EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 24, Problem 44P

(a)

To determine

ToFind: Expressions for the electric field and energy density as a function of the distance R from the axis.

(a)

Expert Solution
Check Mark

Answer to Problem 44P

  ER>R2=0

  uR>R2=0

Explanation of Solution

Given information :

Radius of the wire = R1

Length of the wire = L

Charge = +Q

Inner radius of cylindrical shell = R2

Length of the cylindrical shell = L

Charge = Q

Formula used :

From Gauss’s law:

  E(2πRL)=Qinside0

  E=λ2π0R

Where, E electric field, Qinside is the enclosed charge, R is the radius, L is the length, λ is the charge density and 0 is the absolute permittivity.

Energy density is given by:

  u=120E2

Calculation:

Using Gauss’s law, find the electric field for cylindrical surfaces of radii R<R1,R1<R<R2 , and R>R2

  EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 24, Problem 44P

For R<R1 :

  ER<R1(2πRL)=Qinside0=0

  ER<R1=0

  E=0 for R<R1

  uR<R1=0

For R1<R<R2 :

  ER1<R<R2(2πRL)=Qinside0=λL0

  λ= The linear charge density

  ER1<R<R2=λ2π0R=2kQRL

The energy density in the region R1<R<R2 is

  uR1<R<R2=120ER1<R<R22

  uR1<R<R2=120( 2kλ R)2uR1<R<R2=120( 2kQ RL)2uR1<R<R2=2k20Q2R2L2

For R>R2 :

  ER>R2(2πRL)=Qinside0=0

  ER>R2=0

Because E=0 for R>R2

  uR>R2=0

Conclusion:

Electric field

  ER>R2=0

Energy density

  uR>R2=0

(b)

To determine

ToCalculate: The amount ofenergy that resides in a region between the conductors that has a radius R , a thickness dR , and a volume 2πrLdR .

(b)

Expert Solution
Check Mark

Answer to Problem 44P

  dU=kQ2RLdR

Explanation of Solution

Given information:

Radius of the wire = R1

Length of the wire = L

Charge = +Q

Inner radius of cylindrical shell = R2

Length of the cylindrical shell = L

Charge = Q

Formula used:

The energy residing in a cylindrical shell between the conductors of radius R , thickness dR , and volume 2πRLdR :

  dU=2πRLu(R)dR

Calculation:

  dU=2πRLu(R)dRdU=2πRL( 2 k 2 0 Q 2 R 2 L 2 )drdU=kQ2RLdR

Conclusion:

The amount ofenergy is:

  dU=kQ2RLdR

(c)

To determine

To Calculate: The total energy stored in the capacitor.

(c)

Expert Solution
Check Mark

Answer to Problem 44P

  U=kQ2Lln(R2R1)

Explanation of Solution

Given information:

Radius of the wire = R1

Length of the wire = L

Charge = +Q

Inner radius of cylindrical shell = R2

Length of the cylindrical shell = L

Charge = Q

Formula used:

Energy stored in the capacitor:

  U=12CV2

Where, C is the capacitance and V is the potential difference.

Calculation:

Integrate dU from R=R1 to R=R2 to obtain:

  U=kQ2LR1R2dRR=kQ2Lln( R 2 R 1 )

  U=12CV2U=12Q2CU=Q22( 2π 0 L ln( R 2 R 1 ) )U=kQ2Lln( R 2 R 1 )

Conclusion:

The total energy stored in the capacitor is:

  U=kQ2Lln(R2R1)

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Chapter 24 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

Ch. 24 - Prob. 11PCh. 24 - Prob. 12PCh. 24 - Prob. 13PCh. 24 - Prob. 14PCh. 24 - Prob. 15PCh. 24 - Prob. 16PCh. 24 - Prob. 17PCh. 24 - Prob. 18PCh. 24 - Prob. 19PCh. 24 - Prob. 20PCh. 24 - Prob. 21PCh. 24 - Prob. 22PCh. 24 - Prob. 23PCh. 24 - Prob. 24PCh. 24 - Prob. 25PCh. 24 - Prob. 26PCh. 24 - Prob. 27PCh. 24 - Prob. 28PCh. 24 - Prob. 29PCh. 24 - Prob. 30PCh. 24 - Prob. 31PCh. 24 - Prob. 32PCh. 24 - Prob. 33PCh. 24 - Prob. 34PCh. 24 - Prob. 35PCh. 24 - Prob. 36PCh. 24 - Prob. 37PCh. 24 - Prob. 38PCh. 24 - Prob. 39PCh. 24 - Prob. 40PCh. 24 - Prob. 41PCh. 24 - Prob. 42PCh. 24 - Prob. 43PCh. 24 - Prob. 44PCh. 24 - Prob. 45PCh. 24 - Prob. 46PCh. 24 - Prob. 47PCh. 24 - Prob. 48PCh. 24 - Prob. 49PCh. 24 - Prob. 50PCh. 24 - Prob. 51PCh. 24 - Prob. 52PCh. 24 - Prob. 53PCh. 24 - Prob. 54PCh. 24 - Prob. 55PCh. 24 - Prob. 56PCh. 24 - Prob. 57PCh. 24 - Prob. 58PCh. 24 - Prob. 59PCh. 24 - Prob. 60PCh. 24 - Prob. 61PCh. 24 - Prob. 62PCh. 24 - Prob. 63PCh. 24 - Prob. 64PCh. 24 - Prob. 65PCh. 24 - Prob. 66PCh. 24 - Prob. 67PCh. 24 - Prob. 68PCh. 24 - Prob. 69PCh. 24 - Prob. 70PCh. 24 - Prob. 71PCh. 24 - Prob. 72PCh. 24 - Prob. 73PCh. 24 - Prob. 74PCh. 24 - Prob. 75PCh. 24 - Prob. 76PCh. 24 - Prob. 77PCh. 24 - Prob. 78PCh. 24 - Prob. 79PCh. 24 - Prob. 80PCh. 24 - Prob. 81PCh. 24 - Prob. 82PCh. 24 - Prob. 83PCh. 24 - Prob. 84PCh. 24 - Prob. 85PCh. 24 - Prob. 86PCh. 24 - Prob. 87PCh. 24 - Prob. 88PCh. 24 - Prob. 89PCh. 24 - Prob. 90PCh. 24 - Prob. 91PCh. 24 - Prob. 92PCh. 24 - Prob. 93PCh. 24 - Prob. 94P
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