ToCalculate : The potential difference between the cylinders.

Question

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Answer 1

Question

Chapter 24, Problem 90P

(a)

To determine

ToCalculate: The potential difference between the cylinders.

(a)

Expert Solution

Answer to Problem 90P

$V=2kQln(b/a)κL$

Explanation of Solution

Given information :

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used :

Electric potential:

$V=QC$

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

$C=2π∈0κLln(b/a)$

Where, $κ$ is the dielectric constant, $∈0$ is the absolute permittivity and L is the length of the capacitor, a and b are radii of the

Calculation:

The potential difference between the cylinders to their charge and capacitance is,

$V=QC$

The capacitance of a cylindrical capacitor as a function of its radii a and b and length L:

$C=2π∈0κLln(b/a)$

$V=Qln( b/a )2π∈0κLV=2kQln( b/a )κL$

Conclusion:

The potential difference between the cylinders is $V=2kQln(b/a)κL$ .

(b)

To determine

ToCalculate: The density of the free charge of on the inner cylinder and the outer cylinder.

(b)

Expert Solution

Answer to Problem 90P

$σf(a)=Q2πaL$

$σf(b)=−Q2πbL$

Explanation of Solution

Given information:

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used:

Charge density:

$σ=Q2πrL$

Where, Q is the charge, r is the radius and L is the length of the cylinder.

Calculation:

Surface charge density is:

$σf(a)=Q2πaL$

$σf(b)=−Q2πbL$

Conclusion:

The density of the free charge of on the inner cylinder and the outer cylinder are:

$σf(a)=Q2πaL$

$σf(b)=−Q2πbL$

(c)

To determine

ToCalculate: The bound charge density $σb$ on the inner cylindrical surface of the dielectric and on the outer cylindrical surface of the dielectric.

(c)

Expert Solution

Answer to Problem 90P

$σb(a)=−Q(κ−1)2πaLκ$

$σb(b)=−Q(κ−1)2πbLκ$

Explanation of Solution

Given information :

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used :

Bound charge can be expressed as:

$Qb=Q(κ−1)κ$

Where, $κ$ is the dielectric constant.

Bound charge density:

$σb=QbA$

Where, A is the area.

Calculation:

$Qb(a)=−Q(κ−1)κ$

$Qb(b)=Q(κ−1)κ$

$σb(a)=Qb(a)Aσb(a)= −Q( κ−1 )κ2πaLσb(a)=−Q( κ−1)2πaLκ$

$σb(b)=Qb(b)Aσb(b)= −Q( κ−1 )κ2πbLσb(b)=−Q( κ−1)2πbLκ$

Conclusion:

The bound charge density $σb$ on the inner cylindrical surface of the dielectric and on the outer cylindrical surface of the dielectric.

$σb(a)=−Q(κ−1)2πaLκ$

$σb(b)=−Q(κ−1)2πbLκ$

(d)

To determine

ToCalculate: The total stored energy.

(d)

Expert Solution

Answer to Problem 90P

$U=kQ2ln(b/a)κL$

Explanation of Solution

Given information:

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used:

The energy stored in the capacitor:

$U=12QV$

Where, Q is the charge and V is the electric potential.

Calculation:

The potential difference between the cylinders is $V=2kQln(b/a)κL$ .

The total stored energy in terms of the charge stored and the potential difference between the cylinders:

$U=12QVU=12Q[2kQln( b/a )κL]U=kQ2ln( b/a )κL$

Conclusion:

The total stored energy is,

$U=kQ2ln(b/a)κL$

(e)

To determine

ToCalculate: Mechanical work that is required to remove the dielectric cylindrical shell.

(e)

Expert Solution

Answer to Problem 90P

$W=kQ2(κ−1)ln(b/a)κL$

Explanation of Solution

Given information:

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used:

Work done in terms of potential energy of the system:

$W=ΔU$

Calculation:

The work required to remove the dielectric cylindrical shell in terms of the change in the potential energy of the system:

$W=ΔUW=U′−U$

The potential energy of the system with the dielectric shell in place is,

$U′=κU$

$W=κU−UW=U(κ−1)W=kQ2( κ−1)ln( b/a )κL$

Conclusion:

Mechanical work that is required to remove the dielectric cylindrical shell is $W=kQ2(κ−1)ln(b/a)κL$ .

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Answer 2

Question

Chapter 24, Problem 90P

(a)

To determine

ToCalculate: The potential difference between the cylinders.

(a)

Expert Solution

Answer to Problem 90P

$V=2kQln(b/a)κL$

Explanation of Solution

Given information :

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used :

Electric potential:

$V=QC$

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

$C=2π∈0κLln(b/a)$

Where, $κ$ is the dielectric constant, $∈0$ is the absolute permittivity and L is the length of the capacitor, a and b are radii of the

Calculation:

The potential difference between the cylinders to their charge and capacitance is,

$V=QC$

The capacitance of a cylindrical capacitor as a function of its radii a and b and length L:

$C=2π∈0κLln(b/a)$

$V=Qln( b/a )2π∈0κLV=2kQln( b/a )κL$

Conclusion:

The potential difference between the cylinders is $V=2kQln(b/a)κL$ .

(b)

To determine

ToCalculate: The density of the free charge of on the inner cylinder and the outer cylinder.

(b)

Expert Solution

Answer to Problem 90P

$σf(a)=Q2πaL$

$σf(b)=−Q2πbL$

Explanation of Solution

Given information:

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used:

Charge density:

$σ=Q2πrL$

Where, Q is the charge, r is the radius and L is the length of the cylinder.

Calculation:

Surface charge density is:

$σf(a)=Q2πaL$

$σf(b)=−Q2πbL$

Conclusion:

The density of the free charge of on the inner cylinder and the outer cylinder are:

$σf(a)=Q2πaL$

$σf(b)=−Q2πbL$

(c)

To determine

ToCalculate: The bound charge density $σb$ on the inner cylindrical surface of the dielectric and on the outer cylindrical surface of the dielectric.

(c)

Expert Solution

Answer to Problem 90P

$σb(a)=−Q(κ−1)2πaLκ$

$σb(b)=−Q(κ−1)2πbLκ$

Explanation of Solution

Given information :

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used :

Bound charge can be expressed as:

$Qb=Q(κ−1)κ$

Where, $κ$ is the dielectric constant.

Bound charge density:

$σb=QbA$

Where, A is the area.

Calculation:

$Qb(a)=−Q(κ−1)κ$

$Qb(b)=Q(κ−1)κ$

$σb(a)=Qb(a)Aσb(a)= −Q( κ−1 )κ2πaLσb(a)=−Q( κ−1)2πaLκ$

$σb(b)=Qb(b)Aσb(b)= −Q( κ−1 )κ2πbLσb(b)=−Q( κ−1)2πbLκ$

Conclusion:

The bound charge density $σb$ on the inner cylindrical surface of the dielectric and on the outer cylindrical surface of the dielectric.

$σb(a)=−Q(κ−1)2πaLκ$

$σb(b)=−Q(κ−1)2πbLκ$

(d)

To determine

ToCalculate: The total stored energy.

(d)

Expert Solution

Answer to Problem 90P

$U=kQ2ln(b/a)κL$

Explanation of Solution

Given information:

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used:

The energy stored in the capacitor:

$U=12QV$

Where, Q is the charge and V is the electric potential.

Calculation:

The potential difference between the cylinders is $V=2kQln(b/a)κL$ .

The total stored energy in terms of the charge stored and the potential difference between the cylinders:

$U=12QVU=12Q[2kQln( b/a )κL]U=kQ2ln( b/a )κL$

Conclusion:

The total stored energy is,

$U=kQ2ln(b/a)κL$

(e)

To determine

ToCalculate: Mechanical work that is required to remove the dielectric cylindrical shell.

(e)

Expert Solution

Answer to Problem 90P

$W=kQ2(κ−1)ln(b/a)κL$

Explanation of Solution

Given information:

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used:

Work done in terms of potential energy of the system:

$W=ΔU$

Calculation:

The work required to remove the dielectric cylindrical shell in terms of the change in the potential energy of the system:

$W=ΔUW=U′−U$

The potential energy of the system with the dielectric shell in place is,

$U′=κU$

$W=κU−UW=U(κ−1)W=kQ2( κ−1)ln( b/a )κL$

Conclusion:

Mechanical work that is required to remove the dielectric cylindrical shell is $W=kQ2(κ−1)ln(b/a)κL$ .

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Answer 3

Question

Chapter 24, Problem 90P

(a)

To determine

ToCalculate: The potential difference between the cylinders.

(a)

Expert Solution

Answer to Problem 90P

$V=2kQln(b/a)κL$

Explanation of Solution

Given information :

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used :

Electric potential:

$V=QC$

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

$C=2π∈0κLln(b/a)$

Where, $κ$ is the dielectric constant, $∈0$ is the absolute permittivity and L is the length of the capacitor, a and b are radii of the

Calculation:

The potential difference between the cylinders to their charge and capacitance is,

$V=QC$

The capacitance of a cylindrical capacitor as a function of its radii a and b and length L:

$C=2π∈0κLln(b/a)$

$V=Qln( b/a )2π∈0κLV=2kQln( b/a )κL$

Conclusion:

The potential difference between the cylinders is $V=2kQln(b/a)κL$ .

(b)

To determine

ToCalculate: The density of the free charge of on the inner cylinder and the outer cylinder.

(b)

Expert Solution

Answer to Problem 90P

$σf(a)=Q2πaL$

$σf(b)=−Q2πbL$

Explanation of Solution

Given information:

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used:

Charge density:

$σ=Q2πrL$

Where, Q is the charge, r is the radius and L is the length of the cylinder.

Calculation:

Surface charge density is:

$σf(a)=Q2πaL$

$σf(b)=−Q2πbL$

Conclusion:

The density of the free charge of on the inner cylinder and the outer cylinder are:

$σf(a)=Q2πaL$

$σf(b)=−Q2πbL$

(c)

To determine

ToCalculate: The bound charge density $σb$ on the inner cylindrical surface of the dielectric and on the outer cylindrical surface of the dielectric.

(c)

Expert Solution

Answer to Problem 90P

$σb(a)=−Q(κ−1)2πaLκ$

$σb(b)=−Q(κ−1)2πbLκ$

Explanation of Solution

Given information :

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used :

Bound charge can be expressed as:

$Qb=Q(κ−1)κ$

Where, $κ$ is the dielectric constant.

Bound charge density:

$σb=QbA$

Where, A is the area.

Calculation:

$Qb(a)=−Q(κ−1)κ$

$Qb(b)=Q(κ−1)κ$

$σb(a)=Qb(a)Aσb(a)= −Q( κ−1 )κ2πaLσb(a)=−Q( κ−1)2πaLκ$

$σb(b)=Qb(b)Aσb(b)= −Q( κ−1 )κ2πbLσb(b)=−Q( κ−1)2πbLκ$

Conclusion:

The bound charge density $σb$ on the inner cylindrical surface of the dielectric and on the outer cylindrical surface of the dielectric.

$σb(a)=−Q(κ−1)2πaLκ$

$σb(b)=−Q(κ−1)2πbLκ$

(d)

To determine

ToCalculate: The total stored energy.

(d)

Expert Solution

Answer to Problem 90P

$U=kQ2ln(b/a)κL$

Explanation of Solution

Given information:

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used:

The energy stored in the capacitor:

$U=12QV$

Where, Q is the charge and V is the electric potential.

Calculation:

The potential difference between the cylinders is $V=2kQln(b/a)κL$ .

The total stored energy in terms of the charge stored and the potential difference between the cylinders:

$U=12QVU=12Q[2kQln( b/a )κL]U=kQ2ln( b/a )κL$

Conclusion:

The total stored energy is,

$U=kQ2ln(b/a)κL$

(e)

To determine

ToCalculate: Mechanical work that is required to remove the dielectric cylindrical shell.

(e)

Expert Solution

Answer to Problem 90P

$W=kQ2(κ−1)ln(b/a)κL$

Explanation of Solution

Given information:

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used:

Work done in terms of potential energy of the system:

$W=ΔU$

Calculation:

The work required to remove the dielectric cylindrical shell in terms of the change in the potential energy of the system:

$W=ΔUW=U′−U$

The potential energy of the system with the dielectric shell in place is,

$U′=κU$

$W=κU−UW=U(κ−1)W=kQ2( κ−1)ln( b/a )κL$

Conclusion:

Mechanical work that is required to remove the dielectric cylindrical shell is $W=kQ2(κ−1)ln(b/a)κL$ .

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Answer 4

Question

Chapter 24, Problem 90P

(a)

To determine

ToCalculate: The potential difference between the cylinders.

(a)

Expert Solution

Answer to Problem 90P

$V=2kQln(b/a)κL$

Explanation of Solution

Given information :

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used :

Electric potential:

$V=QC$

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

$C=2π∈0κLln(b/a)$

Where, $κ$ is the dielectric constant, $∈0$ is the absolute permittivity and L is the length of the capacitor, a and b are radii of the

Calculation:

The potential difference between the cylinders to their charge and capacitance is,

$V=QC$

The capacitance of a cylindrical capacitor as a function of its radii a and b and length L:

$C=2π∈0κLln(b/a)$

$V=Qln( b/a )2π∈0κLV=2kQln( b/a )κL$

Conclusion:

The potential difference between the cylinders is $V=2kQln(b/a)κL$ .

(b)

To determine

ToCalculate: The density of the free charge of on the inner cylinder and the outer cylinder.

(b)

Expert Solution

Answer to Problem 90P

$σf(a)=Q2πaL$

$σf(b)=−Q2πbL$

Explanation of Solution

Given information:

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used:

Charge density:

$σ=Q2πrL$

Where, Q is the charge, r is the radius and L is the length of the cylinder.

Calculation:

Surface charge density is:

$σf(a)=Q2πaL$

$σf(b)=−Q2πbL$

Conclusion:

The density of the free charge of on the inner cylinder and the outer cylinder are:

$σf(a)=Q2πaL$

$σf(b)=−Q2πbL$

(c)

To determine

ToCalculate: The bound charge density $σb$ on the inner cylindrical surface of the dielectric and on the outer cylindrical surface of the dielectric.

(c)

Expert Solution

Answer to Problem 90P

$σb(a)=−Q(κ−1)2πaLκ$

$σb(b)=−Q(κ−1)2πbLκ$

Explanation of Solution

Given information :

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used :

Bound charge can be expressed as:

$Qb=Q(κ−1)κ$

Where, $κ$ is the dielectric constant.

Bound charge density:

$σb=QbA$

Where, A is the area.

Calculation:

$Qb(a)=−Q(κ−1)κ$

$Qb(b)=Q(κ−1)κ$

$σb(a)=Qb(a)Aσb(a)= −Q( κ−1 )κ2πaLσb(a)=−Q( κ−1)2πaLκ$

$σb(b)=Qb(b)Aσb(b)= −Q( κ−1 )κ2πbLσb(b)=−Q( κ−1)2πbLκ$

Conclusion:

The bound charge density $σb$ on the inner cylindrical surface of the dielectric and on the outer cylindrical surface of the dielectric.

$σb(a)=−Q(κ−1)2πaLκ$

$σb(b)=−Q(κ−1)2πbLκ$

(d)

To determine

ToCalculate: The total stored energy.

(d)

Expert Solution

Answer to Problem 90P

$U=kQ2ln(b/a)κL$

Explanation of Solution

Given information:

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used:

The energy stored in the capacitor:

$U=12QV$

Where, Q is the charge and V is the electric potential.

Calculation:

The potential difference between the cylinders is $V=2kQln(b/a)κL$ .

The total stored energy in terms of the charge stored and the potential difference between the cylinders:

$U=12QVU=12Q[2kQln( b/a )κL]U=kQ2ln( b/a )κL$

Conclusion:

The total stored energy is,

$U=kQ2ln(b/a)κL$

(e)

To determine

ToCalculate: Mechanical work that is required to remove the dielectric cylindrical shell.

(e)

Expert Solution

Answer to Problem 90P

$W=kQ2(κ−1)ln(b/a)κL$

Explanation of Solution

Given information:

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used:

Work done in terms of potential energy of the system:

$W=ΔU$

Calculation:

The work required to remove the dielectric cylindrical shell in terms of the change in the potential energy of the system:

$W=ΔUW=U′−U$

The potential energy of the system with the dielectric shell in place is,

$U′=κU$

$W=κU−UW=U(κ−1)W=kQ2( κ−1)ln( b/a )κL$

Conclusion:

Mechanical work that is required to remove the dielectric cylindrical shell is $W=kQ2(κ−1)ln(b/a)κL$ .

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Answer 5

Chapter 24, Problem 90P

(a)

To determine

ToCalculate: The potential difference between the cylinders.

(a)

Expert Solution

Answer to Problem 90P

$V=2kQln(b/a)κL$

Explanation of Solution

Given information :

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used :

Electric potential:

$V=QC$

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

$C=2π∈0κLln(b/a)$

Where, $κ$ is the dielectric constant, $∈0$ is the absolute permittivity and L is the length of the capacitor, a and b are radii of the

Calculation:

The potential difference between the cylinders to their charge and capacitance is,

$V=QC$

The capacitance of a cylindrical capacitor as a function of its radii a and b and length L:

$C=2π∈0κLln(b/a)$

$V=Qln( b/a )2π∈0κLV=2kQln( b/a )κL$

Conclusion:

The potential difference between the cylinders is $V=2kQln(b/a)κL$ .

(b)

To determine

ToCalculate: The density of the free charge of on the inner cylinder and the outer cylinder.

(b)

Expert Solution

Answer to Problem 90P

$σf(a)=Q2πaL$

$σf(b)=−Q2πbL$

Explanation of Solution

Given information:

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used:

Charge density:

$σ=Q2πrL$

Where, Q is the charge, r is the radius and L is the length of the cylinder.

Calculation:

Surface charge density is:

$σf(a)=Q2πaL$

$σf(b)=−Q2πbL$

Conclusion:

The density of the free charge of on the inner cylinder and the outer cylinder are:

$σf(a)=Q2πaL$

$σf(b)=−Q2πbL$

(c)

To determine

ToCalculate: The bound charge density $σb$ on the inner cylindrical surface of the dielectric and on the outer cylindrical surface of the dielectric.

(c)

Expert Solution

Answer to Problem 90P

$σb(a)=−Q(κ−1)2πaLκ$

$σb(b)=−Q(κ−1)2πbLκ$

Explanation of Solution

Given information :

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used :

Bound charge can be expressed as:

$Qb=Q(κ−1)κ$

Where, $κ$ is the dielectric constant.

Bound charge density:

$σb=QbA$

Where, A is the area.

Calculation:

$Qb(a)=−Q(κ−1)κ$

$Qb(b)=Q(κ−1)κ$

$σb(a)=Qb(a)Aσb(a)= −Q( κ−1 )κ2πaLσb(a)=−Q( κ−1)2πaLκ$

$σb(b)=Qb(b)Aσb(b)= −Q( κ−1 )κ2πbLσb(b)=−Q( κ−1)2πbLκ$

Conclusion:

The bound charge density $σb$ on the inner cylindrical surface of the dielectric and on the outer cylindrical surface of the dielectric.

$σb(a)=−Q(κ−1)2πaLκ$

$σb(b)=−Q(κ−1)2πbLκ$

(d)

To determine

ToCalculate: The total stored energy.

(d)

Expert Solution

Answer to Problem 90P

$U=kQ2ln(b/a)κL$

Explanation of Solution

Given information:

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used:

The energy stored in the capacitor:

$U=12QV$

Where, Q is the charge and V is the electric potential.

Calculation:

The potential difference between the cylinders is $V=2kQln(b/a)κL$ .

The total stored energy in terms of the charge stored and the potential difference between the cylinders:

$U=12QVU=12Q[2kQln( b/a )κL]U=kQ2ln( b/a )κL$

Conclusion:

The total stored energy is,

$U=kQ2ln(b/a)κL$

(e)

To determine

ToCalculate: Mechanical work that is required to remove the dielectric cylindrical shell.

(e)

Expert Solution

Answer to Problem 90P

$W=kQ2(κ−1)ln(b/a)κL$

Explanation of Solution

Given information:

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used:

Work done in terms of potential energy of the system:

$W=ΔU$

Calculation:

The work required to remove the dielectric cylindrical shell in terms of the change in the potential energy of the system:

$W=ΔUW=U′−U$

The potential energy of the system with the dielectric shell in place is,

$U′=κU$

$W=κU−UW=U(κ−1)W=kQ2( κ−1)ln( b/a )κL$

Conclusion:

Mechanical work that is required to remove the dielectric cylindrical shell is $W=kQ2(κ−1)ln(b/a)κL$ .

Answer 6

Chapter 24, Problem 90P

(a)

To determine

ToCalculate: The potential difference between the cylinders.

(a)

Expert Solution

Answer to Problem 90P

$V=2kQln(b/a)κL$

Explanation of Solution

Given information :

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used :

Electric potential:

$V=QC$

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

$C=2π∈0κLln(b/a)$

Where, $κ$ is the dielectric constant, $∈0$ is the absolute permittivity and L is the length of the capacitor, a and b are radii of the

Calculation:

The potential difference between the cylinders to their charge and capacitance is,

$V=QC$

The capacitance of a cylindrical capacitor as a function of its radii a and b and length L:

$C=2π∈0κLln(b/a)$

$V=Qln( b/a )2π∈0κLV=2kQln( b/a )κL$

Conclusion:

The potential difference between the cylinders is $V=2kQln(b/a)κL$ .

Answer 7

Chapter 24, Problem 90P

(a)

To determine

ToCalculate: The potential difference between the cylinders.

Answer 8

(a)

Expert Solution

Answer to Problem 90P

$V=2kQln(b/a)κL$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given information :

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used :

Electric potential:

$V=QC$

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

$C=2π∈0κLln(b/a)$

Where, $κ$ is the dielectric constant, $∈0$ is the absolute permittivity and L is the length of the capacitor, a and b are radii of the

Calculation:

The potential difference between the cylinders to their charge and capacitance is,

$V=QC$

The capacitance of a cylindrical capacitor as a function of its radii a and b and length L:

$C=2π∈0κLln(b/a)$

$V=Qln( b/a )2π∈0κLV=2kQln( b/a )κL$

Conclusion:

The potential difference between the cylinders is $V=2kQln(b/a)κL$ .

Answer 13

(b)

To determine

ToCalculate: The density of the free charge of on the inner cylinder and the outer cylinder.

(b)

Expert Solution

Answer to Problem 90P

$σf(a)=Q2πaL$

$σf(b)=−Q2πbL$

Explanation of Solution

Given information:

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used:

Charge density:

$σ=Q2πrL$

Where, Q is the charge, r is the radius and L is the length of the cylinder.

Calculation:

Surface charge density is:

$σf(a)=Q2πaL$

$σf(b)=−Q2πbL$

Conclusion:

The density of the free charge of on the inner cylinder and the outer cylinder are:

$σf(a)=Q2πaL$

$σf(b)=−Q2πbL$

Answer 14

(b)

To determine

ToCalculate: The density of the free charge of on the inner cylinder and the outer cylinder.

Answer 15

(b)

Expert Solution

Answer to Problem 90P

$σf(a)=Q2πaL$

$σf(b)=−Q2πbL$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given information:

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used:

Charge density:

$σ=Q2πrL$

Where, Q is the charge, r is the radius and L is the length of the cylinder.

Calculation:

Surface charge density is:

$σf(a)=Q2πaL$

$σf(b)=−Q2πbL$

Conclusion:

The density of the free charge of on the inner cylinder and the outer cylinder are:

$σf(a)=Q2πaL$

$σf(b)=−Q2πbL$

Answer 20

(c)

To determine

ToCalculate: The bound charge density $σb$ on the inner cylindrical surface of the dielectric and on the outer cylindrical surface of the dielectric.

(c)

Expert Solution

Answer to Problem 90P

$σb(a)=−Q(κ−1)2πaLκ$

$σb(b)=−Q(κ−1)2πbLκ$

Explanation of Solution

Given information :

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used :

Bound charge can be expressed as:

$Qb=Q(κ−1)κ$

Where, $κ$ is the dielectric constant.

Bound charge density:

$σb=QbA$

Where, A is the area.

Calculation:

$Qb(a)=−Q(κ−1)κ$

$Qb(b)=Q(κ−1)κ$

$σb(a)=Qb(a)Aσb(a)= −Q( κ−1 )κ2πaLσb(a)=−Q( κ−1)2πaLκ$

$σb(b)=Qb(b)Aσb(b)= −Q( κ−1 )κ2πbLσb(b)=−Q( κ−1)2πbLκ$

Conclusion:

The bound charge density $σb$ on the inner cylindrical surface of the dielectric and on the outer cylindrical surface of the dielectric.

$σb(a)=−Q(κ−1)2πaLκ$

$σb(b)=−Q(κ−1)2πbLκ$

Answer 21

(c)

To determine

ToCalculate: The bound charge density $σb$ on the inner cylindrical surface of the dielectric and on the outer cylindrical surface of the dielectric.

Answer 22

(c)

Expert Solution

Answer to Problem 90P

$σb(a)=−Q(κ−1)2πaLκ$

$σb(b)=−Q(κ−1)2πbLκ$

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given information :

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used :

Bound charge can be expressed as:

$Qb=Q(κ−1)κ$

Where, $κ$ is the dielectric constant.

Bound charge density:

$σb=QbA$

Where, A is the area.

Calculation:

$Qb(a)=−Q(κ−1)κ$

$Qb(b)=Q(κ−1)κ$

$σb(a)=Qb(a)Aσb(a)= −Q( κ−1 )κ2πaLσb(a)=−Q( κ−1)2πaLκ$

$σb(b)=Qb(b)Aσb(b)= −Q( κ−1 )κ2πbLσb(b)=−Q( κ−1)2πbLκ$

Conclusion:

The bound charge density $σb$ on the inner cylindrical surface of the dielectric and on the outer cylindrical surface of the dielectric.

$σb(a)=−Q(κ−1)2πaLκ$

$σb(b)=−Q(κ−1)2πbLκ$

Answer 27

(d)

To determine

ToCalculate: The total stored energy.

(d)

Expert Solution

Answer to Problem 90P

$U=kQ2ln(b/a)κL$

Explanation of Solution

Given information:

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used:

The energy stored in the capacitor:

$U=12QV$

Where, Q is the charge and V is the electric potential.

Calculation:

The potential difference between the cylinders is $V=2kQln(b/a)κL$ .

The total stored energy in terms of the charge stored and the potential difference between the cylinders:

$U=12QVU=12Q[2kQln( b/a )κL]U=kQ2ln( b/a )κL$

Conclusion:

The total stored energy is,

$U=kQ2ln(b/a)κL$

Answer 28

(d)

To determine

ToCalculate: The total stored energy.

Answer 29

(d)

Expert Solution

Answer to Problem 90P

$U=kQ2ln(b/a)κL$

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given information:

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used:

The energy stored in the capacitor:

$U=12QV$

Where, Q is the charge and V is the electric potential.

Calculation:

The potential difference between the cylinders is $V=2kQln(b/a)κL$ .

The total stored energy in terms of the charge stored and the potential difference between the cylinders:

$U=12QVU=12Q[2kQln( b/a )κL]U=kQ2ln( b/a )κL$

Conclusion:

The total stored energy is,

$U=kQ2ln(b/a)κL$

Answer 34

(e)

To determine

ToCalculate: Mechanical work that is required to remove the dielectric cylindrical shell.

(e)

Expert Solution

Answer to Problem 90P

$W=kQ2(κ−1)ln(b/a)κL$

Explanation of Solution

Given information:

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used:

Work done in terms of potential energy of the system:

$W=ΔU$

Calculation:

The work required to remove the dielectric cylindrical shell in terms of the change in the potential energy of the system:

$W=ΔUW=U′−U$

The potential energy of the system with the dielectric shell in place is,

$U′=κU$

$W=κU−UW=U(κ−1)W=kQ2( κ−1)ln( b/a )κL$

Conclusion:

Mechanical work that is required to remove the dielectric cylindrical shell is $W=kQ2(κ−1)ln(b/a)κL$ .

Answer 35

(e)

To determine

ToCalculate: Mechanical work that is required to remove the dielectric cylindrical shell.

Answer 36

(e)

Expert Solution

Answer to Problem 90P

$W=kQ2(κ−1)ln(b/a)κL$

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Given information:

Radii of coaxial conducting thin cylindrical shells = $a & b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $−Q$

Formula used:

Work done in terms of potential energy of the system:

$W=ΔU$

Calculation:

The work required to remove the dielectric cylindrical shell in terms of the change in the potential energy of the system:

$W=ΔUW=U′−U$

The potential energy of the system with the dielectric shell in place is,

$U′=κU$

$W=κU−UW=U(κ−1)W=kQ2( κ−1)ln( b/a )κL$

Conclusion:

Mechanical work that is required to remove the dielectric cylindrical shell is $W=kQ2(κ−1)ln(b/a)κL$ .

Answer 41

Ch. 24 - Prob. 1P Ch. 24 - Prob. 2P Ch. 24 - Prob. 3P Ch. 24 - Prob. 4P Ch. 24 - Prob. 5P Ch. 24 - Prob. 6P Ch. 24 - Prob. 7P Ch. 24 - Prob. 8P Ch. 24 - Prob. 9P Ch. 24 - Prob. 10P

ToCalculate : The potential difference between the cylinders.

Concept explainers

Videos

ToCalculate: The potential difference between the cylinders.

Answer to Problem 90P

Explanation of Solution

ToCalculate: The density of the free charge of on the inner cylinder and the outer cylinder.

Answer to Problem 90P

Explanation of Solution

ToCalculate: The bound charge density $σb$ on the inner cylindrical surface of the dielectric and on the outer cylindrical surface of the dielectric.

Answer to Problem 90P

Explanation of Solution

ToCalculate: The total stored energy.

Answer to Problem 90P

Explanation of Solution

ToCalculate: Mechanical work that is required to remove the dielectric cylindrical shell.

Answer to Problem 90P

Explanation of Solution

Want to see more full solutions like this?

Chapter 24 Solutions

ToCalculate : The potential difference between the cylinders.

Concept explainers

Videos

ToCalculate: The potential difference between the cylinders.

Answer to Problem 90P

Explanation of Solution

ToCalculate: The density of the free charge of on the inner cylinder and the outer cylinder.

Answer to Problem 90P

Explanation of Solution

ToCalculate: The bound charge density σb<m:math><m:mrow><m:msub><m:mi>σ</m:mi><m:mi>b</m:mi></m:msub></m:mrow></m:math> on the inner cylindrical surface of the dielectric and on the outer cylindrical surface of the dielectric.

Answer to Problem 90P

Explanation of Solution

ToCalculate: The total stored energy.

Answer to Problem 90P

Explanation of Solution

ToCalculate: Mechanical work that is required to remove the dielectric cylindrical shell.

Answer to Problem 90P

Explanation of Solution

Want to see more full solutions like this?

Chapter 24 Solutions

ToCalculate: The bound charge density $σb$ on the inner cylindrical surface of the dielectric and on the outer cylindrical surface of the dielectric.