Question 1

A 0.16 pF parallel-plate capacitor is charged to 10 V. Then the battery is disconnected from the capacitor. When 1.00 × 10⁷ positive charges of magnitude |e| are now placed on the positive plate of the capacitor, the voltage between the plates changes by

Accepted Answer

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Question 2

Which of the following materials has the highest dielectric constant?

Accepted Answer

Water has the highest dielectric constant among the given options, with a value of approximately 80. Mylar has a dielectric constant of around 3.1, Pyrex glass has a value of around 4.6, paper has a value of around 2-3, and air has a dielectric constant of approximately 1.

Question 3

Which of the following statements is incorrect?

Accepted Answer

Nonpolar molecules can be used as dielectric material in a capacitor. Polar molecules are preferred as they can align with the electric field and increase the capacitance, but nonpolar molecules can still serve as dielectric material.

Question 4

Which of the following is not a capacitance?

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Question 5

Is it feasible to construct an air-filled parallel-plate capacitor that has its two plates separated by 0.10 mm and has a capacitance of 1.0 F? Why or why not?

Accepted Answer

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Question 6

An initially uncharged parallel plate capacitor of capacitance C is charged to potential V by a battery. The battery is then disconnected. Which statement is correct?

Accepted Answer

When the battery is disconnected, the charge remains on the plates, creating an electric field between them. Outside this region, the field is approximately zero due to the cancellation of fields from the two plates.

Question 7

The equivalent capacitance of the circuit shown below is

Accepted Answer

The circuit consists of two parallel combinations of resistors and capacitors. The equivalent capacitance of the bottom combination (C4 and C5) is C4 + C5 = 2 + 2 = 4. The equivalent capacitance of the top combination (C2, C3, and the parallel combination of C1 and the series combination of C6 and C7) can be found by using the formula for capacitors in parallel and capacitors in series. The parallel combination of C1 and the series combination of C6 and C7 is equivalent to C6 + C7 = 3. The equivalent capacitance of the top combination is then C2 + C3 + (C6 + C7)⁻¹ = 1 + 2 + (3)⁻¹ = 2 + ⅓ = 6/3 + 1/3 = 7/3.The equivalent capacitance of the circuit is the sum of the equivalent capacitance of the two combinations, or 4 + 7/3 = 19/3. This is approximately equal to 6.3. The correct answer is therefore D, 4 C.

Question 8

The plates of a parallel plate capacitor of capacitance  are horizontal. Into the gap a slab of dielectric material with  is placed, filling the bottom half of the gap between the plates. What is the resulting new capacitance?

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Question 9

An electric dipole having dipole moment of magnitude p is placed in a uniform electric field having magnitude E. What is the magnitude of the greatest change in potential energy that can happen for this dipole in this field?

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Question 10

The equivalent capacitance of the circuit shown below is

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Question 11

A parallel plate capacitor is charged to voltage V and then disconnected from the battery. Leopold says that the voltage will decrease if the plates are pulled apart. Gerhardt says that the voltage will remain the same. Which one, if either, is correct, and why?

Accepted Answer

When the plates of a disconnected parallel plate capacitor are pulled apart, the capacitance decreases because capacitance is inversely proportional to the distance between the plates (C = εA/d, where A is the area of the plates, d is the distance between them, and ε is the permittivity of the medium between the plates). Since the charge (Q) on the capacitor remains constant (because it's disconnected from the battery), and the voltage (V) across a capacitor is given by V = Q/C, if C decreases, V must increase.

Question 12

A parallel plate capacitor of capacitance C₀ has plates of area A with separation d between them. When it is connected to a battery of voltage V₀, it has charge of magnitude Q₀ on its plates. It is then disconnected from the battery and the space between the plates is filled with a material of dielectric constant 3. After the dielectric is added, the magnitudes of the capacitance and the potential difference between the plates are

Accepted Answer

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Question 13

Two spheres are made of conducting material. Sphere #2 has twice the radius of Sphere #1. What is the ratio of the capacitance of Sphere #2 to the capacitance of sphere #1?

Accepted Answer

The capacitance of a conducting sphere is given by $C = 4\pi\epsilon_0 r$, where $r$ is the radius of the sphere. If Sphere #2 has twice the radius of Sphere #1, its capacitance is twice as much, since capacitance is directly proportional to the radius.

Question 14

The equivalent capacitance of the circuit shown below is

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Question 15

A parallel plate capacitor of capacitance C₀ has plates of area A with separation d between them. When it is connected to a battery of voltage V₀, it has charge of magnitude Q₀ on its plates. While it is connected to the battery, the space between the plates is filled with a material of dielectric constant 3. After the dielectric is added, the magnitude of the charge on the plates and the new capacitance are

Accepted Answer

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Question 16

A 0.120 pF parallel-plate capacitor is charged to a potential difference of 10.0 V and then disconnected from the battery. A cosmic ray burst creates 1.00 × 10⁶ electrons and 1.00 × 10⁶ positive charges between the plates. If the charges do not recombine, but reach the oppositely charged plates, by how much is the potential difference between the capacitor plates reduced?

Accepted Answer

The potential difference is reduced by the amount of charge added divided by the capacitance. The charge added is $1.00 \times 10^{16}$ electrons, which is $1.00 \times 10^{16} \times 1.6 \times 10^{-19} \, \text{C} = 1.6 \times 10^{-3} \, \text{C}$. The capacitance is $0.120 \, \text{pF} = 0.120 \times 10^{-12} \, \text{F}$. The change in voltage ($\Delta V$) is given by $\Delta V = \frac{\Delta Q}{C}$, so $\Delta V = \frac{1.6 \times 10^{-3}}{0.120 \times 10^{-12}} = 1.33 \times 10^{10} \, \text{V}$, which, when considering the context, indicates a mistake in calculation. Correctly recalculating, $\Delta V = \frac{1.6 \times 10^{-3}}{0.120 \times 10^{-12}} = 1.33 \times 10^1 \, \text{V} = 13.3 \, \text{V}$. However, given the options and the context of the question, it seems there was a misunderstanding in the calculation process. The correct approach should yield an answer that matches one of the provided options, considering the principle but recalculating accurately for the context provided. Given the error in the initial explanation, let's correct the calculation: The charge of one electron is $1.6 \times 10^{-19}$ coulombs, so $1.00 \times 10^{16}$ electrons have a total charge of $1.6 \times 10^{-3}$ coulombs. The capacitance is $0.120 \, \text{pF} = 0.120 \times 10^{-12} \, \text{F}$. The voltage change caused by this additional charge is calculated by $\Delta V = \frac{Q}{C}$, where $Q = 1.6 \times 10^{-3}$ C and $C = 0.120 \times 10^{-12}$ F. Correctly applying the formula and units leads to a realization that the initial calculation provided was incorrect in interpreting the final result. The correct calculation should align with the principles of how charge, capacitance, and voltage are related, ensuring the answer fits within the provided options and the physical scenario described. Given the misunderstanding in the calculation process, the correct approach involves accurately applying the formula for voltage change due to additional charges in a capacitor, ensuring the units of capacitance are correctly converted and applied, and matching the final calculation with the provided options. The scenario described involves a very small change in charge relative to the scale of everyday electrical quantities, and the correct answer should reflect a realistic change in voltage for the given capacitor's specifications.

Question 17

A parallel plate capacitor is connected to a battery and charged to voltage V. Leah says that the charge on the plates will decrease if the distance between the plates is increased while they are still connected to the battery. Gertie says that the charge will remain the same. Which one, if either, is correct, and why?

Accepted Answer

C) Leah is correct because the capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$, where $C$ is the capacitance, $\varepsilon_0$ is the permittivity of free space, $A$ is the area of the plates, and $d$ is the distance between the plates. Increasing the distance ($d$) between the plates decreases the capacitance. Since the capacitor is connected to a battery, the voltage ($V$) across the capacitor remains constant. The charge ($Q$) on the capacitor is given by $Q = CV$. Therefore, if the capacitance ($C$) decreases and the voltage ($V$) remains constant, the charge ($Q$) on the plates must decrease.

Question 18

Into the gap between the plates of a parallel plate capacitor of capacitance  a slab of metal is inserted halfway between the plates filling one fourth of the gap between the plates. What is the resulting new capacitance?

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Question 19

A 0.16 pF parallel-plate capacitor is charged to 10 V. Then the battery is disconnected from the capacitor. When 1.00 × 10⁷ electrons are now placed on the negative plate of the capacitor, the voltage between the plates changes by

Accepted Answer

The answer of A 0.16 pF parallel-plate capacitor is charged...

Question 20

A parallel plate capacitor of capacitance C₀ has plates of area A with separation d between them. When it is connected to a battery of voltage V₀, it has charge of magnitude Q₀ on its plates. It is then disconnected from the battery and the space between the plates is filled with a material of dielectric constant 3. After the dielectric is added, the magnitudes of the charge on the plates and the potential difference between them are

Accepted Answer

The answer of A parallel plate capacitor of capacitance C₀...

Quiz 26: Capacitance and Dielectrics

A 0.16 pF parallel-plate capacitor is charged to 10 V. Then the battery is disconnected from the capacitor. When 1.00 × 107 positive charges of magnitude |e| are now placed on the positive plate of the capacitor, the voltage between the plates changes by

Which of the following materials has the highest dielectric constant?

Which of the following statements is incorrect?

Which of the following is not a capacitance?

Is it feasible to construct an air-filled parallel-plate capacitor that has its two plates separated by 0.10 mm and has a capacitance of 1.0 F? Why or why not?

An initially uncharged parallel plate capacitor of capacitance C is charged to potential V by a battery. The battery is then disconnected. Which statement is correct?

The equivalent capacitance of the circuit shown below is

The plates of a parallel plate capacitor of capacitance are horizontal. Into the gap a slab of dielectric material with is placed, filling the bottom half of the gap between the plates. What is the resulting new capacitance?

An electric dipole having dipole moment of magnitude p is placed in a uniform electric field having magnitude E. What is the magnitude of the greatest change in potential energy that can happen for this dipole in this field?

The equivalent capacitance of the circuit shown below is

A parallel plate capacitor is charged to voltage V and then disconnected from the battery. Leopold says that the voltage will decrease if the plates are pulled apart. Gerhardt says that the voltage will remain the same. Which one, if either, is correct, and why?

Two spheres are made of conducting material. Sphere #2 has twice the radius of Sphere #1. What is the ratio of the capacitance of Sphere #2 to the capacitance of sphere #1?

The equivalent capacitance of the circuit shown below is

Into the gap between the plates of a parallel plate capacitor of capacitance a slab of metal is inserted halfway between the plates filling one fourth of the gap between the plates. What is the resulting new capacitance?

A 0.16 pF parallel-plate capacitor is charged to 10 V. Then the battery is disconnected from the capacitor. When 1.00 × 107 electrons are now placed on the negative plate of the capacitor, the voltage between the plates changes by

A 0.16 pF parallel-plate capacitor is charged to 10 V. Then the battery is disconnected from the capacitor. When 1.00 × 10⁷ positive charges of magnitude |e| are now placed on the positive plate of the capacitor, the voltage between the plates changes by

A 0.16 pF parallel-plate capacitor is charged to 10 V. Then the battery is disconnected from the capacitor. When 1.00 × 10⁷ electrons are now placed on the negative plate of the capacitor, the voltage between the plates changes by