With what velocity does the electron strike the positive plate if a cylindrical capacitor is made of two coaxial conductors—the outer one’s diameter is 20.2 mm and the inner one’s diameter is 0.2 mm; the inner conductor is 1000-V positive with respect to the outer conductor?

Question

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Answer 1

Question

Chapter 2, Problem 2.18P

(a)

To determine

With what velocity does the electron strike the positive plate if a cylindrical capacitor is made of two coaxial conductors—the outer one’s diameter is 20.2 mm and the inner one’s diameter is 0.2 mm; the inner conductor is 1000-V positive with respect to the outer conductor?

(a)

Expert Solution

Explanation of Solution

Given info:.

V=1000Vd=0.01mm=9.11×10−31kgq=1.602×10−19C.

Formula used:.

Kinetic energy

KE=12mv2.

Potential energy

PE=qV.

Calculation:.

We have,

12mv2=qVv=2qVm=2×1.602×10−19×10009.11×10−31=1.9×107m/s.

Conclusion:.

Velocity=1.9×107m/s

(b)

To determine

How long does it take the electron to travel the 1-cm distance? Compare the results to those of Problem 2.17.

(b)

Expert Solution

Explanation of Solution

Given info:.

V=1000Vd=0.01mm=9.11×10−31kgq=1.602×10−19Crb=20.22×10−3m=1.01×10−2mra=1×10−4m.

Formula used:.

Force

F = ma = qE

Calculation:.

We have force,

F=ma=qEa=qEm.

For a cylindrical capacitor,

E=1rVln(rbra).

We have,

v=at=drdt=qmVrln(rbra)trdr=qmVln(rbra)tdt∫rarbrdr=qmVln(rbra)∫0ttdt12(rb2−ra2)=qmVln(rbra)t22t=(rb2−ra2)qmVln(rbra)=((1.01×10−2)2−(1×10−4)2)1.602×10−199.11×10−311000ln(1.01×10−21×10−4)=1.65×10−9s.

This value is higher than the value obtained in Problem 2.17

Conclusion:.

Time taken=1.65×10−9s

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Answer 2

Question

Chapter 2, Problem 2.18P

(a)

To determine

With what velocity does the electron strike the positive plate if a cylindrical capacitor is made of two coaxial conductors—the outer one’s diameter is 20.2 mm and the inner one’s diameter is 0.2 mm; the inner conductor is 1000-V positive with respect to the outer conductor?

(a)

Expert Solution

Explanation of Solution

Given info:.

V=1000Vd=0.01mm=9.11×10−31kgq=1.602×10−19C.

Formula used:.

Kinetic energy

KE=12mv2.

Potential energy

PE=qV.

Calculation:.

We have,

12mv2=qVv=2qVm=2×1.602×10−19×10009.11×10−31=1.9×107m/s.

Conclusion:.

Velocity=1.9×107m/s

(b)

To determine

How long does it take the electron to travel the 1-cm distance? Compare the results to those of Problem 2.17.

(b)

Expert Solution

Explanation of Solution

Given info:.

V=1000Vd=0.01mm=9.11×10−31kgq=1.602×10−19Crb=20.22×10−3m=1.01×10−2mra=1×10−4m.

Formula used:.

Force

F = ma = qE

Calculation:.

We have force,

F=ma=qEa=qEm.

For a cylindrical capacitor,

E=1rVln(rbra).

We have,

v=at=drdt=qmVrln(rbra)trdr=qmVln(rbra)tdt∫rarbrdr=qmVln(rbra)∫0ttdt12(rb2−ra2)=qmVln(rbra)t22t=(rb2−ra2)qmVln(rbra)=((1.01×10−2)2−(1×10−4)2)1.602×10−199.11×10−311000ln(1.01×10−21×10−4)=1.65×10−9s.

This value is higher than the value obtained in Problem 2.17

Conclusion:.

Time taken=1.65×10−9s

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Answer 3

Question

Chapter 2, Problem 2.18P

(a)

To determine

With what velocity does the electron strike the positive plate if a cylindrical capacitor is made of two coaxial conductors—the outer one’s diameter is 20.2 mm and the inner one’s diameter is 0.2 mm; the inner conductor is 1000-V positive with respect to the outer conductor?

(a)

Expert Solution

Explanation of Solution

Given info:.

V=1000Vd=0.01mm=9.11×10−31kgq=1.602×10−19C.

Formula used:.

Kinetic energy

KE=12mv2.

Potential energy

PE=qV.

Calculation:.

We have,

12mv2=qVv=2qVm=2×1.602×10−19×10009.11×10−31=1.9×107m/s.

Conclusion:.

Velocity=1.9×107m/s

(b)

To determine

How long does it take the electron to travel the 1-cm distance? Compare the results to those of Problem 2.17.

(b)

Expert Solution

Explanation of Solution

Given info:.

V=1000Vd=0.01mm=9.11×10−31kgq=1.602×10−19Crb=20.22×10−3m=1.01×10−2mra=1×10−4m.

Formula used:.

Force

F = ma = qE

Calculation:.

We have force,

F=ma=qEa=qEm.

For a cylindrical capacitor,

E=1rVln(rbra).

We have,

v=at=drdt=qmVrln(rbra)trdr=qmVln(rbra)tdt∫rarbrdr=qmVln(rbra)∫0ttdt12(rb2−ra2)=qmVln(rbra)t22t=(rb2−ra2)qmVln(rbra)=((1.01×10−2)2−(1×10−4)2)1.602×10−199.11×10−311000ln(1.01×10−21×10−4)=1.65×10−9s.

This value is higher than the value obtained in Problem 2.17

Conclusion:.

Time taken=1.65×10−9s

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Answer 4

Question

Chapter 2, Problem 2.18P

(a)

To determine

With what velocity does the electron strike the positive plate if a cylindrical capacitor is made of two coaxial conductors—the outer one’s diameter is 20.2 mm and the inner one’s diameter is 0.2 mm; the inner conductor is 1000-V positive with respect to the outer conductor?

(a)

Expert Solution

Explanation of Solution

Given info:.

V=1000Vd=0.01mm=9.11×10−31kgq=1.602×10−19C.

Formula used:.

Kinetic energy

KE=12mv2.

Potential energy

PE=qV.

Calculation:.

We have,

12mv2=qVv=2qVm=2×1.602×10−19×10009.11×10−31=1.9×107m/s.

Conclusion:.

Velocity=1.9×107m/s

(b)

To determine

How long does it take the electron to travel the 1-cm distance? Compare the results to those of Problem 2.17.

(b)

Expert Solution

Explanation of Solution

Given info:.

V=1000Vd=0.01mm=9.11×10−31kgq=1.602×10−19Crb=20.22×10−3m=1.01×10−2mra=1×10−4m.

Formula used:.

Force

F = ma = qE

Calculation:.

We have force,

F=ma=qEa=qEm.

For a cylindrical capacitor,

E=1rVln(rbra).

We have,

v=at=drdt=qmVrln(rbra)trdr=qmVln(rbra)tdt∫rarbrdr=qmVln(rbra)∫0ttdt12(rb2−ra2)=qmVln(rbra)t22t=(rb2−ra2)qmVln(rbra)=((1.01×10−2)2−(1×10−4)2)1.602×10−199.11×10−311000ln(1.01×10−21×10−4)=1.65×10−9s.

This value is higher than the value obtained in Problem 2.17

Conclusion:.

Time taken=1.65×10−9s

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Answer 5

Chapter 2, Problem 2.18P

(a)

To determine

With what velocity does the electron strike the positive plate if a cylindrical capacitor is made of two coaxial conductors—the outer one’s diameter is 20.2 mm and the inner one’s diameter is 0.2 mm; the inner conductor is 1000-V positive with respect to the outer conductor?

(a)

Expert Solution

Explanation of Solution

Given info:.

V=1000Vd=0.01mm=9.11×10−31kgq=1.602×10−19C.

Formula used:.

Kinetic energy

KE=12mv2.

Potential energy

PE=qV.

Calculation:.

We have,

12mv2=qVv=2qVm=2×1.602×10−19×10009.11×10−31=1.9×107m/s.

Conclusion:.

Velocity=1.9×107m/s

(b)

To determine

How long does it take the electron to travel the 1-cm distance? Compare the results to those of Problem 2.17.

(b)

Expert Solution

Explanation of Solution

Given info:.

V=1000Vd=0.01mm=9.11×10−31kgq=1.602×10−19Crb=20.22×10−3m=1.01×10−2mra=1×10−4m.

Formula used:.

Force

F = ma = qE

Calculation:.

We have force,

F=ma=qEa=qEm.

For a cylindrical capacitor,

E=1rVln(rbra).

We have,

v=at=drdt=qmVrln(rbra)trdr=qmVln(rbra)tdt∫rarbrdr=qmVln(rbra)∫0ttdt12(rb2−ra2)=qmVln(rbra)t22t=(rb2−ra2)qmVln(rbra)=((1.01×10−2)2−(1×10−4)2)1.602×10−199.11×10−311000ln(1.01×10−21×10−4)=1.65×10−9s.

This value is higher than the value obtained in Problem 2.17

Conclusion:.

Time taken=1.65×10−9s

Answer 6

Chapter 2, Problem 2.18P

(a)

To determine

With what velocity does the electron strike the positive plate if a cylindrical capacitor is made of two coaxial conductors—the outer one’s diameter is 20.2 mm and the inner one’s diameter is 0.2 mm; the inner conductor is 1000-V positive with respect to the outer conductor?

(a)

Expert Solution

Explanation of Solution

Given info:.

V=1000Vd=0.01mm=9.11×10−31kgq=1.602×10−19C.

Formula used:.

Kinetic energy

KE=12mv2.

Potential energy

PE=qV.

Calculation:.

We have,

12mv2=qVv=2qVm=2×1.602×10−19×10009.11×10−31=1.9×107m/s.

Conclusion:.

Velocity=1.9×107m/s

Answer 7

Chapter 2, Problem 2.18P

(a)

To determine

With what velocity does the electron strike the positive plate if a cylindrical capacitor is made of two coaxial conductors—the outer one’s diameter is 20.2 mm and the inner one’s diameter is 0.2 mm; the inner conductor is 1000-V positive with respect to the outer conductor?

Answer 8

(a)

Expert Solution

Explanation of Solution

Given info:.

V=1000Vd=0.01mm=9.11×10−31kgq=1.602×10−19C.

Formula used:.

Kinetic energy

KE=12mv2.

Potential energy

PE=qV.

Calculation:.

We have,

12mv2=qVv=2qVm=2×1.602×10−19×10009.11×10−31=1.9×107m/s.

Conclusion:.

Velocity=1.9×107m/s

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

How long does it take the electron to travel the 1-cm distance? Compare the results to those of Problem 2.17.

(b)

Expert Solution

Explanation of Solution

Given info:.

V=1000Vd=0.01mm=9.11×10−31kgq=1.602×10−19Crb=20.22×10−3m=1.01×10−2mra=1×10−4m.

Formula used:.

Force

F = ma = qE

Calculation:.

We have force,

F=ma=qEa=qEm.

For a cylindrical capacitor,

E=1rVln(rbra).

We have,

v=at=drdt=qmVrln(rbra)trdr=qmVln(rbra)tdt∫rarbrdr=qmVln(rbra)∫0ttdt12(rb2−ra2)=qmVln(rbra)t22t=(rb2−ra2)qmVln(rbra)=((1.01×10−2)2−(1×10−4)2)1.602×10−199.11×10−311000ln(1.01×10−21×10−4)=1.65×10−9s.

This value is higher than the value obtained in Problem 2.17

Conclusion:.

Time taken=1.65×10−9s

Answer 13

(b)

To determine

How long does it take the electron to travel the 1-cm distance? Compare the results to those of Problem 2.17.

Answer 14

(b)

Expert Solution

Explanation of Solution

Given info:.

V=1000Vd=0.01mm=9.11×10−31kgq=1.602×10−19Crb=20.22×10−3m=1.01×10−2mra=1×10−4m.

Formula used:.

Force

F = ma = qE

Calculation:.

We have force,

F=ma=qEa=qEm.

For a cylindrical capacitor,

E=1rVln(rbra).

We have,

v=at=drdt=qmVrln(rbra)trdr=qmVln(rbra)tdt∫rarbrdr=qmVln(rbra)∫0ttdt12(rb2−ra2)=qmVln(rbra)t22t=(rb2−ra2)qmVln(rbra)=((1.01×10−2)2−(1×10−4)2)1.602×10−199.11×10−311000ln(1.01×10−21×10−4)=1.65×10−9s.