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 the formula $U = \frac{1}{2}CV^2$, where $U$ is the energy, $C$ is the capacitance, and $V$ is the voltage across the capacitor. When the capacitor is disconnected from the battery and the plate separation is increased, the capacitance $C = \frac{\varepsilon_0A}{d}$ decreases because $d$ (the distance between the plates) increases, and $A$ (the area of the plates) and $\varepsilon_0$ (the permittivity of free space) remain constant. Since the charge $Q$ remains constant (no charge leaks off), and $V = \frac{Q}{C}$, the voltage across the capacitor increases as $C$ decreases. Therefore, because the energy $U$ depends on the square of the voltage ($V^2$), the energy stored in the capacitor increases as the separation between the plates increases.

Question 2

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 charge and C is capacitance), the potential difference increases as the capacitance decreases.

Question 3

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

The capacitance of a capacitor is defined as the charge per unit voltage (C = Q/V), so if the charge is doubled and the potential difference also doubles (as per B), the capacitance remains constant (as per C).

Question 4

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 5

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 across the capacitor. Since the charge on the capacitor remains constant, doubling the capacitance means that the voltage across the capacitor will halve (as Q=CV, where Q is the charge). Thus, the energy stored in the capacitor will decrease to half of its original value (since V^2 is halved).

Question 6

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 $C = \frac{\varepsilon_0 A}{d}$, 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 $C' = \frac{\varepsilon_0 (2A)}{(d/2)} = 4 \times \frac{\varepsilon_0 A}{d} = 4C$.

Question 7

The potential (relative to infinity)at the midpoint of a square is 3.0 V when a point charge of +Q is located at one of the corners of the square.What is the potential (relative to infinity)at the center when each of the other corners is also contains a point charge of +Q?

Accepted Answer

Since the potential at the midpoint of the square is proportional to the charge (V ∝ Q) and the charge at each corner is the same, the total charge at the four corners is 4Q. 
When there is only one charge at one corner, the potential at the midpoint of a square is 3.0 V. 
When there is one charge Q at each corner, the potential at the midpoint due to each charge is the same and they all contribute to the total potential at the center. 
Using the formula V = kQ/r, the potential due to one charge at a distance r from the center is V = kQ/r. Since the diagonal of the square is twice the length of the side, the distance from the corner charges to the center is r = d/2 = s√2/2. Thus, the potential at the midpoint of the square due to one of the corner charges is:

V1 = kQ/(s√2/2) = kQs√2/2

The potential at the center of the square due to all four corner charges is:

Vtotal = 4V1 = 4kQs√2/2 = 2kQs√2

where k is the Coulomb constant.

So the potential at the center of the square is proportional to Q and s√2. Since the charge and the distance are the same as in the one-charge case, the potential at the center due to all four corner charges is:

Vtotal = 4V1 = 4(3.0 V) = 12 V

Question 8

The electric potential at a distance of 4 m from a certain point charge is 200 V relative to infinity.What is the potential (relative to infinity)at a distance of 2 m from the same charge?

Accepted Answer

The electric potential $V$ due to a point charge varies inversely with the distance $r$ from the charge, according to the formula $V = k\frac{Q}{r}$, where $k$ is a constant and $Q$ is the charge. If the potential at 4 m is 200 V, then halving the distance to 2 m doubles the potential to 400 V, because the relationship between potential and distance is directly inverse.

Question 9

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. Therefore, if one capacitor has twice the plate area of another, its capacitance will also be twice as much. Therefore, the larger capacitor has capacitance 2C.

Question 10

A 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 given by V = Q/C, where Q is the charge on the plates and C is the capacitance. The capacitance of a parallel plate capacitor is given by C = ε₀A/d, where ε₀ is the permittivity of free space, A is the area of the plates, and d is the separation between the plates. When the separation (d) is doubled, the new capacitance (C') becomes ε₀A/(2d), which is half of the original capacitance. Since the charge (Q) remains the same (because the capacitor is disconnected from the battery), the new potential difference (V') is twice the original potential difference (V), because V' = Q/C' and C' is half of C.

Question 11

If the $\vec { E }$ field between the plates of a given air-filled capacitor is weakened by removing charge from the plates,the capacitance of that capacitor

Accepted Answer

The capacitance of a capacitor is determined by its physical characteristics (area of the plates, distance between the plates, and the dielectric material between the plates) and not by the amount of charge on the plates or the electric field strength.

Question 12

Which statements must be true about the surface of a charged conductor in which no charge is moving? (There could be more than one correct choice.)

Accepted Answer

The electric potential is constant over the surface of a conductor in electrostatic equilibrium, and the electric field is perpendicular to the surface.

Question 13

If the result of your calculations for a quantity has SI units of C² ∙ s²/(kg ∙ m²),that quantity could be

Accepted Answer

The answer of If the result of your calculations for...

Question 14

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 magnitude of charge on the plates of a capacitor is directly proportional to the potential difference between the plates and inversely proportional to the distance between them. Since the potential difference across the plates is constant, doubling the separation distance will result in halving the magnitude of charge on the plates. Therefore, the correct answer is C) be cut in half.

Question 15

The plates of a parallel-plate capacitor are maintained with constant voltage by a battery as they are pulled apart.What happens to the magnitude of the $\vec { E }$ field between the plates during this process?

Accepted Answer

As the plates are pulled apart, the distance between them increases, resulting in a decrease in capacitance. Since the voltage across the plates is held constant, the charge on the plates remains the same. According to the equation $C=\dfrac{Q}{V}$, a decrease in capacitance results in an increase in electric field magnitude for a given charge on the plates. Therefore, the magnitude of the electric field between the plates increases as they are pulled apart.

Question 16

A region of space contains a uniform electric field,directed toward the right,as shown in the figure.Which statement about this situation is correct?

Accepted Answer

The answer of A region of space contains a uniform...

Question 17

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

Accepted Answer

The energy stored in a capacitor is proportional to the square of the potential difference, so doubling the potential difference will result in the energy being quadrupled (2^2=4). Therefore, the correct choice is A.

Question 18

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 plate area and inversely proportional to the plate separation. Therefore, increasing the plate area and decreasing the plate separation will increase the capacitance.

Question 19

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 parallel-plate capacitor is given by:
$$ U = \frac{1}{2}\epsilon_0 \frac{\pi}{4} D^2 \frac{V^2}{d} $$
where $\epsilon_0$ is the permittivity of free space, $D$ is the diameter of the plates, $d$ is the distance between the plates, and $V$ is the potential difference between the plates.

To triple the energy stored, we need to increase $U$ by a factor of 3. Therefore, we can write:
$$ \frac{1}{2}\epsilon_0 \frac{\pi}{4} D^2 \frac{(3V)^2}{d} = 3U $$
Simplifying and solving for $D$, we get:
$$ D = \sqrt{3} \sqrt[4]{\frac{24U d}{\pi \epsilon_0 V^2}} $$
We want to change only the diameter of the plates, so $d$ and $V$ must be constant. Therefore, we can see that the only option that satisfies this condition is choice C), where $D = D\sqrt{3}$.

Question 20

Four charged particles (two having a charge +Q and two having a charge -Q)are arranged in the xy-plane,as shown in the figure.These particles are all equidistant from the origin.The electric potential (relative to infinity)at point P on the z-axis due to these particles,is

Accepted Answer

The answer of Four charged particles (two having a charge...

Quiz 16: The Electric Field

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

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?

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.)

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

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

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?

The potential (relative to infinity) at the midpoint of a square is 3.0 V when a point charge of +Q is located at one of the corners of the square.What is the potential (relative to infinity) at the center when each of the other corners is also contains a point charge of +Q?

The electric potential at a distance of 4 m from a certain point charge is 200 V relative to infinity.What is the potential (relative to infinity) at a distance of 2 m from the same charge?

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

If the $\vec { E }$ field between the plates of a given air-filled capacitor is weakened by removing charge from the plates,the capacitance of that capacitor

Which statements must be true about the surface of a charged conductor in which no charge is moving? (There could be more than one correct choice.)

If the result of your calculations for a quantity has SI units of C² ∙ s²/(kg ∙ m²) ,that quantity could be

The plates of a parallel-plate capacitor are maintained with constant voltage by a battery as they are pulled apart.What happens to the magnitude of the $\vec { E }$ field between the plates during this process?

A region of space contains a uniform electric field,directed toward the right,as shown in the figure.Which statement about this situation is correct?

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

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

Four charged particles (two having a charge +Q and two having a charge -Q) are arranged in the xy-plane,as shown in the figure.These particles are all equidistant from the origin.The electric potential (relative to infinity) at point P on the z-axis due to these particles,is

Quiz 16: The Electric Field

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

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?

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.)

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

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

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?

The potential (relative to infinity) at the midpoint of a square is 3.0 V when a point charge of +Q is located at one of the corners of the square.What is the potential (relative to infinity) at the center when each of the other corners is also contains a point charge of +Q?

The electric potential at a distance of 4 m from a certain point charge is 200 V relative to infinity.What is the potential (relative to infinity) at a distance of 2 m from the same charge?

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

If the E⃗\vec { E }E field between the plates of a given air-filled capacitor is weakened by removing charge from the plates,the capacitance of that capacitor

Which statements must be true about the surface of a charged conductor in which no charge is moving? (There could be more than one correct choice.)

If the result of your calculations for a quantity has SI units of C2 ∙ s2/(kg ∙ m2) ,that quantity could be

The plates of a parallel-plate capacitor are maintained with constant voltage by a battery as they are pulled apart.What happens to the magnitude of the E⃗\vec { E }E field between the plates during this process?

A region of space contains a uniform electric field,directed toward the right,as shown in the figure.Which statement about this situation is correct?

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

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

Four charged particles (two having a charge +Q and two having a charge -Q) are arranged in the xy-plane,as shown in the figure.These particles are all equidistant from the origin.The electric potential (relative to infinity) at point P on the z-axis due to these particles,is

If the $\vec { E }$ field between the plates of a given air-filled capacitor is weakened by removing charge from the plates,the capacitance of that capacitor

If the result of your calculations for a quantity has SI units of C² ∙ s²/(kg ∙ m²) ,that quantity could be

The plates of a parallel-plate capacitor are maintained with constant voltage by a battery as they are pulled apart.What happens to the magnitude of the $\vec { E }$ field between the plates during this process?