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Solutions

MCQ

1. A

Here, you must recall that the formula for the capacitance of a parallel plate capacitor goes like 1/d. When d gets larger, C gets smaller by the same factor. Thus you know right away that the capacitance goes to half of its initial value. The potential across the plates remains constant because the capacitor is connected to a battery. This means that if the capacitance changes, the charge on the plates must change as well. From Q = CV, you can see that the charge goes like the capacitance. So, Q must also be decreased by one-half.

2. D

Potential is a linear function between capacitor plates. Point P is one-fourth of the way from the bottom plate, at 2 V, to the top plate, at 10 V. There is a voltage of 8 V difference between the plates, so P would be one-fourth of this difference, or 2 V, of the way from the starting potential of 2 V. That is, P will be at 4 V.

3. C

Simply apply the formula V = kQ/r = 9*10^9 N*m^2/C^2 * 5*10^-8 C/ 0.50 m = 9*10^2 V

4. A

According to V = kQ/r, the smaller r is, the bigger the potential. Besides, potential is a scalar, so the we simply look at which summation of potentials yields the largest result. D and E are the same, but the distance is greater than A, so the potential at those points is smaller. B and C are farther than A, so the potential at those points is also smaller.

5. C

When you see this question, you should immediately think of the equation C = e A/d, instead of Q=CV. Recall that the first equation characterizes the capacitance of a capacitor, while the latter one is simply a linear relationship between Q and V, with the CONSTANT ratio C. Thus, b and d cannot be the right answer. As we can see from the first equation, C varies directly with A. So the answer is C, increasing the area.

6. A

Electron volt is defined as the energy acquired by a particle carrying a charge whose magnitude equals that on the electron as a result of moving through a potential difference of 1V.

MCQ

1. A

Here, you must recall that the formula for the capacitance of a parallel plate capacitor goes like 1/d. When d gets larger, C gets smaller by the same factor. Thus you know right away that the capacitance goes to half of its initial value. The potential across the plates remains constant because the capacitor is connected to a battery. This means that if the capacitance changes, the charge on the plates must change as well. From Q = CV, you can see that the charge goes like the capacitance. So, Q must also be decreased by one-half.

2. D

Potential is a linear function between capacitor plates. Point P is one-fourth of the way from the bottom plate, at 2 V, to the top plate, at 10 V. There is a voltage of 8 V difference between the plates, so P would be one-fourth of this difference, or 2 V, of the way from the starting potential of 2 V. That is, P will be at 4 V.

3. C

Simply apply the formula V = kQ/r = 9*10^9 N*m^2/C^2 * 5*10^-8 C/ 0.50 m = 9*10^2 V

4. A

According to V = kQ/r, the smaller r is, the bigger the potential. Besides, potential is a scalar, so the we simply look at which summation of potentials yields the largest result. D and E are the same, but the distance is greater than A, so the potential at those points is smaller. B and C are farther than A, so the potential at those points is also smaller.

5. C

When you see this question, you should immediately think of the equation C = e A/d, instead of Q=CV. Recall that the first equation characterizes the capacitance of a capacitor, while the latter one is simply a linear relationship between Q and V, with the CONSTANT ratio C. Thus, b and d cannot be the right answer. As we can see from the first equation, C varies directly with A. So the answer is C, increasing the area.

6. A

Electron volt is defined as the energy acquired by a particle carrying a charge whose magnitude equals that on the electron as a result of moving through a potential difference of 1V.