯ = Q (As + B) ¯E where A and B are constants. What is the capacitance of this capacitor? Express your answer in terms of parameters defined in the problem, and physical or mathematical constants. .
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The capacitance can be found from its definition C=
Q ∆V
once the potential difference between the conductors is determined. Since the electric field is known along the path i, the potential difference can be found. Note that the electric field points opposite the integration path. Z
Z ∆V = −
d
~ =− ~ · ds E 0
· 2 ¸d · 2 ¸ s d Q (As + B) cos 180◦ ds = Q A + Bs = Q A + Bd 2 2 0
So C=
Q Q 2 ¢= = ¡ d2 2 + 2Bd ∆V Ad Q A 2 + Bd
1. (6 points) In the problem above, how does the electric potential difference calculated along the electric field line marked ii compare to that calculated along the electric field line marked i? . . . . . . . . . . . . . . . . . . . . . . . The electric potential within a conductor at equilibrium is zero, so the potential difference between two conductors must be the same, no matter what path it is calculated over. Electric potential difference along ii is the same as that along i.
Quiz #3 Solutions Page 2 of 5
III. (16 points) The battery in the illustrated circuit has emf E. Each of the four capacitors has the same capacitance, C. Once the circuit has been connected for a long time, what energy is stored in the bottom-most capacitor, marked with an asterisk, with respect to zero energy at zero charge? Express your answer in terms of parameters defined in the problem, and physical or mathematical constants. . . . . . . . . . . . . . . . . . The top and right-most capacitors are in series (same charge). Their equivalent capacitance can be found. 1 1 1 = + Ceq1 C C
⇒
Ceq1 =
C 2
The two bottom capacitors are in parallel (same potential difference). Their equivalent capacitance can be found. Ceq2 = C + C = 2C These two new equivalent capacitances are in series. The equivalent capacitance for the entire circuit can be found. 1 1 1 = + Ceq C/2 2C
⇒
Ceq =
2C 5
The charge delivered by the battery is µ Q = Ceq ∆V =
2C 5
¶ E
This charge is on the 2C/5 equivalent capacitor, and on each of the C/2 and 2C equivalent capacitors (since they are in series). The potential difference across the 2C equivalent capacitor is ∆V =
Q (2C/5) E E = = Ceq2 2C 5
Since the two bottom capacitors are in parallel, this is also the potential difference across each of them individually. The energy stored in the bottom-most capacitor, then, is U=
1 2C
2
(∆V ) =
1 2C
µ ¶2 E CE 2 = 5 50
2. (6 points) If the energy found in the problem above is U0 , what energy would be stored in that same capacitor if the battery were replaced by one with twice the emf, 2E? When the battery has emf 2E, the energy stored in the bottom-most capacitor is . . . . . . . . . . . . . . . . . . . . . .
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The stored energy is proportional to the square of the potential across the capacitor. In a circuit with only one battery, the energy stored in each capacitor must be proportional to the square of the battery’s emf. The energy stored in the bottom-most capacitor is 4U0 .
Quiz #3 Solutions Page 3 of 5
3. (10 points) Electric potential is graphed as a function of position. Where is the electric field magnitude greatest? At that location, in which direction does the electric field point? . . . . . . . . . . . . . . . . . . . . . . . Since the electric field is related to the electric potential by ~ = − δV E δs the greatest field magnitude will be where the slope of the graph has greatest magnitude, and the electric field points toward lower potential.
Electric field has greatest magnitude between 0 and 2 cm. It points in the negative direction.
4. (10 points) Wire 1 and wire 2 are completely separate, unconnected, wires made from the same metal. Wire 1 has twice the diameter and half the electric field of wire 2. How does the current I1 in wire 1 compare to the current I2 in wire 2? . . . . . . . . . . . . . . . . . . . . . . . Because the wires are made from the same metal, they have the same conductivity. ~ = σ J~ E
⇒
E=σ
I A
⇒
σ=
E1 A1 E2 A2 = I1 I2
Then µ I1 = I2
E1 π4 d21 E2 π4 d22
Ã
¶ = I2
E2 2
(2d2 ) E2 d22
2
!
= 2I2
Quiz #3 Solutions Page 4 of 5
5. (10 points) A cylindrical resistor is constructed with radius r and length L. The current through it is related to the electric potential difference between the ends, as graphed below. Another resistor is constructed of the same material. It has the same radius, but half the length. What is the resistance of this second resistor? .
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.
.
.
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The first resistor has resistance R=
10 V ∆V = = 0.5 Ω I 20 A
Resistance of an object is related to its resistivity and geometry by R=ρ
L A
The second resistor has the same resistivity as the first one, because it is made of the same material. It has the same cross-sectional area because it has the same radius. Because it has half the length, it must have half the resistance. 0.25 Ω
6. (10 points) The bulb is about one meter from the battery. Once the switch is closed, how long will it take for electrons from the battery to reach the bulb? . . . . . . . . . . . . . . Typical electron drift speeds in circuits are on the order of a fraction of a millimeter per second. Therefore, it will take thousands of seconds for electrons from the battery to reach the bulb. About an hour.
Quiz #3 Solutions Page 5 of 5
Quiz #3
Solutions
²0 Permittivity constant mp Mass of a proton µ0 Permeability constant me Mass of an electron e Fundamental charge c Speed of light K Coulomb constant g Acceleration due to gravity Unless otherwise directed, friction, drag, and gravity should be neglected, and all batteries and wires are ideal. Any integrals in free-response problems must be evaluated. I. (16 points) A cylindrical wire with radius R carries a total current I0 . The current density magnitude in the wire is given by 2 ¯ ¯ ¯J~¯ = J0 r R where J0 is a constant and r is the distance from the center of the wire. What is J0 , in terms of R, I0 , and physical or mathematical constants? . . . . . . . . . . . . . . . . . . . . . . . Remember that current density is an area density. Z I=
~ J~ · dA
~ that are small in the radial direction, Since the current density varies radially, choose area elements dA namely, rings of area 2πr dr. These rings have an area vector (perpendicular to their surface) that is parallel to the current density. Z
R
I0 = 0
r2 J0 J0 cos 0◦ 2πr dr = 2π R R
Z
R 0
· ¸R J0 r4 J0 4 r dr = 2π R =π R 4 0 2R 3
So J0 =
2I0 πR3
Quiz #3 Solutions Page 1 of 5
II. (16 points) Two conductors carrying equal and opposite charges ±Q constitute a capacitor (not a parallel-plate capacitor!). The conductors are separated by a distance d along the electric field line marked i. The electric field magnitude along this line depends on the charge Q and the distance s from the negatively-charged conductor according to ¯ ¯ ~ ¯ = Q (As + B) ¯E where A and B are constants. What is the capacitance of this capacitor? Express your answer in terms of parameters defined in the problem, and physical or mathematical constants. .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
The capacitance can be found from its definition C=
Q ∆V
once the potential difference between the conductors is determined. Since the electric field is known along the path i, the potential difference can be found. Note that the electric field points opposite the integration path. Z
Z ∆V = −
d
~ =− ~ · ds E 0
· 2 ¸d · 2 ¸ s d Q (As + B) cos 180◦ ds = Q A + Bs = Q A + Bd 2 2 0
So C=
Q Q 2 ¢= = ¡ d2 2 + 2Bd ∆V Ad Q A 2 + Bd
1. (6 points) In the problem above, how does the electric potential difference calculated along the electric field line marked ii compare to that calculated along the electric field line marked i? . . . . . . . . . . . . . . . . . . . . . . . The electric potential within a conductor at equilibrium is zero, so the potential difference between two conductors must be the same, no matter what path it is calculated over. Electric potential difference along ii is the same as that along i.
Quiz #3 Solutions Page 2 of 5
III. (16 points) The battery in the illustrated circuit has emf E. Each of the four capacitors has the same capacitance, C. Once the circuit has been connected for a long time, what energy is stored in the bottom-most capacitor, marked with an asterisk, with respect to zero energy at zero charge? Express your answer in terms of parameters defined in the problem, and physical or mathematical constants. . . . . . . . . . . . . . . . . . The top and right-most capacitors are in series (same charge). Their equivalent capacitance can be found. 1 1 1 = + Ceq1 C C
⇒
Ceq1 =
C 2
The two bottom capacitors are in parallel (same potential difference). Their equivalent capacitance can be found. Ceq2 = C + C = 2C These two new equivalent capacitances are in series. The equivalent capacitance for the entire circuit can be found. 1 1 1 = + Ceq C/2 2C
⇒
Ceq =
2C 5
The charge delivered by the battery is µ Q = Ceq ∆V =
2C 5
¶ E
This charge is on the 2C/5 equivalent capacitor, and on each of the C/2 and 2C equivalent capacitors (since they are in series). The potential difference across the 2C equivalent capacitor is ∆V =
Q (2C/5) E E = = Ceq2 2C 5
Since the two bottom capacitors are in parallel, this is also the potential difference across each of them individually. The energy stored in the bottom-most capacitor, then, is U=
1 2C
2
(∆V ) =
1 2C
µ ¶2 E CE 2 = 5 50
2. (6 points) If the energy found in the problem above is U0 , what energy would be stored in that same capacitor if the battery were replaced by one with twice the emf, 2E? When the battery has emf 2E, the energy stored in the bottom-most capacitor is . . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
The stored energy is proportional to the square of the potential across the capacitor. In a circuit with only one battery, the energy stored in each capacitor must be proportional to the square of the battery’s emf. The energy stored in the bottom-most capacitor is 4U0 .
Quiz #3 Solutions Page 3 of 5
3. (10 points) Electric potential is graphed as a function of position. Where is the electric field magnitude greatest? At that location, in which direction does the electric field point? . . . . . . . . . . . . . . . . . . . . . . . Since the electric field is related to the electric potential by ~ = − δV E δs the greatest field magnitude will be where the slope of the graph has greatest magnitude, and the electric field points toward lower potential.
Electric field has greatest magnitude between 0 and 2 cm. It points in the negative direction.
4. (10 points) Wire 1 and wire 2 are completely separate, unconnected, wires made from the same metal. Wire 1 has twice the diameter and half the electric field of wire 2. How does the current I1 in wire 1 compare to the current I2 in wire 2? . . . . . . . . . . . . . . . . . . . . . . . Because the wires are made from the same metal, they have the same conductivity. ~ = σ J~ E
⇒
E=σ
I A
⇒
σ=
E1 A1 E2 A2 = I1 I2
Then µ I1 = I2
E1 π4 d21 E2 π4 d22
Ã
¶ = I2
E2 2
(2d2 ) E2 d22
2
!
= 2I2
Quiz #3 Solutions Page 4 of 5
5. (10 points) A cylindrical resistor is constructed with radius r and length L. The current through it is related to the electric potential difference between the ends, as graphed below. Another resistor is constructed of the same material. It has the same radius, but half the length. What is the resistance of this second resistor? .
.
.
.
.
.
.
.
.
The first resistor has resistance R=
10 V ∆V = = 0.5 Ω I 20 A
Resistance of an object is related to its resistivity and geometry by R=ρ
L A
The second resistor has the same resistivity as the first one, because it is made of the same material. It has the same cross-sectional area because it has the same radius. Because it has half the length, it must have half the resistance. 0.25 Ω
6. (10 points) The bulb is about one meter from the battery. Once the switch is closed, how long will it take for electrons from the battery to reach the bulb? . . . . . . . . . . . . . . Typical electron drift speeds in circuits are on the order of a fraction of a millimeter per second. Therefore, it will take thousands of seconds for electrons from the battery to reach the bulb. About an hour.
Quiz #3 Solutions Page 5 of 5
