T/F questions. You NEEED to solve all of them unless doesn\'t worth it. 1.( ) On

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T/F questions. You NEEED to solve all of them unless doesn't worth it.

1.(           ) Only charges can be stored inside a capacitor, but not electrical energy.
2.(           ) Both charges and electrical energy can be stored inside a capacitor.
3(           ) It is safe to open the case of a home theater system and touch inside, once the system is unplugged from the outlet.
4(           ) It is not safe to open the case of a home theater system and touch inside, even when the system is unplugged from the outlet, because capacitors inside may have been charged and stored large amount of energy, which can give you a “shock” while discharging.
5(`          ) When a capacitor is connected to a battery, it is gradually charged, and the energy from the battery is used to do work, which gradually increased the potential energy inside the capacitor. It is similar to slowly stretching a spring. Work was done to the spring, and potential energy is stored in the spring.
6(           ) When a spring’s length is stretched up to Dx, the magnitude of the spring force is |F|=kDx, hence the total work needed to stretch from the equilibrium point is then FDx=k Dx2.
7(           ) When a spring’s length is stretched up to Dx, the magnitude of the spring force is |F|=kDx, the total work needed to stretch from the equilibrium point is not FDx=k Dx2.   Because Dx is the maximum extension and kDx is the maximum force. It is not a constant force and the work is not equal to F times distance. If we integrate F dx for any small amount of dx at any location, the total work from F= 0 to F= kDx is the area of a triangle in the F vs. x plot, that is ½ k Dx2. Hence, the energy stored in the spring is ½ k Dx2.
8(           ) When a capacitor is charged to have potential difference DV, the amount of charge is Q=CDV, hence, the total work needed to charge it from Q=0 is then QDV=CDV2.
9(           ) When a capacitor is charged to have potential difference DV, the total amount of charge is Q=CDV. hence, the total work needed to charge it from Q=0 is NOT QDV=CDV2. Because Dx is the maximum voltage difference and CDV is the maximum charge which only happens at the end. It is not a constant voltage difference; work is not equal to total Q times DV. If we integrate DV dq for any small amount of charge dq, the total work needed is the area of a triangle in the DV vs. Q plot, that is ½ Q DV =½ C DV2. Hence, the energy stored in the capacitor is ½ C DV2.
10(           ) The energy stored in the capacitor is ½ QDV, which is also equal to ½ CDV2 and ½ Q2/C.
11(           ) The energy stored in the capacitor is ½ QDV, which is also equal to ½ CQ2 and ½ DV2/C.
12(           ) For two capacitors having the same charge, the one with smaller capacitance has less energy stored.
13(           ) For two capacitors having the same charge, the one with smaller capacitance needs more DV and has more energy stored.
14(           ) For two capacitors having the same voltage difference, the one with larger capacitance has less energy stored.
15(           ) For two capacitors having the same voltage difference, the one with larger capacitance has more Q and has more energy stored.
16(           ) If a capacitor is kept being connected with a battery, its voltage difference remains the same and to separate the two plates further will leads to a larger capacitance, hence the stored energy increases.
17(           ) If a capacitor is kept being connected with a battery, its voltage difference remains the same and to separate the two plates further will leads to a smaller capacitance, hence the stored energy reduces.
18(           ) If a capacitor is kept being connected with a battery, its voltage difference remains the same and to separate the two plates further will leads to a smaller E field in between (E=DV/d), that means less amount of charge density and total charge, hence the hence the stored energy decreases (It was partially discharged).
19(           ) If a charged capacitor is disconnected with any battery or circuit, its voltage difference remains the same and to separate the two plates further will leads to a smaller capacitance, hence the stored energy reduces.
20(           ) If a charged capacitor is disconnected with any battery or circuit, its charge Q remains the same, and to separate the two plates further will leads to a larger capacitance, hence the stored energy decreases.
21(           ) If a charged capacitor is disconnected with any battery or circuit, its charge Q remains the same, and to separate the two plates further will leads to a larger capacitance, hence the stored energy increases.
22(           ) If a charged capacitor is disconnected with any battery or circuit, its charge Q remains the same, and to separate the two plates further will leads to a smaller capacitance, hence the stored energy increases.
23(           ) If a charged capacitor is disconnected with any battery or circuit, its charge Q remains the same, hence the charge density and E field strength remains, and to separate the two plates further will leads to a large voltage difference (DV=Ed) , hence the stored energy increases.
24(           ) If a charged capacitor is disconnected with any battery or circuit, its charge Q remains the same, and to separate the two plates further will leads to a smaller capacitance and a larger DV, hence the stored energy increases. The extra energy must come from nowhere because it is disconnected with any battery or circuit.
25(           ) If a charged capacitor is disconnected with any battery or circuit, its charge Q remains the same, and to separate the two plates further will leads to a smaller capacitance and a larger DV, hence the stored energy increases. The extra energy must come from the external agent’s work. The positively and negatively charged plates attracts each other, to separate them further requires work done by the external agent, similar to stretching a spring.

26(           ) Energy density uE is the electrical energy per unit volume. It is U/volume. In the parallel plate capacitor case it is U/(Ad).
27(           ) Energy density uE has unit Joule.
28(           ) Energy density uE has unit J/m3.
29(           ) In the parallel plate capacitor, U= ½ QDV, the energy density is uE=U/(Ad) = ½ (Q/A) (DV/d)= ½ (s) (E).
30(           ) In the parallel plate capacitor, U= ½ QDV, the energy density is uE=U/(Ad) = ½ (Q/A) (DV/d)= ½ (s) (E). Recall E= s/e0, we have s= e0E. Hence, the energy density is uE= ½ s E = ½ e0E2
31(           ) In the parallel plate capacitor, U= ½ QDV, the energy density is uE=U/(Ad) = ½ (Q/A) (DV/d)= ½ (s) (E). Recall s= E/ e0. Hence, the energy density is uE= ½ s E = ½ E2 / e0
32(           ) The electrical energy density at any location is proportional to the E field strength there.
33(           ) The electrical energy density at any location is proportional to the E field square there.
34(           ) To double the E field strength will make the electrical energy density to double.
35(           ) To double the E field strength will increase the electrical energy density by a factor of four.

36(           ) When a charged capacitor is disconnected with battery, its Q remains unchanged. Now when a dielectric material is added in between the two plates, the E field in between is reduced by a factor or k, so does the voltage difference DV. Hence, the capacitance is C=Q/ DV is increased by a factor of k, where k is called the dielectric constant.
37(           ) When a charged capacitor is disconnected with battery, its Q remains unchanged. Now when a dielectric material is added in between the two plates, the capacitance is C=Q/ DV is reduced by a factor of k, where k is called the dielectric constant.
38(           ) When a charged capacitor is connected with battery, its DV remains unchanged. Now when a dielectric material is added in between the two plates, to maintain the actual E field inside the dielectric unchanged (E= DV/d), the battery needs to add more charges to the conducting plates, hence, the capacitance is C=Q/ DV is increased by a factor of k, where k is called the dielectric constant. (see the bottom notation on page 791)
39(           ) k is the dielectric constant, which varies for different materials. Vacuum and dry air has k about equal to 1, and other materials have k >1.
40(           ) k is the dielectric constant, which varies for different materials. Vacuum and dry air has k about equal to 1, and other materials have k <1.
41(           ) Without dielectric C=e0A/d. In order to get a large capacitance, we need either a large Area or a very small distance d.
42(           ) Without dielectric C=e0A/d. In order to get a large capacitance, we need either a small Area or a very large distance d.
43(           ) With dielectric, C=ke0A/d, we can use a smaller area to obtain the same value of capacitance, because k>1.

44(           ) When the E field in the capacitor is more than the breaking strength of the materials in between, the material in between the capacitor will start conducting, destroying the capacitor.
45(           ) Table 26.1 says that the breaking strength of dry air is 3 x 106 V/m. That means if I have a capacitor filled with dry air and d=0.001m =1mm, the maximum voltage that can be applied to this capacitor is 3000 V, above which the capacitor is destroyed.
46(           ) Table 26.1 says that the breaking strength of Teflon is 60 x 106 V/m. That means if I have a capacitor filled with Teflon and d=0.001m =1mm, the maximum voltage that can be applied to this capacitor is 60 x 106 V, below which the capacitor is safe.
47(           ) Table 26.1 says that the breaking strength of Teflon is 60 x 106 V/m. The breaking strength of dry air is 3 x 106 V/m. The dielectric constant of Teflon is 2.1. That of dry air is about 1.00. That means if I have a capacitor filled with Teflon instead of dry air, the maximum voltage it can handle will increase by a factor of 20.
48(           ) Table 26.1 says that the breaking strength of Teflon is 60 x 106 V/m. The breaking strength of dry air is 3 x 106 V/m. The dielectric constant of Teflon is 2.1. That of dry air is about 1.00. That means if I have a capacitor filled with Teflon instead of dry air, the capacitance will increase by a factor of 20.
49(           ) Table 26.1 says that the breaking strength of Teflon is 60 x 106 V/m. The breaking strength of dry air is 3 x 106 V/m. The dielectric constant of Teflon is 2.1. That of dry air is about 1.00. That means if I have a capacitor filled with Teflon instead of dry air, the maximum voltage it can handle will increase by a factor of 2.1.
50(           ) Table 26.1 says that the breaking strength of Teflon is 60 x 106 V/m. The breaking strength of dry air is 3 x 106 V/m. The dielectric constant of Teflon is 2.1. That of dry air is about 1.00. That means if I have a capacitor filled with Teflon instead of dry air, the capacitance will increase by a factor of 2.1.

51(           ) Table 26.1 says that the breaking strength of Teflon is 60 x 106 V/m. The breaking strength of dry air is 3 x 106 V/m. Table 26.1 says that the dielectric constant of Teflon is 2.1. That of dry air is about 1.00. That means if I have a capacitor filled with Teflon instead of dry air, the maximum voltage it can handle will increase by a factor of 20.
52(           ) Table 26.1 says that the dielectric constant of Teflon is 2.1. That of dry air is about 1.00. That means if I have a capacitor filled with Teflon instead of dry air, the maximum voltage it can handle will increase by a factor of 20.

53(           ) Adding dielectric into the capacitor will increase the capacitance. U=½ Q2/C, if Q is unchanged, the total stored U will increases.
54(           ) Adding dielectric into the capacitor will increase the capacitance. U=½ Q2/C, if Q is unchanged, the total stored U will decrease.
55(           ) Adding dielectric into the capacitor will increase the capacitance. U=½ C DV2, if DV is unchanged, the total stored U will increases.
56(           ) Adding dielectric into the capacitor will increase the capacitance. U=½ C DV2, if DV is unchanged, the total stored U will decrease.
57(           ) Adding dielectric into the capacitor will decrease the capacitance. U=½ Q2/C, if Q is unchanged, the total stored U will increases.
58(           ) Adding dielectric into the capacitor will decrease the capacitance. U=½ C DV2, if DV is unchanged, the total stored U will decreases.
Bonus question:
59(           ) With dielectric, for fixed values of E and DV=Ed, the energy density uE=U/(Ad) = ½ C DV2/(Ad). Because C is increased by a factor of k, while DV, A and d are not changed, uE will increase by a factor of k. Hence uE= ½ ke0 E2, instead of ½ e0 E2. If we keep DV and E unchanged, adding dielectric will increase the stored potential energy U, and the energy per unit volume.

60. You have three capacitors and a battery. In which of the following combinations of the three capacitors is the maximum possible energy stored when the combination is attached to the battery? (a) series (b) parallel (c) no difference because both combinations store the same amount of energy

Explanation / Answer

1) Both charges and electrical energy can be stored inside a capacitor. Hence the statement is false

2) This statement is true

3) It is not safe to open the case of a home theater system and touch inside even if the system is unplugged because capacitors inside may have been charged, which can give shock. Hence the statement is false

4) Above explanation tells the statement is true

5)True because energy stored in a capacitor is (1/2)CV^2 and energy stored in a spring is (1/2)Kx^2

6)When a spring’s length is stretched up to Dx, Dx is the maximum extension and kDx is the maximum force. It is not a constant force and the work is not equal to F times distance. If we integrate F dx for any small amount of dx at any location, the total work from F= 0 to F= kDx is the area of a triangle in the F vs. x plot, that is ½ k Dx2. Hence, the energy stored in the spring is ½ k Dx2. Hence the statement is false

7) above explanation says that this statement is true

8)

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