COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 16, Problem 89QAP
To determine

(a)

An expression for the electric field in the radial region ri.

Expert Solution
Check Mark

Answer to Problem 89QAP

An expression for the electric field in the radial region ri is, Er<Ri=0

Explanation of Solution

Given:

Radius of inner sphere, Ri

Surface Charge density for inner sphere, +σi

Radius of outer sphere, Ro

Surface Charge density for outer sphere, σo

Formula used:

The electric force is given by,
  E=σ2ε0

Where,
  E =Electric field
  σ = Surface charges density
  ε0 =Permittivity of free space

Calculation:

The sphere is conducting. The charge enclosed by the sphere is zero. Thus, the electric field inside the sphere is zero.
  Er<Ri=0

To determine

(b)

An expression for the electric field in the radial region Rio.

Expert Solution
Check Mark

Answer to Problem 89QAP

An expression for the electric field in the radial region Rio is, ERi<r<Ro=σiRi2ε0r2 and directed outside from the sphere.

Explanation of Solution

Given:

Radius of inner sphere, Ri

Surface Charge density for inner sphere, +σi

Radius of outer sphere, Ro

Surface Charge density for outer sphere, σo

Formula used:

The electric force is given by,
  E=Q4πε0r2

Where,
  E =Electric field
  Q = Charge
  ε0 =Permittivity of free space
r=Distance

Calculation:

The electric force is given by,
  E=Q4πε0r2

But, Q=σiA

  E=σi4πRi24πε0r2ERi<r<Ro=σiRi2ε0r2

It is directed outside from the sphere.

To determine

(c)

An expression for the electric field in the radial region r>Ro.

Expert Solution
Check Mark

Answer to Problem 89QAP

An expression for the electric field in the radial region r>Rois, E=σiRi2σoRo2ε0r2

Explanation of Solution

Given:

Radius of inner sphere, Ri

Surface Charge density for inner sphere, +σi

Radius of outer sphere, Ro

Surface Charge density for outer sphere, σo

Formula used:

The electric force is given by,
  E=Q4πε0r2

Where,
  E =Electric field
  Q = Charge
  ε0 =Permittivity of free space
r=DistanceCalculation:

The electric field is given by,
  E=Q4πε0r2

But, Q=σiA

  Ei=σi4πRi24πε0r2Ei=σiRi2ε0r2

And
  Eo=σi4πRo24πε0r2Eo=σiRo2ε0r2

The electric field is given by,
  E=EiEoE=σiRi2ε0r2σoRo2ε0r2E=σiRi2σoRo2ε0r2

If the value of σiRi2σoRo2>1 The field is directed outwards.

If the value of σiRi2σoRo2<1 The field is directed inwards.

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Chapter 16 Solutions

COLLEGE PHYSICS

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