COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 16, Problem 66QAP
To determine

(a)

The electric flux out of each face.

Expert Solution
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Answer to Problem 66QAP

The electric flux out of each face is, ϕI=0Vm

  ϕII=48Vm

  ϕIII=48Vm

  ϕIV=48Vm

  ϕV=0Vm

Explanation of Solution

Given:

Height of prism, h=40cm

Depth of prism, d=30cm

Length of prism, l=80cm

Electric field, E=500N/C

Formula used:

The electric flux is given by,
  ϕ=EAcosθ

Where,
  ϕ =Electric flux
  E = Electric field
  A =Area of prism

Calculation:

The electric flux is given by,
  ϕ=EAcosθ

For surface I: θ=90°C

Thus,
  ϕI=EAcos90ϕI=0Vm

For surface II: θ=0°C

Thus,
  ϕII=EAcos0ϕII=500×0.4×0.3×0.8×1VmϕII=48Vm

For surface III: θ=0°C

Thus,
  ϕIII=EAcos0ϕIII=500×0.4×0.3×0.8×1VmϕIII=48Vm

For surface IV: θ=0°C

Thus,
  ϕIV=EAcos0ϕIV=500×0.4×0.3×0.8×1VmϕIV=48Vm

For surface V: θ=90°C

Thus,
  ϕV=EAcos90ϕV=0Vm

To determine

(b)

The net electric flux.

Expert Solution
Check Mark

Answer to Problem 66QAP

The net electric flux is, Φ=164Vm

Explanation of Solution

Height of prism, h=40cm

Depth of prism, d=30cm

Length of prism, l=80cm

Electric field, E=500N/C

Formula used:

The electric flux is given by,
  ϕ=EAcosθ

Where,
  ϕ =Electric flux
  E = Electric field
  A =Area of prism

Calculation:

The net electric flux is,
  Φ=ϕI+ϕII+ϕIII+ϕIV+ϕVΦ=0+48+48+48+0Φ=164Vm

To determine

(c)

The electric field the prism enclosed.

Expert Solution
Check Mark

Answer to Problem 66QAP

The electric field enclosed by the prism is, Enet=199500N/C

Explanation of Solution

Given:

Height of prism, h=40cm

Depth of prism, d=30cm

Length of prism, l=80cm

Electric field, E=500N/C

Radius of solid sphere, R

Charge, Q=2.00μC

Formula used:

The electric flux is given by,
  E=QAε0

Where,
  E =Electric flux
  Q = Charge
A=Surface area of prism

Calculation:

The electric flux is given by,
  EQ=QAε0

Area of the prism is,
  A=3(l×h)+(h×d)A=3(0.8×0.4)+(0.4×0.3)A=1.08m2

Thus,
  EQ=2.00× 10 61.08×8.85× 10 12EQ=2.09×105N/C

So, electric field of the prism is,

Enet=E+EQEnet=5002.00×105Enet=199500N/C

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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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