ToCalculate : The energy stored in the capacitor.

Question

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Answer 1

Question

Chapter 24, Problem 85P

(a)

To determine

ToCalculate: The energy stored in the capacitor.

(a)

Expert Solution

Answer to Problem 85P

U=Q2d2∈0a[(κ−1)x+a]

Explanation of Solution

Given information :

Charge of an electrically isolated capacitor =Q

Distance between plated =d

Lengths of capacitor plates =a and b

Formula used :

Energy stored in capacitor is:

U=12Q2C

Where, Q is the charge stored and C is the capacitance of the capacitor.

Equivalent capacitance of parallel plate capacitor:

Ceq=C1+C2+...

Calculation:

The energy stored in the capacitor as a function of the equivalent capacitance Ceq is,

U=12Q2Ceq

The capacitances of the two capacitors are,

C1=κ∈0axd

And

C2=∈0a(a−x)d

Ceq=C1+C2Ceq=κ∈0axd+∈0a(a−x)dCeq=∈0ad(κx+a−x)Ceq=∈0ad[(κ−1)x+a]

U=Q2d2∈0a[(κ−1)x+a]

Conclusion:

The energy stored in the capacitor is, U=Q2d2∈0a[(κ−1)x+a]

(b)

To determine

ToCalculate: The force by examining how the stored energy varies with x .

(b)

Expert Solution

Answer to Problem 85P

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Explanation of Solution

Given information:

Charge of an electrically isolated capacitor =Q

Distance between plated =d

Lengths of capacitor plates =a and b

Energy stored in the capacitor, U=Q2d2∈0a[(κ−1)x+a]

Formula used:

Electric force:

F=−dUdx

U is the energy stored.

Calculation:

F=−dUdxF=−ddx[12Q2∈0a[(κ−1)x+a]]F=Q2d2∈0addx{[(κ−1)x+a]−1}F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Conclusion:

The force by examining how the stored energy varies with x is,

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

(c)

To determine

ToCalculate: The force in terms of the capacitance and potential difference V between the plates.

(c)

Expert Solution

Answer to Problem 85P

F=(κ−1)a∈0V22d

Explanation of Solution

Given information :

Charge of an electrically isolated capacitor =Q

Distance between plated =d

Lengths of capacitor plates =a&b

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Ceq=∈0ad[(κ−1)x+a]

Formula Used:

The capacitance can be obtained by:

C=∈0Ad

Where, ∈0 is the absolute permittivity, A is the area of the plates and d is the distance between the plates.

Charge stored in capacitor, Q=CV

Where, C is the capacitance and V is the voltage.

Calculation:

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Multiply and divide by a∈od2 .

F=(κ−1)Q2(a∈0d)2(a∈0d)2[(κ−1)x+a]2F=(κ−1)Q2(a∈0d)2Ceq2Q=CV∵C=∈0AdQ=(∈0Ad)VF=(κ−1)a∈0V22d

This expression is independent of x.

Conclusion:

The force in terms of the capacitance and potential difference V between the plates is,

F=(κ−1)a∈0V22d

(d)

To determine

ToFind: The position from where the force originates.

(d)

Expert Solution

Answer to Problem 85P

The force originates from the fringing fields around the edges of the capacitor.

Explanation of Solution

Introduction:

The fringe field is the magnetic field at the edge. It occurs outside the center of the magnet. This depends on the magnet’s core.

The force originates from the condenser edges of fringing fields. The effect of force is to draw the polarized dielectric in between the plates of the condenser.

Conclusion:

Hence, the force originates from the fringing fields around the edges of the capacitor.

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Answer 2

Question

Chapter 24, Problem 85P

(a)

To determine

ToCalculate: The energy stored in the capacitor.

(a)

Expert Solution

Answer to Problem 85P

U=Q2d2∈0a[(κ−1)x+a]

Explanation of Solution

Given information :

Charge of an electrically isolated capacitor =Q

Distance between plated =d

Lengths of capacitor plates =a and b

Formula used :

Energy stored in capacitor is:

U=12Q2C

Where, Q is the charge stored and C is the capacitance of the capacitor.

Equivalent capacitance of parallel plate capacitor:

Ceq=C1+C2+...

Calculation:

The energy stored in the capacitor as a function of the equivalent capacitance Ceq is,

U=12Q2Ceq

The capacitances of the two capacitors are,

C1=κ∈0axd

And

C2=∈0a(a−x)d

Ceq=C1+C2Ceq=κ∈0axd+∈0a(a−x)dCeq=∈0ad(κx+a−x)Ceq=∈0ad[(κ−1)x+a]

U=Q2d2∈0a[(κ−1)x+a]

Conclusion:

The energy stored in the capacitor is, U=Q2d2∈0a[(κ−1)x+a]

(b)

To determine

ToCalculate: The force by examining how the stored energy varies with x .

(b)

Expert Solution

Answer to Problem 85P

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Explanation of Solution

Given information:

Charge of an electrically isolated capacitor =Q

Distance between plated =d

Lengths of capacitor plates =a and b

Energy stored in the capacitor, U=Q2d2∈0a[(κ−1)x+a]

Formula used:

Electric force:

F=−dUdx

U is the energy stored.

Calculation:

F=−dUdxF=−ddx[12Q2∈0a[(κ−1)x+a]]F=Q2d2∈0addx{[(κ−1)x+a]−1}F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Conclusion:

The force by examining how the stored energy varies with x is,

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

(c)

To determine

ToCalculate: The force in terms of the capacitance and potential difference V between the plates.

(c)

Expert Solution

Answer to Problem 85P

F=(κ−1)a∈0V22d

Explanation of Solution

Given information :

Charge of an electrically isolated capacitor =Q

Distance between plated =d

Lengths of capacitor plates =a&b

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Ceq=∈0ad[(κ−1)x+a]

Formula Used:

The capacitance can be obtained by:

C=∈0Ad

Where, ∈0 is the absolute permittivity, A is the area of the plates and d is the distance between the plates.

Charge stored in capacitor, Q=CV

Where, C is the capacitance and V is the voltage.

Calculation:

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Multiply and divide by a∈od2 .

F=(κ−1)Q2(a∈0d)2(a∈0d)2[(κ−1)x+a]2F=(κ−1)Q2(a∈0d)2Ceq2Q=CV∵C=∈0AdQ=(∈0Ad)VF=(κ−1)a∈0V22d

This expression is independent of x.

Conclusion:

The force in terms of the capacitance and potential difference V between the plates is,

F=(κ−1)a∈0V22d

(d)

To determine

ToFind: The position from where the force originates.

(d)

Expert Solution

Answer to Problem 85P

The force originates from the fringing fields around the edges of the capacitor.

Explanation of Solution

Introduction:

The fringe field is the magnetic field at the edge. It occurs outside the center of the magnet. This depends on the magnet’s core.

The force originates from the condenser edges of fringing fields. The effect of force is to draw the polarized dielectric in between the plates of the condenser.

Conclusion:

Hence, the force originates from the fringing fields around the edges of the capacitor.

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Answer 3

Question

Chapter 24, Problem 85P

(a)

To determine

ToCalculate: The energy stored in the capacitor.

(a)

Expert Solution

Answer to Problem 85P

U=Q2d2∈0a[(κ−1)x+a]

Explanation of Solution

Given information :

Charge of an electrically isolated capacitor =Q

Distance between plated =d

Lengths of capacitor plates =a and b

Formula used :

Energy stored in capacitor is:

U=12Q2C

Where, Q is the charge stored and C is the capacitance of the capacitor.

Equivalent capacitance of parallel plate capacitor:

Ceq=C1+C2+...

Calculation:

The energy stored in the capacitor as a function of the equivalent capacitance Ceq is,

U=12Q2Ceq

The capacitances of the two capacitors are,

C1=κ∈0axd

And

C2=∈0a(a−x)d

Ceq=C1+C2Ceq=κ∈0axd+∈0a(a−x)dCeq=∈0ad(κx+a−x)Ceq=∈0ad[(κ−1)x+a]

U=Q2d2∈0a[(κ−1)x+a]

Conclusion:

The energy stored in the capacitor is, U=Q2d2∈0a[(κ−1)x+a]

(b)

To determine

ToCalculate: The force by examining how the stored energy varies with x .

(b)

Expert Solution

Answer to Problem 85P

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Explanation of Solution

Given information:

Charge of an electrically isolated capacitor =Q

Distance between plated =d

Lengths of capacitor plates =a and b

Energy stored in the capacitor, U=Q2d2∈0a[(κ−1)x+a]

Formula used:

Electric force:

F=−dUdx

U is the energy stored.

Calculation:

F=−dUdxF=−ddx[12Q2∈0a[(κ−1)x+a]]F=Q2d2∈0addx{[(κ−1)x+a]−1}F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Conclusion:

The force by examining how the stored energy varies with x is,

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

(c)

To determine

ToCalculate: The force in terms of the capacitance and potential difference V between the plates.

(c)

Expert Solution

Answer to Problem 85P

F=(κ−1)a∈0V22d

Explanation of Solution

Given information :

Charge of an electrically isolated capacitor =Q

Distance between plated =d

Lengths of capacitor plates =a&b

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Ceq=∈0ad[(κ−1)x+a]

Formula Used:

The capacitance can be obtained by:

C=∈0Ad

Where, ∈0 is the absolute permittivity, A is the area of the plates and d is the distance between the plates.

Charge stored in capacitor, Q=CV

Where, C is the capacitance and V is the voltage.

Calculation:

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Multiply and divide by a∈od2 .

F=(κ−1)Q2(a∈0d)2(a∈0d)2[(κ−1)x+a]2F=(κ−1)Q2(a∈0d)2Ceq2Q=CV∵C=∈0AdQ=(∈0Ad)VF=(κ−1)a∈0V22d

This expression is independent of x.

Conclusion:

The force in terms of the capacitance and potential difference V between the plates is,

F=(κ−1)a∈0V22d

(d)

To determine

ToFind: The position from where the force originates.

(d)

Expert Solution

Answer to Problem 85P

The force originates from the fringing fields around the edges of the capacitor.

Explanation of Solution

Introduction:

The fringe field is the magnetic field at the edge. It occurs outside the center of the magnet. This depends on the magnet’s core.

The force originates from the condenser edges of fringing fields. The effect of force is to draw the polarized dielectric in between the plates of the condenser.

Conclusion:

Hence, the force originates from the fringing fields around the edges of the capacitor.

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Answer 4

Question

Chapter 24, Problem 85P

(a)

To determine

ToCalculate: The energy stored in the capacitor.

(a)

Expert Solution

Answer to Problem 85P

U=Q2d2∈0a[(κ−1)x+a]

Explanation of Solution

Given information :

Charge of an electrically isolated capacitor =Q

Distance between plated =d

Lengths of capacitor plates =a and b

Formula used :

Energy stored in capacitor is:

U=12Q2C

Where, Q is the charge stored and C is the capacitance of the capacitor.

Equivalent capacitance of parallel plate capacitor:

Ceq=C1+C2+...

Calculation:

The energy stored in the capacitor as a function of the equivalent capacitance Ceq is,

U=12Q2Ceq

The capacitances of the two capacitors are,

C1=κ∈0axd

And

C2=∈0a(a−x)d

Ceq=C1+C2Ceq=κ∈0axd+∈0a(a−x)dCeq=∈0ad(κx+a−x)Ceq=∈0ad[(κ−1)x+a]

U=Q2d2∈0a[(κ−1)x+a]

Conclusion:

The energy stored in the capacitor is, U=Q2d2∈0a[(κ−1)x+a]

(b)

To determine

ToCalculate: The force by examining how the stored energy varies with x .

(b)

Expert Solution

Answer to Problem 85P

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Explanation of Solution

Given information:

Charge of an electrically isolated capacitor =Q

Distance between plated =d

Lengths of capacitor plates =a and b

Energy stored in the capacitor, U=Q2d2∈0a[(κ−1)x+a]

Formula used:

Electric force:

F=−dUdx

U is the energy stored.

Calculation:

F=−dUdxF=−ddx[12Q2∈0a[(κ−1)x+a]]F=Q2d2∈0addx{[(κ−1)x+a]−1}F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Conclusion:

The force by examining how the stored energy varies with x is,

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

(c)

To determine

ToCalculate: The force in terms of the capacitance and potential difference V between the plates.

(c)

Expert Solution

Answer to Problem 85P

F=(κ−1)a∈0V22d

Explanation of Solution

Given information :

Charge of an electrically isolated capacitor =Q

Distance between plated =d

Lengths of capacitor plates =a&b

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Ceq=∈0ad[(κ−1)x+a]

Formula Used:

The capacitance can be obtained by:

C=∈0Ad

Where, ∈0 is the absolute permittivity, A is the area of the plates and d is the distance between the plates.

Charge stored in capacitor, Q=CV

Where, C is the capacitance and V is the voltage.

Calculation:

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Multiply and divide by a∈od2 .

F=(κ−1)Q2(a∈0d)2(a∈0d)2[(κ−1)x+a]2F=(κ−1)Q2(a∈0d)2Ceq2Q=CV∵C=∈0AdQ=(∈0Ad)VF=(κ−1)a∈0V22d

This expression is independent of x.

Conclusion:

The force in terms of the capacitance and potential difference V between the plates is,

F=(κ−1)a∈0V22d

(d)

To determine

ToFind: The position from where the force originates.

(d)

Expert Solution

Answer to Problem 85P

The force originates from the fringing fields around the edges of the capacitor.

Explanation of Solution

Introduction:

The fringe field is the magnetic field at the edge. It occurs outside the center of the magnet. This depends on the magnet’s core.

The force originates from the condenser edges of fringing fields. The effect of force is to draw the polarized dielectric in between the plates of the condenser.

Conclusion:

Hence, the force originates from the fringing fields around the edges of the capacitor.

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Answer 5

Chapter 24, Problem 85P

(a)

To determine

ToCalculate: The energy stored in the capacitor.

(a)

Expert Solution

Answer to Problem 85P

U=Q2d2∈0a[(κ−1)x+a]

Explanation of Solution

Given information :

Charge of an electrically isolated capacitor =Q

Distance between plated =d

Lengths of capacitor plates =a and b

Formula used :

Energy stored in capacitor is:

U=12Q2C

Where, Q is the charge stored and C is the capacitance of the capacitor.

Equivalent capacitance of parallel plate capacitor:

Ceq=C1+C2+...

Calculation:

The energy stored in the capacitor as a function of the equivalent capacitance Ceq is,

U=12Q2Ceq

The capacitances of the two capacitors are,

C1=κ∈0axd

And

C2=∈0a(a−x)d

Ceq=C1+C2Ceq=κ∈0axd+∈0a(a−x)dCeq=∈0ad(κx+a−x)Ceq=∈0ad[(κ−1)x+a]

U=Q2d2∈0a[(κ−1)x+a]

Conclusion:

The energy stored in the capacitor is, U=Q2d2∈0a[(κ−1)x+a]

(b)

To determine

ToCalculate: The force by examining how the stored energy varies with x .

(b)

Expert Solution

Answer to Problem 85P

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Explanation of Solution

Given information:

Charge of an electrically isolated capacitor =Q

Distance between plated =d

Lengths of capacitor plates =a and b

Energy stored in the capacitor, U=Q2d2∈0a[(κ−1)x+a]

Formula used:

Electric force:

F=−dUdx

U is the energy stored.

Calculation:

F=−dUdxF=−ddx[12Q2∈0a[(κ−1)x+a]]F=Q2d2∈0addx{[(κ−1)x+a]−1}F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Conclusion:

The force by examining how the stored energy varies with x is,

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

(c)

To determine

ToCalculate: The force in terms of the capacitance and potential difference V between the plates.

(c)

Expert Solution

Answer to Problem 85P

F=(κ−1)a∈0V22d

Explanation of Solution

Given information :

Charge of an electrically isolated capacitor =Q

Distance between plated =d

Lengths of capacitor plates =a&b

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Ceq=∈0ad[(κ−1)x+a]

Formula Used:

The capacitance can be obtained by:

C=∈0Ad

Where, ∈0 is the absolute permittivity, A is the area of the plates and d is the distance between the plates.

Charge stored in capacitor, Q=CV

Where, C is the capacitance and V is the voltage.

Calculation:

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Multiply and divide by a∈od2 .

F=(κ−1)Q2(a∈0d)2(a∈0d)2[(κ−1)x+a]2F=(κ−1)Q2(a∈0d)2Ceq2Q=CV∵C=∈0AdQ=(∈0Ad)VF=(κ−1)a∈0V22d

This expression is independent of x.

Conclusion:

The force in terms of the capacitance and potential difference V between the plates is,

F=(κ−1)a∈0V22d

(d)

To determine

ToFind: The position from where the force originates.

(d)

Expert Solution

Answer to Problem 85P

The force originates from the fringing fields around the edges of the capacitor.

Explanation of Solution

Introduction:

The fringe field is the magnetic field at the edge. It occurs outside the center of the magnet. This depends on the magnet’s core.

The force originates from the condenser edges of fringing fields. The effect of force is to draw the polarized dielectric in between the plates of the condenser.

Conclusion:

Hence, the force originates from the fringing fields around the edges of the capacitor.

Answer 6

Chapter 24, Problem 85P

(a)

To determine

ToCalculate: The energy stored in the capacitor.

(a)

Expert Solution

Answer to Problem 85P

U=Q2d2∈0a[(κ−1)x+a]

Explanation of Solution

Given information :

Charge of an electrically isolated capacitor =Q

Distance between plated =d

Lengths of capacitor plates =a and b

Formula used :

Energy stored in capacitor is:

U=12Q2C

Where, Q is the charge stored and C is the capacitance of the capacitor.

Equivalent capacitance of parallel plate capacitor:

Ceq=C1+C2+...

Calculation:

The energy stored in the capacitor as a function of the equivalent capacitance Ceq is,

U=12Q2Ceq

The capacitances of the two capacitors are,

C1=κ∈0axd

And

C2=∈0a(a−x)d

Ceq=C1+C2Ceq=κ∈0axd+∈0a(a−x)dCeq=∈0ad(κx+a−x)Ceq=∈0ad[(κ−1)x+a]

U=Q2d2∈0a[(κ−1)x+a]

Conclusion:

The energy stored in the capacitor is, U=Q2d2∈0a[(κ−1)x+a]

Answer 7

Chapter 24, Problem 85P

(a)

To determine

ToCalculate: The energy stored in the capacitor.

Answer 8

(a)

Expert Solution

Answer to Problem 85P

U=Q2d2∈0a[(κ−1)x+a]

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given information :

Charge of an electrically isolated capacitor =Q

Distance between plated =d

Lengths of capacitor plates =a and b

Formula used :

Energy stored in capacitor is:

U=12Q2C

Where, Q is the charge stored and C is the capacitance of the capacitor.

Equivalent capacitance of parallel plate capacitor:

Ceq=C1+C2+...

Calculation:

The energy stored in the capacitor as a function of the equivalent capacitance Ceq is,

U=12Q2Ceq

The capacitances of the two capacitors are,

C1=κ∈0axd

And

C2=∈0a(a−x)d

Ceq=C1+C2Ceq=κ∈0axd+∈0a(a−x)dCeq=∈0ad(κx+a−x)Ceq=∈0ad[(κ−1)x+a]

U=Q2d2∈0a[(κ−1)x+a]

Conclusion:

The energy stored in the capacitor is, U=Q2d2∈0a[(κ−1)x+a]

Answer 13

(b)

To determine

ToCalculate: The force by examining how the stored energy varies with x .

(b)

Expert Solution

Answer to Problem 85P

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Explanation of Solution

Given information:

Charge of an electrically isolated capacitor =Q

Distance between plated =d

Lengths of capacitor plates =a and b

Energy stored in the capacitor, U=Q2d2∈0a[(κ−1)x+a]

Formula used:

Electric force:

F=−dUdx

U is the energy stored.

Calculation:

F=−dUdxF=−ddx[12Q2∈0a[(κ−1)x+a]]F=Q2d2∈0addx{[(κ−1)x+a]−1}F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Conclusion:

The force by examining how the stored energy varies with x is,

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Answer 14

(b)

To determine

ToCalculate: The force by examining how the stored energy varies with x .

Answer 15

(b)

Expert Solution

Answer to Problem 85P

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given information:

Charge of an electrically isolated capacitor =Q

Distance between plated =d

Lengths of capacitor plates =a and b

Energy stored in the capacitor, U=Q2d2∈0a[(κ−1)x+a]

Formula used:

Electric force:

F=−dUdx

U is the energy stored.

Calculation:

F=−dUdxF=−ddx[12Q2∈0a[(κ−1)x+a]]F=Q2d2∈0addx{[(κ−1)x+a]−1}F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Conclusion:

The force by examining how the stored energy varies with x is,

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Answer 20

(c)

To determine

ToCalculate: The force in terms of the capacitance and potential difference V between the plates.

(c)

Expert Solution

Answer to Problem 85P

F=(κ−1)a∈0V22d

Explanation of Solution

Given information :

Charge of an electrically isolated capacitor =Q

Distance between plated =d

Lengths of capacitor plates =a&b

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Ceq=∈0ad[(κ−1)x+a]

Formula Used:

The capacitance can be obtained by:

C=∈0Ad

Where, ∈0 is the absolute permittivity, A is the area of the plates and d is the distance between the plates.

Charge stored in capacitor, Q=CV

Where, C is the capacitance and V is the voltage.

Calculation:

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Multiply and divide by a∈od2 .

F=(κ−1)Q2(a∈0d)2(a∈0d)2[(κ−1)x+a]2F=(κ−1)Q2(a∈0d)2Ceq2Q=CV∵C=∈0AdQ=(∈0Ad)VF=(κ−1)a∈0V22d

This expression is independent of x.

Conclusion:

The force in terms of the capacitance and potential difference V between the plates is,

F=(κ−1)a∈0V22d

Answer 21

(c)

To determine

ToCalculate: The force in terms of the capacitance and potential difference V between the plates.

Answer 22

(c)

Expert Solution

Answer to Problem 85P

F=(κ−1)a∈0V22d

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given information :

Charge of an electrically isolated capacitor =Q

Distance between plated =d

Lengths of capacitor plates =a&b

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Ceq=∈0ad[(κ−1)x+a]

Formula Used:

The capacitance can be obtained by:

C=∈0Ad

Where, ∈0 is the absolute permittivity, A is the area of the plates and d is the distance between the plates.

Charge stored in capacitor, Q=CV

Where, C is the capacitance and V is the voltage.

Calculation:

F=(κ−1)Q2d2∈0a[(κ−1)x+a]2

Multiply and divide by a∈od2 .

F=(κ−1)Q2(a∈0d)2(a∈0d)2[(κ−1)x+a]2F=(κ−1)Q2(a∈0d)2Ceq2Q=CV∵C=∈0AdQ=(∈0Ad)VF=(κ−1)a∈0V22d

This expression is independent of x.

Conclusion:

The force in terms of the capacitance and potential difference V between the plates is,

F=(κ−1)a∈0V22d

Answer 27

(d)

To determine

ToFind: The position from where the force originates.

(d)

Expert Solution

Answer to Problem 85P

The force originates from the fringing fields around the edges of the capacitor.

Explanation of Solution

Introduction:

The fringe field is the magnetic field at the edge. It occurs outside the center of the magnet. This depends on the magnet’s core.

The force originates from the condenser edges of fringing fields. The effect of force is to draw the polarized dielectric in between the plates of the condenser.

Conclusion:

Hence, the force originates from the fringing fields around the edges of the capacitor.

Answer 28

(d)

To determine

ToFind: The position from where the force originates.

Answer 29

(d)

Expert Solution

Answer to Problem 85P

The force originates from the fringing fields around the edges of the capacitor.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Introduction:

The fringe field is the magnetic field at the edge. It occurs outside the center of the magnet. This depends on the magnet’s core.

The force originates from the condenser edges of fringing fields. The effect of force is to draw the polarized dielectric in between the plates of the condenser.

Conclusion:

Hence, the force originates from the fringing fields around the edges of the capacitor.