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 86P

(a)

To determine

ToCalculate: The electric field in the regions a<r<12(a + b) and 12(a + b)<r<b .

(a)

Expert Solution
Check Mark

Answer to Problem 86P

  E1=kQκ1r2E2=kQκ2r2

Explanation of Solution

Given information :

The radius of spherical capacitor =a

Charge =+Q

Inner radius of conducting spherical shell =b

Charge =Q

Dielectric constants =κ1&κ2

Formula used :

Electric field:

  E=kQκr2

Where, k is the Coulomb’s constant, Q is the charge, κ is the dielectric constant and r is the distance between the plates.

Calculation:

The electric fields in the regions is a<r<12(a + b) and 12(a + b)<r<b is:

  E1=kQκ1r2E2=kQκ2r2

Conclusion:

The electric field in the regions a<r<12(a + b) and 12(a + b)<r<b is,

  E1=kQκ1r2E2=kQκ2r2

(b)

To determine

ToCalculate: TheIntegrate expression of dV=Edl .

(b)

Expert Solution
Check Mark

Answer to Problem 86P

  |V|=kQ(ba)(a+b)κ1a+κ2bκ1κ2ab

Explanation of Solution

Given information:

The radius of spherical capacitor =a

Charge =+Q

Inner radius of conducting spherical shell =b

Charge =Q

Dielectric constants =κ1&κ2

Formula used:

  |V|=|V1V2|

Calculation:

Because the electric field is directed radially away from the center, the potential of the outer spherical shell is less than the potential of the inner sphere. Hence, the difference in potential between the spheres is given by:

  |V|=|V1V2||V|=|a12(a+b)E1dr(12(a+b)bE2dr)||V|=|kQκ1a12(a+b)r2dr+kQκ212(a+b)br2dr|

  |V|=|[kQκ11r]a12(a+b)+kQκ21r|12(a+b)b|V|=|kQκ1(2a+b1a)+kQκ2(1b2a+b)||V|=|kQ(ab)(a+b)κ1a+κ2bκ1κ2ab||V|=kQ(ba)(a+b)κ1a+κ2bκ1κ2ab

Conclusion:

TheIntegrate expression of dV=Edl is |V|=kQ(ba)(a+b)κ1a+κ2bκ1κ2ab .

(c)

To determine

ToCalculate: An expression for the capacitance of the system.

(c)

Expert Solution
Check Mark

Answer to Problem 86P

  C=QkQ(ab)a+bκ1b+κ2bκ1κ2ab

Explanation of Solution

Given information :

The radius of spherical capacitor =a

Charge =+Q

Inner radius of conducting spherical shell =b

Charge =Q

Dielectric constants =κ1&κ2

Formula used :

The capacitance of the capacitor:

  C=Q|V|

Q is the charge stored and V is the voltage.

Calculation:

  C=Q|V|C=QkQ(ab)a+bκ1b+κ2bκ1κ2ab

Conclusion:

An expression for the capacitance of the system is,

  C=QkQ(ab)a+bκ1b+κ2bκ1κ2ab

(d)

To determine

ToShow:The answer from Part (c) simplifies to the expected one if the κ1 equals κ2 .

(d)

Expert Solution
Check Mark

Explanation of Solution

Given information:

The radius of spherical capacitor =a

Charge =+Q

Inner radius of conducting spherical shell =b

Charge =Q

Dielectric constants =κ1&κ2

Formula used:

The charge stored in the capacitor:

  Q=CV

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

Calculation:

  |V|=kQ(ba)(a+b)κ1a+κ2bκ1κ2ab

  Q=CVC=QV

  κ1=κ2=κ

  C=κκab(a+b)k(ba)(κa+κb)C=κabk(ba)C=4π0κabba

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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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