FlipIt for College Physics (Algebra Version - Six Months Access)
FlipIt for College Physics (Algebra Version - Six Months Access)
17th Edition
ISBN: 9781319032432
Author: Todd Ruskell
Publisher: W.H. Freeman & Co
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Chapter 16, Problem 66QAP
To determine

(a)

The electric flux out of each face.

Expert Solution
Check Mark

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

FlipIt for College Physics (Algebra Version - Six Months Access)

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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