Question 1

The equivalent capacitance of the circuit shown below is

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Question 2

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

The capacitance of a parallel-plate capacitor with a dielectric between its plates is given by $C = \kappa \epsilon_0 A/d$, where $\kappa$ is the dielectric constant, $\epsilon_0$ is the vacuum permittivity, $A$ is the area of the plates, and $d$ is the separation between the plates.

After the dielectric is added, the capacitance becomes $C' = 3C_0$.

Since the battery is ideal, the potential difference between the plates remains the same before and after the dielectric is added, i.e., $V' = V_0$.

Thus, the correct answer is (D) 3C0, 11eb8c7e_f590_6189_8371_e97dcabfcd5a_TB8847_11 V0.

Question 3

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 what?

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Question 4

Which of the following is not a capacitance?

Accepted Answer

All of the choices are given in the format of a hexadecimal UUID, which is a unique identifier. None of these UUIDs represent a capacitance value. However, option E has a different format than the others with an extra set of digits in the middle, making it the odd one out.

Question 5

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 charge on the capacitor before the burst is Q = CV = (0.120 pF)(10.0 V) = 1.20 × 10^-12 C. After the burst, the total charge on the capacitor is Q' = Q + 1.00 × 10^6e, where e is the magnitude of the charge on an electron. The electric field between the plates is then E = Q'/C = (Q + 1.00 × 10^6e)/(0.120 pF) = 8.33 × 10^7 V/m. The potential difference between the plates is then V' = Ed = (8.33 × 10^7 V/m)(d), where d is the distance between the plates. Since the charges do not recombine and reach the opposite plates, the potential difference is reduced by the work done by the charges moving through the electric field, which is W = QV' = (1.00 × 10^6)(-1.60 × 10^-19 C)(8.33 × 10^7 V/m)(0.001 m) = -1.33 V. Therefore, the new potential difference is 10.0 V - 1.33 V = 8.67 V.

Question 6

The equivalent capacitance of the circuit shown below is

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Question 7

The equivalent capacitance of the circuit shown below is

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Question 8

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

When a dielectric is added between the plates, the capacitance increases by a factor of the dielectric constant, i.e. C = C0*k, where k = 3. Therefore, Q = CV = C0*k*V0. Since the battery is disconnected, the charge on the plates remains constant, i.e. Q = Q0. Therefore, V = Q/C = Q0/(C0*k) = Q0/C, which is equal to V0/3. So, the correct answer is option B.

Question 9

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

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Question 10

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 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

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 directly proportional to its radius. Since Sphere #2 has twice the radius of Sphere #1, its capacitance is also twice that of Sphere #1.

Question 13

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 14

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?

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Question 15

Which of the following statements is incorrect?

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...

Question 16

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?

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Question 17

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 18

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

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Question 19

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 20

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.

Quiz 27: Capacitors and Batteries

The equivalent capacitance of the circuit shown below is

Which of the following is not a capacitance?

The equivalent capacitance of the circuit shown below is

The equivalent capacitance of the circuit shown below is

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?

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?

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?

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?

Which of the following statements is incorrect?

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?

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?

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

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?