Question 1

A parallel-plate capacitor is connected to a battery and becomes fully charged. The capacitor is then disconnected, and the separation between the plates is increased in such a way that no charge leaks off. As the plates are being separated, the energy stored in this capacitor

Accepted Answer

The energy stored in a capacitor is given by $U=\frac{1}{2}CV^2$  , where $C$ is the capacitance and $V$ is the voltage. As the plates of the capacitor are separated, the capacitance decreases, but the voltage across the capacitor remains the same. Therefore, the energy stored in the capacitor increases according to the formula $U=\frac{1}{2}C\left(\frac{Q}{C}\right)^2=\frac{1}{2}\frac{Q^2}{C}$.

Question 2

Which of the following changes will increase the capacitance of a parallel-plate capacitor? (There could be more than one correct choice.)

Accepted Answer

Introducing a dielectric material between the plates increases capacitance by a factor of the material's dielectric constant (κ). Decreasing the separation between the plates increases capacitance since the capacitance is inversely proportional to the distance between the plates. Increasing the charge on the plates (A) increases the potential difference, which can increase capacitance, but it is not directly related to capacitance. Decreasing the potential between the plates (B) and increasing the potential between the plates (C) are contradictory statements and do not make sense in relation to increasing capacitance.

Question 3

An ideal parallel-plate capacitor having circular plates of diameter D that are a distance d apart stores energy U when it is connected across a fixed potential difference. If you want to triple the amount of energy stored in this capacitor by changing only the size of its plates, the diameter should be changed to

Accepted Answer

The energy stored in a capacitor is proportional to the capacitance, which for a parallel-plate capacitor is proportional to the area of the plates. To triple the energy, the area must be tripled, which means the diameter should be increased by a factor of $\sqrt{3}$.

Question 4

At a distance d from a point charge Q, the energy density in its electric field is u. If we double the charge, what is the energy density at the same point?

Accepted Answer

The energy density in an electric field is proportional to the square of the electric field, which is proportional to the square of the charge. Doubling the charge quadruples the energy density.

Question 5

Which of the following will increase the capacitance of a parallel-plate capacitor? (There could be more than one correct choice.)

Accepted Answer

The capacitance of a parallel-plate capacitor is directly proportional to the area of the plates and inversely proportional to the distance between them. Therefore, increasing the plate area and decreasing the plate separation will increase the capacitance. Option A is incorrect as it suggests a decrease in the plate area, which will actually decrease the capacitance. Option B is incorrect as the potential difference between the plates does not affect the capacitance. Option C is incorrect as an increase in potential difference does not affect the capacitance but will increase the stored charge. Option E is incorrect as an increase in charge on the plates will not affect the capacitance but will increase the potential difference between the plates.

Question 6

An ideal parallel-plate capacitor consists of two parallel plates of area A separated by a distance d. This capacitor is connected to a battery that maintains a constant potential difference across the plates. If the separation between the plates is now doubled, the amount of electrical energy stored on the capacitor will

Accepted Answer

The electrical energy stored on a capacitor is given by the equation E = (1/2)CV^2, where C is the capacitance and V is the potential difference across the capacitor. The capacitance of a parallel-plate capacitor is given by C = εA/d, where ε is the permittivity of the material between the plates, A is the area of the plates, and d is the distance between the plates. Thus, if the separation between the plates is doubled, the capacitance is halved. Since the potential difference across the plates is held constant by the battery, the amount of electrical energy stored on the capacitor is proportional to the capacitance, and will also be halved. Therefore, the correct answer is C) be cut in half.

Question 7

A parallel-plate capacitor consists of a set of two parallel plates of area A separated by a distance d. This capacitor is connected to a battery that maintains a constant potential difference across the plates. A slab of a dielectric material is inserted in the region between the plates and completely fills it. What changes would you observe as the dielectric is inserted? (There could be more than one correct choice.)

Accepted Answer

When a dielectric material is inserted between the plates of a capacitor, the electric field between the plates is reduced. This results in an increase in the capacitance of the capacitor. The amount of charge stored on the plates of the capacitor also increases, as the potential difference across the plates remains constant but the capacitance increases. Therefore, both the charge on the plates of the capacitor and its capacitance would change. Choice D is incorrect because the potential difference across the plates remains constant. Choice E is incorrect because inserting a dielectric material between the plates of a capacitor does cause changes, as explained above.

Question 8

Doubling the potential across a given capacitor causes the energy stored in that capacitor to

Accepted Answer

The energy stored in a capacitor is given by the formula E= 1/2 * C * V^2, where C is the capacitance and V is the potential across the capacitor. Doubling the potential (V) will result in the energy stored (E) increasing by a factor of 4 (2^2), as the potential term is squared in the formula. Therefore, the correct choice is A.

Question 9

A battery charges a parallel-plate capacitor fully and then is removed. The plates are then slowly pulled apart. What happens to the potential difference between the plates as they are being separated?

Accepted Answer

When the plates of a capacitor are pulled apart, the capacitance decreases because capacitance is inversely proportional to the distance between the plates. Since the charge remains constant (the battery is removed, so no charge can flow in or out), and the potential difference (V) is given by V = Q/C (where Q is the charge and C is the capacitance), the potential difference must increase as the capacitance decreases.

Question 10

The plates of a parallel-plate capacitor are maintained with constant voltage by a battery as they are pulled apart. What happens to the strength of the electric field between the plates during this process?

Accepted Answer

As the plates are pulled apart, the distance between them increases, and the capacitance of the capacitor decreases. As the voltage remains constant, according to the equation C=Q/V, the charge on the plates also remains constant. Therefore, as the capacitance decreases, the electric field strength E=V/d between the plates must also decrease.

Question 11

An ideal parallel-plate capacitor consists of two parallel plates of area A separated by a distance d. This capacitor is connected to a battery and charged until its plates carry charges +Q and -Q, and the battery is then disconnected. If the separation between the plates is now doubled, the electrical energy stored in the capacitor will

Accepted Answer

The answer of An ideal parallel-plate capacitor consists of two...

Question 12

When a dielectric material is introduced between the plates of a parallel-plate capacitor and completely fills the space, the capacitance increases by a factor of 4. What is the dielectric constant of the material that was introduced?

Accepted Answer

The capacitance of a parallel-plate capacitor is given by $C=\frac{\epsilon A}{d}$, where $\epsilon$ is the permittivity of the medium between the plates, $A$ is the area of the plates, and $d$ is the separation between the plates. When a dielectric material is introduced, the capacitance becomes $C'=\frac{\epsilon_r \epsilon A}{d}$, where $\epsilon_r$ is the dielectric constant of the material. We are told that the capacitance increases by a factor of 4, so $C'=4C$. Substituting in the expressions for $C$ and $C'$, we get $\frac{\epsilon_r \epsilon A}{d}=4\frac{\epsilon A}{d}$, which simplifies to $\epsilon_r=4$. Therefore, the dielectric constant of the material is $\boxed{	extbf{(D) }4}$.

Question 13

Two ideal parallel-plate capacitors are identical in every respect except that one has twice the plate area of the other. If the smaller capacitor has capacitance C, the larger one has capacitance

Accepted Answer

The capacitance of a parallel-plate capacitor is directly proportional to the plate area, so if one capacitor has twice the plate area of another, it will have twice the capacitance. Therefore, the larger capacitor in this scenario will have a capacitance of 2C.

Question 14

A dielectric material such as paper is inserted between the plates of a capacitor as the capacitor holds a fixed charge on its plates. What happens to the electric field between the plates as the dielectric is inserted?

Accepted Answer

When a dielectric material is inserted between the plates of a capacitor, the electric field between the plates will become weaker. This is because the dielectric material reduces the electric field strength by decreasing the potential difference between the plates, while the charge remains the same. The amount of weakening depends on the permittivity of the dielectric material.

Question 15

When a certain capacitor carries charges of ±10 µC on its plates, the potential difference cross the plates is 25 V. Which of the following statements about this capacitor are true? (There could be more than one correct choice.)

Accepted Answer

Capacitance is defined as the charge stored per unit potential difference, $C = \frac{Q}{V}$. Doubling the charge while keeping the capacitance constant will double the potential difference (B is correct). The capacitance of a capacitor is determined by its physical characteristics (geometry and dielectric material) and does not change with the amount of charge or the potential difference (C is correct).

Question 16

The plates of a parallel-plate capacitor are maintained with constant potential by a battery as they are pulled apart. During this process, the amount of charge on the plates

Accepted Answer

As the plates are pulled apart, the distance between them increases, which reduces the capacitance of the capacitor. Since the voltage across the plates is maintained constant by the battery, the charge on the plates must decrease to maintain the constant potential difference. Therefore, the amount of charge on the plates must decrease during the process.

Question 17

An ideal parallel-plate capacitor has a capacitance of C. If the area of the plates is doubled and the distance between the plates is halved, what is the new capacitance?

Accepted Answer

The capacitance of a parallel-plate capacitor is given by $C = \epsilon_0 \frac{A}{d}$, where $A$ is the area of the plates and $d$ is the distance between them. Doubling the area and halving the distance results in a new capacitance of $C' = \epsilon_0 \frac{2A}{d/2} = 4\epsilon_0 \frac{A}{d} = 4C$.

Question 18

An capacitor consists of two large parallel plates of area A separated by a very small distance d. This capacitor is connected to a battery and charged until its plates carry charges +Q and -Q, and then disconnected from the battery. If the separation between the plates is now doubled, the potential difference between the plates will

Accepted Answer

The potential difference (V) between the plates of a capacitor is directly proportional to the separation (d) between the plates when the charge (Q) on the plates is kept constant. Therefore, if the separation is doubled, the potential difference will also double.

Question 19

An ideal parallel-plate capacitor consists of two parallel plates of area A separated by a distance d. This capacitor is connected across a battery that maintains a constant potential difference between the plates. If the separation between the plates is now doubled, the magnitude of the charge on the plates will

Accepted Answer

The capacitance of a parallel-plate capacitor is given by C = ε0A/d, where ε0 is the permittivity of free space. If the separation between the plates is doubled, then the capacitance is halved. Since the potential difference across the plates is constant, the charge on the plates is given by Q = CV. If C is halved, then Q must also be halved. Therefore, the magnitude of the charge on the plates is cut in half.

Question 20

Doubling the capacitance of a capacitor that is holding a constant charge causes the energy stored in that capacitor to

Accepted Answer

The energy stored in a capacitor is given by E=1/2CV^2, where C is the capacitance and V is the voltage. If the capacitance is doubled while holding the charge constant, the voltage across the capacitor will be halved. Therefore, the energy stored will decrease to one-half.

Quiz 21: Electric Potential