Figure 25-18 shows plots of charge versus potential difference for three parallel-plate capacitors that have the plate areas and separations given in the table. Which plot goes with which capacitor? Figure 25-18 Question 1. Capacitor Area Separation 1 A d 2 2 A d 3 A 2d

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Answer 1

Textbook Question

Chapter 25, Problem 1Q

Figure 25-18 shows plots of charge versus potential difference for three parallel-plate capacitors that have the plate areas and separations given in the table. Which plot goes with which capacitor?

Chapter 25, Problem 1Q, Figure 25-18 shows plots of charge versus potential difference for three parallel-plate capacitors

Figure 25-18 Question 1.

Capacitor	Area	Separation
1	A	d
2	2A	d
3	A	2d

Expert Solution & Answer

To determine

To find:

Which plot goes with which capacitor.

Answer to Problem 1Q

Solution:

Plot a goes with the capacitor 2. The plot b goes with capacitor 1 and plot c goes with capacitor 3.

Explanation of Solution

1) Concept:

Using Eq.25-9, we can find capacitance of each capacitor from the given values of area and plate separation. Then using Eq.25-1, we can compare the capacitance from the slopes of the plots. Then, comparing the predictions about the capacitance yielded from both the equations, we can find out which plot goes with which capacitor.

2) Formulae:

i) From Eq.25-1, the charge q is

q=CV

ii) From Eq.25-9, the capacitance is

C=ε0Ad

3) Given:

i) For capacitor 1, area is A

ii) For capacitor 2, area is 2A

iii) For capacitor 3, area is A

iv) For capacitor 1, separation is d

v) For capacitor 2, separation is d

vi) For capacitor 3, separation is 2d

4) Calculations:

From Eq.25-1, the charge q is

q=CV

where C is the capacitance and V is the potential difference.

Therefore, the capacitance is given by

C=qV=slope………………………….……………………………………………………(1)

From Eq.25-9, the capacitance is

C=ε0Ad

where ε0 is the permittivity, A is the area of the plates and d is the separation between the plates.

Let, A=1 and d=1. Putting it in Eq.25-9 we get,

Capacitor 1 has capacitance,

C1=ε011=ε0

Similarly, Capacitor 2 has capacitance,

C2=ε021=2ε0

And Capacitor 3 has capacitance,

C3=ε012=12ε0

From this, we can interpret that capacitance of capacitor 2 has the greatest value and capacitor 1 has the greater capacitance than that of capacitor 3.

From the graph, we can infer that the plot a has the greatest slope. Hence, the capacitance and the slope of plot b is greater than that of plot c. Hence, corresponding capacitance.

Therefore, we can conclude that, Plot a goes with capacitor 2, plot b goes with capacitor 1 and plot c goes with capacitor 3.

Conclusion:

We can predict about the plots between the charge and the voltage corresponding to capacitors having some area and separation between the plates of capacitor from the slope of the graph and the formula for capacitance.

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03:20

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A planar double pendulum consists of two point masses \[m_1 = 1.00~\mathrm{kg}, \qquad m_2 = 1.00~\mathrm{kg}\]connected by massless, rigid rods of lengths \[L_1 = 1.00~\mathrm{m}, \qquad L_2 = 1.20~\mathrm{m}.\]The upper rod is hinged to a fixed pivot; gravity acts vertically downward with\[g = 9.81~\mathrm{m\,s^{-2}}.\]Define the generalized coordinates \(\theta_1,\theta_2\) as the angles each rod makes with thedownward vertical (positive anticlockwise, measured in radians unless stated otherwise).At \(t=0\) the system is released from rest with \[\theta_1(0)=120^{\circ}, \qquad\theta_2(0)=-10^{\circ}, \qquad\dot{\theta}_1(0)=\dot{\theta}_2(0)=0 .\]Using the exact nonlinear equations of motion (no small-angle or planar-pendulumapproximations) and assuming the rods never stretch or slip, determine the angle\(\theta_2\) at the instant\[t = 10.0~\mathrm{s}.\]Give the result in degrees, in the interval \((-180^{\circ},180^{\circ}]\).

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Answer 2

Textbook Question

Chapter 25, Problem 1Q

Figure 25-18 shows plots of charge versus potential difference for three parallel-plate capacitors that have the plate areas and separations given in the table. Which plot goes with which capacitor?

Chapter 25, Problem 1Q, Figure 25-18 shows plots of charge versus potential difference for three parallel-plate capacitors

Figure 25-18 Question 1.

Capacitor	Area	Separation
1	A	d
2	2A	d
3	A	2d

Expert Solution & Answer

To determine

To find:

Which plot goes with which capacitor.

Answer to Problem 1Q

Solution:

Plot a goes with the capacitor 2. The plot b goes with capacitor 1 and plot c goes with capacitor 3.

Explanation of Solution

1) Concept:

Using Eq.25-9, we can find capacitance of each capacitor from the given values of area and plate separation. Then using Eq.25-1, we can compare the capacitance from the slopes of the plots. Then, comparing the predictions about the capacitance yielded from both the equations, we can find out which plot goes with which capacitor.

2) Formulae:

i) From Eq.25-1, the charge q is

q=CV

ii) From Eq.25-9, the capacitance is

C=ε0Ad

3) Given:

i) For capacitor 1, area is A

ii) For capacitor 2, area is 2A

iii) For capacitor 3, area is A

iv) For capacitor 1, separation is d

v) For capacitor 2, separation is d

vi) For capacitor 3, separation is 2d

4) Calculations:

From Eq.25-1, the charge q is

q=CV

where C is the capacitance and V is the potential difference.

Therefore, the capacitance is given by

C=qV=slope………………………….……………………………………………………(1)

From Eq.25-9, the capacitance is

C=ε0Ad

where ε0 is the permittivity, A is the area of the plates and d is the separation between the plates.

Let, A=1 and d=1. Putting it in Eq.25-9 we get,

Capacitor 1 has capacitance,

C1=ε011=ε0

Similarly, Capacitor 2 has capacitance,

C2=ε021=2ε0

And Capacitor 3 has capacitance,

C3=ε012=12ε0

From this, we can interpret that capacitance of capacitor 2 has the greatest value and capacitor 1 has the greater capacitance than that of capacitor 3.

From the graph, we can infer that the plot a has the greatest slope. Hence, the capacitance and the slope of plot b is greater than that of plot c. Hence, corresponding capacitance.

Therefore, we can conclude that, Plot a goes with capacitor 2, plot b goes with capacitor 1 and plot c goes with capacitor 3.

Conclusion:

We can predict about the plots between the charge and the voltage corresponding to capacitors having some area and separation between the plates of capacitor from the slope of the graph and the formula for capacitance.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

03:20

Students have asked these similar questions

For each of the actions depicted below, a magnet and/or metal loop moves with velocity v→ (v→ is constant and has the same magnitude in all parts). Determine whether a current is induced in the metal loop. If so, indicate the direction of the current in the loop, either clockwise or counterclockwise when seen from the right of the loop. The axis of the magnet is lined up with the center of the loop. For the action depicted in (Figure 5), indicate the direction of the induced current in the loop (clockwise, counterclockwise or zero, when seen from the right of the loop). I know that the current is clockwise, I just dont understand why. Please fully explain why it's clockwise, Thank you

A planar double pendulum consists of two point masses \[m_1 = 1.00~\mathrm{kg}, \qquad m_2 = 1.00~\mathrm{kg}\]connected by massless, rigid rods of lengths \[L_1 = 1.00~\mathrm{m}, \qquad L_2 = 1.20~\mathrm{m}.\]The upper rod is hinged to a fixed pivot; gravity acts vertically downward with\[g = 9.81~\mathrm{m\,s^{-2}}.\]Define the generalized coordinates \(\theta_1,\theta_2\) as the angles each rod makes with thedownward vertical (positive anticlockwise, measured in radians unless stated otherwise).At \(t=0\) the system is released from rest with \[\theta_1(0)=120^{\circ}, \qquad\theta_2(0)=-10^{\circ}, \qquad\dot{\theta}_1(0)=\dot{\theta}_2(0)=0 .\]Using the exact nonlinear equations of motion (no small-angle or planar-pendulumapproximations) and assuming the rods never stretch or slip, determine the angle\(\theta_2\) at the instant\[t = 10.0~\mathrm{s}.\]Give the result in degrees, in the interval \((-180^{\circ},180^{\circ}]\).

What are the expected readings of the ammeter and voltmeter for the circuit in the figure below? (R = 5.60 Ω, ΔV = 6.30 V) ammeter I =

Answer 3

Textbook Question

Chapter 25, Problem 1Q

Figure 25-18 shows plots of charge versus potential difference for three parallel-plate capacitors that have the plate areas and separations given in the table. Which plot goes with which capacitor?

Chapter 25, Problem 1Q, Figure 25-18 shows plots of charge versus potential difference for three parallel-plate capacitors

Figure 25-18 Question 1.

Capacitor	Area	Separation
1	A	d
2	2A	d
3	A	2d

Expert Solution & Answer

To determine

To find:

Which plot goes with which capacitor.

Answer to Problem 1Q

Solution:

Plot a goes with the capacitor 2. The plot b goes with capacitor 1 and plot c goes with capacitor 3.

Explanation of Solution

1) Concept:

Using Eq.25-9, we can find capacitance of each capacitor from the given values of area and plate separation. Then using Eq.25-1, we can compare the capacitance from the slopes of the plots. Then, comparing the predictions about the capacitance yielded from both the equations, we can find out which plot goes with which capacitor.

2) Formulae:

i) From Eq.25-1, the charge q is

q=CV

ii) From Eq.25-9, the capacitance is

C=ε0Ad

3) Given:

i) For capacitor 1, area is A

ii) For capacitor 2, area is 2A

iii) For capacitor 3, area is A

iv) For capacitor 1, separation is d

v) For capacitor 2, separation is d

vi) For capacitor 3, separation is 2d

4) Calculations:

From Eq.25-1, the charge q is

q=CV

where C is the capacitance and V is the potential difference.

Therefore, the capacitance is given by

C=qV=slope………………………….……………………………………………………(1)

From Eq.25-9, the capacitance is

C=ε0Ad

where ε0 is the permittivity, A is the area of the plates and d is the separation between the plates.

Let, A=1 and d=1. Putting it in Eq.25-9 we get,

Capacitor 1 has capacitance,

C1=ε011=ε0

Similarly, Capacitor 2 has capacitance,

C2=ε021=2ε0

And Capacitor 3 has capacitance,

C3=ε012=12ε0

From this, we can interpret that capacitance of capacitor 2 has the greatest value and capacitor 1 has the greater capacitance than that of capacitor 3.

From the graph, we can infer that the plot a has the greatest slope. Hence, the capacitance and the slope of plot b is greater than that of plot c. Hence, corresponding capacitance.

Therefore, we can conclude that, Plot a goes with capacitor 2, plot b goes with capacitor 1 and plot c goes with capacitor 3.

Conclusion:

We can predict about the plots between the charge and the voltage corresponding to capacitors having some area and separation between the plates of capacitor from the slope of the graph and the formula for capacitance.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

03:20

Answer 4

Textbook Question

Chapter 25, Problem 1Q

Figure 25-18 shows plots of charge versus potential difference for three parallel-plate capacitors that have the plate areas and separations given in the table. Which plot goes with which capacitor?

Chapter 25, Problem 1Q, Figure 25-18 shows plots of charge versus potential difference for three parallel-plate capacitors

Figure 25-18 Question 1.

Capacitor	Area	Separation
1	A	d
2	2A	d
3	A	2d

Expert Solution & Answer

To determine

To find:

Which plot goes with which capacitor.

Answer to Problem 1Q

Solution:

Plot a goes with the capacitor 2. The plot b goes with capacitor 1 and plot c goes with capacitor 3.

Explanation of Solution

1) Concept:

Using Eq.25-9, we can find capacitance of each capacitor from the given values of area and plate separation. Then using Eq.25-1, we can compare the capacitance from the slopes of the plots. Then, comparing the predictions about the capacitance yielded from both the equations, we can find out which plot goes with which capacitor.

2) Formulae:

i) From Eq.25-1, the charge q is

q=CV

ii) From Eq.25-9, the capacitance is

C=ε0Ad

3) Given:

i) For capacitor 1, area is A

ii) For capacitor 2, area is 2A

iii) For capacitor 3, area is A

iv) For capacitor 1, separation is d

v) For capacitor 2, separation is d

vi) For capacitor 3, separation is 2d

4) Calculations:

From Eq.25-1, the charge q is

q=CV

where C is the capacitance and V is the potential difference.

Therefore, the capacitance is given by

C=qV=slope………………………….……………………………………………………(1)

From Eq.25-9, the capacitance is

C=ε0Ad

where ε0 is the permittivity, A is the area of the plates and d is the separation between the plates.

Let, A=1 and d=1. Putting it in Eq.25-9 we get,

Capacitor 1 has capacitance,

C1=ε011=ε0

Similarly, Capacitor 2 has capacitance,

C2=ε021=2ε0

And Capacitor 3 has capacitance,

C3=ε012=12ε0

From this, we can interpret that capacitance of capacitor 2 has the greatest value and capacitor 1 has the greater capacitance than that of capacitor 3.

From the graph, we can infer that the plot a has the greatest slope. Hence, the capacitance and the slope of plot b is greater than that of plot c. Hence, corresponding capacitance.

Therefore, we can conclude that, Plot a goes with capacitor 2, plot b goes with capacitor 1 and plot c goes with capacitor 3.

Conclusion:

We can predict about the plots between the charge and the voltage corresponding to capacitors having some area and separation between the plates of capacitor from the slope of the graph and the formula for capacitance.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

03:20

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

To find:

Which plot goes with which capacitor.

Answer to Problem 1Q

Solution:

Plot a goes with the capacitor 2. The plot b goes with capacitor 1 and plot c goes with capacitor 3.

Explanation of Solution

1) Concept:

Using Eq.25-9, we can find capacitance of each capacitor from the given values of area and plate separation. Then using Eq.25-1, we can compare the capacitance from the slopes of the plots. Then, comparing the predictions about the capacitance yielded from both the equations, we can find out which plot goes with which capacitor.

2) Formulae:

i) From Eq.25-1, the charge q is

q=CV

ii) From Eq.25-9, the capacitance is

C=ε0Ad

3) Given:

i) For capacitor 1, area is A

ii) For capacitor 2, area is 2A

iii) For capacitor 3, area is A

iv) For capacitor 1, separation is d

v) For capacitor 2, separation is d

vi) For capacitor 3, separation is 2d

4) Calculations:

From Eq.25-1, the charge q is

q=CV

where C is the capacitance and V is the potential difference.

Therefore, the capacitance is given by

C=qV=slope………………………….……………………………………………………(1)

From Eq.25-9, the capacitance is

C=ε0Ad

where ε0 is the permittivity, A is the area of the plates and d is the separation between the plates.

Let, A=1 and d=1. Putting it in Eq.25-9 we get,

Capacitor 1 has capacitance,

C1=ε011=ε0

Similarly, Capacitor 2 has capacitance,

C2=ε021=2ε0

And Capacitor 3 has capacitance,

C3=ε012=12ε0

From this, we can interpret that capacitance of capacitor 2 has the greatest value and capacitor 1 has the greater capacitance than that of capacitor 3.

From the graph, we can infer that the plot a has the greatest slope. Hence, the capacitance and the slope of plot b is greater than that of plot c. Hence, corresponding capacitance.

Therefore, we can conclude that, Plot a goes with capacitor 2, plot b goes with capacitor 1 and plot c goes with capacitor 3.

Conclusion:

We can predict about the plots between the charge and the voltage corresponding to capacitors having some area and separation between the plates of capacitor from the slope of the graph and the formula for capacitance.

Answer 8

Expert Solution & Answer

To determine

To find:

Which plot goes with which capacitor.

Answer to Problem 1Q

Solution:

Plot a goes with the capacitor 2. The plot b goes with capacitor 1 and plot c goes with capacitor 3.

Explanation of Solution

1) Concept:

Using Eq.25-9, we can find capacitance of each capacitor from the given values of area and plate separation. Then using Eq.25-1, we can compare the capacitance from the slopes of the plots. Then, comparing the predictions about the capacitance yielded from both the equations, we can find out which plot goes with which capacitor.

2) Formulae:

i) From Eq.25-1, the charge q is

q=CV

ii) From Eq.25-9, the capacitance is

C=ε0Ad

3) Given:

i) For capacitor 1, area is A

ii) For capacitor 2, area is 2A

iii) For capacitor 3, area is A

iv) For capacitor 1, separation is d

v) For capacitor 2, separation is d

vi) For capacitor 3, separation is 2d

4) Calculations:

From Eq.25-1, the charge q is

q=CV

where C is the capacitance and V is the potential difference.

Therefore, the capacitance is given by

C=qV=slope………………………….……………………………………………………(1)

From Eq.25-9, the capacitance is

C=ε0Ad

where ε0 is the permittivity, A is the area of the plates and d is the separation between the plates.

Let, A=1 and d=1. Putting it in Eq.25-9 we get,

Capacitor 1 has capacitance,

C1=ε011=ε0

Similarly, Capacitor 2 has capacitance,

C2=ε021=2ε0

And Capacitor 3 has capacitance,

C3=ε012=12ε0

From this, we can interpret that capacitance of capacitor 2 has the greatest value and capacitor 1 has the greater capacitance than that of capacitor 3.

From the graph, we can infer that the plot a has the greatest slope. Hence, the capacitance and the slope of plot b is greater than that of plot c. Hence, corresponding capacitance.

Therefore, we can conclude that, Plot a goes with capacitor 2, plot b goes with capacitor 1 and plot c goes with capacitor 3.

Conclusion:

We can predict about the plots between the charge and the voltage corresponding to capacitors having some area and separation between the plates of capacitor from the slope of the graph and the formula for capacitance.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

To find:

Which plot goes with which capacitor.

Answer 12

Answer to Problem 1Q

Solution:

Plot a goes with the capacitor 2. The plot b goes with capacitor 1 and plot c goes with capacitor 3.

Answer 13

Explanation of Solution

1) Concept:

Using Eq.25-9, we can find capacitance of each capacitor from the given values of area and plate separation. Then using Eq.25-1, we can compare the capacitance from the slopes of the plots. Then, comparing the predictions about the capacitance yielded from both the equations, we can find out which plot goes with which capacitor.

2) Formulae:

i) From Eq.25-1, the charge q is

q=CV

ii) From Eq.25-9, the capacitance is

C=ε0Ad

3) Given:

i) For capacitor 1, area is A

ii) For capacitor 2, area is 2A

iii) For capacitor 3, area is A

iv) For capacitor 1, separation is d

v) For capacitor 2, separation is d

vi) For capacitor 3, separation is 2d

4) Calculations:

From Eq.25-1, the charge q is

q=CV

where C is the capacitance and V is the potential difference.

Therefore, the capacitance is given by

C=qV=slope………………………….……………………………………………………(1)

From Eq.25-9, the capacitance is

C=ε0Ad

where ε0 is the permittivity, A is the area of the plates and d is the separation between the plates.

Let, A=1 and d=1. Putting it in Eq.25-9 we get,

Capacitor 1 has capacitance,

C1=ε011=ε0

Similarly, Capacitor 2 has capacitance,

C2=ε021=2ε0

And Capacitor 3 has capacitance,

C3=ε012=12ε0

From this, we can interpret that capacitance of capacitor 2 has the greatest value and capacitor 1 has the greater capacitance than that of capacitor 3.

From the graph, we can infer that the plot a has the greatest slope. Hence, the capacitance and the slope of plot b is greater than that of plot c. Hence, corresponding capacitance.

Therefore, we can conclude that, Plot a goes with capacitor 2, plot b goes with capacitor 1 and plot c goes with capacitor 3.

Conclusion:

We can predict about the plots between the charge and the voltage corresponding to capacitors having some area and separation between the plates of capacitor from the slope of the graph and the formula for capacitance.