Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 22, Problem 77P

a.

To determine

The magnitude and the direction of the electric field at x=0.40m .

The electric field E=203.6kN/C pointing at θ=56.3° from the x axis.

Given:

  Physics for Scientists and Engineers, Chapter 22, Problem 77P , additional homework tip  1

The charges are placed as shown in the figure. The first plane at y=0.6m . The second plane at x=1.0m . The spherical shell centered at the intersection of the planes at point (1.0,-0.6) in the x-y plane.

The surface charge densities are

  σ1=3.0μnC/m2

  σ2=2.0μnC/m2

  σ3=3.0μC/m3

Formula Used:

Electric field

  E=σ2εor^

E is the electric field.

  σ is the surface charge density.

  r^ is the unit vector in the direction normal to the charged plane.

  εo is the permittivity of free space.

The resultant electric field at point is E=E1+E2+Eshpere

Calculations:

The resultant electric field at point is E=E1+E2+Eshpere

  E=σ2εor^

Electric field at point 1 due to sphere.

As the point is inside the sphere the electric field is zero.

  Eshpere=0

Electric field at point 1 due to plane 1

Substituting values

  E1=3.0μnC/m22(8.85×1012C2/N.m2)j ^

  E1=(169.4kN/C)j ^

The electric field at point 1 due to plane 2.

  E2=-2.0μnC/m22(8.85×1012C2/N.m2)( i ^)

  E2=(112.9kN/C) i ^

Substituting in the equation

The resultant electric field at point is E=E1+E2+Eshpere

Substituting

  E=(112.9kN/C) i ^ +(169.4kN/C)j ^ +0

  E=(112.9kN/C) i ^ +(169.4kN/C)j ^

The magnitude of the electric field is

   E=(x2+y2)

  E=(112.9kN/C)2+(169.4kN/C)2

  E=203.6kN/C2

Direction:

  θ=tan1(yx)

  θ=tan1(169.4kN/C112.9kN/C)=56.31°

Conclusion:

The electric field E=203.6kN/C pointing at θ=56.3° from the x axis.

b.

The magnitude and the direction of the electric field at x=2.50m .

The electric field E=263kN/C pointing at θ=153° from the x axis.

Given:

  Physics for Scientists and Engineers, Chapter 22, Problem 77P , additional homework tip  2

The charges are placed as shown in the figure. The first plane at y=0.6m . The second plane at x=1.0m . The spherical shell centered at the intersection of the planes at point (1.0,-0.6) in the x-y plane.

The surface charge densities are

  σ1=3.0μnC/m2

  σ2=2.0μnC/m2

  σ3=3.0μC/m3

Formula Used:

Electric field

  E=σ2εor^

E is the electric field.

  σ is the surface charge density.

  r^ is the unit vector in the direction normal to the charged plane.

  εo is the permittivity of free space.

The resultant electric field at point is E=E1+E2+Eshpere

Calculations:

The resultant electric field at point is E=E1+E2+Eshpere

  E=σ2εor^

Electric field at point 1 due to sphere.

  E =(kQspherer2)r^ where Q is the charge in the sphere.

Where r^ is a unit vector pointing from (1.0m,-0.6m) to (2.50m,0)

  Qsphere=σAsphere

  =4πσR2

  Qsphere=4π(3.0μC/m2)(1.0m)2

  Qsphere=37.30μC

  r^=0.9285 i ^+0.3714j ^

  Esphere=(8.988×109N.m2/C2)(37.70μC)(1.616m)2r^

  Esphere=(129.8kN/C)(0.9285 i ^+0.3714j ^)

  Esphere=(120.5kN/C) i ^+(-48.22kN/C)j ^

Electric field at point 1 due to plane 1

Substituting values in the formula E=σ2εor^

  E1=3.0μnC/m22(8.85×1012C2/N.m2)j ^

  E1=(169.4kN/C)j ^

The electric field at point 1 due to plane 2.

  E2=-2.0μnC/m22(8.85×1012C2/N.m2)( i ^)

  E2=(112.9kN/C) i ^

Substituting in the equation

The resultant electric field at point is E=E1+E2+Eshpere

Substituting

  E=(112.9kN/C) i ^ +(169.4kN/C)j ^ +(120.5kN/C) i ^+(-48.22kN/C)j ^

  E=(233.5kN/C) i ^ +(121.2kN/C)j ^

The magnitude of the electric field is

  E=(x2+y2)

  E=(233.5kN/C)2+(121.2kN/C)2

  E=263kN/C

Direction:

  θ=tan1(yx)

  θ=tan1(121.2kN/C-233.5kN/C)=153°

Conclusion:

The electric field E=263kN/C pointing at θ=153° from the x axis.

a.

Expert Solution
Check Mark

Answer to Problem 77P

The electric field E=203.6kN/C pointing at θ=56.3° from the x axis.

Explanation of Solution

Given:

  Physics for Scientists and Engineers, Chapter 22, Problem 77P , additional homework tip  3

The charges are placed as shown in the figure. The first plane at y=0.6m . The second plane at x=1.0m . The spherical shell centered at the intersection of the planes at point (1.0,-0.6) in the x-y plane.

The surface charge densities are

  σ1=3.0μnC/m2

  σ2=2.0μnC/m2

  σ3=3.0μC/m3

Formula Used:

Electric field

  E=σ2εor^

E is the electric field.

  σ is the surface charge density.

  r^ is the unit vector in the direction normal to the charged plane.

  εo is the permittivity of free space.

The resultant electric field at point is E=E1+E2+Eshpere

Calculations:

The resultant electric field at point is E=E1+E2+Eshpere

  E=σ2εor^

Electric field at point 1 due to sphere.

As the point is inside the sphere the electric field is zero.

  Eshpere=0

Electric field at point 1 due to plane 1

Substituting values

  E1=3.0μnC/m22(8.85×1012C2/N.m2)j ^

  E1=(169.4kN/C)j ^

The electric field at point 1 due to plane 2.

  E2=-2.0μnC/m22(8.85×1012C2/N.m2)( i ^)

  E2=(112.9kN/C) i ^

Substituting in the equation

The resultant electric field at point is E=E1+E2+Eshpere

Substituting

  E=(112.9kN/C) i ^ +(169.4kN/C)j ^ +0

  E=(112.9kN/C) i ^ +(169.4kN/C)j ^

The magnitude of the electric field is

   E=(x2+y2)

  E=(112.9kN/C)2+(169.4kN/C)2

  E=203.6kN/C2

Direction:

  θ=tan1(yx)

  θ=tan1(169.4kN/C112.9kN/C)=56.31°

Conclusion:

The electric field E=203.6kN/C pointing at θ=56.3° from the x axis.

b.

To determine

The magnitude and the direction of the electric field at x=2.50m .

The electric field E=263kN/C pointing at θ=153° from the x axis.

Given:

  Physics for Scientists and Engineers, Chapter 22, Problem 77P , additional homework tip  4

The charges are placed as shown in the figure. The first plane at y=0.6m . The second plane at x=1.0m . The spherical shell centered at the intersection of the planes at point (1.0,-0.6) in the x-y plane.

The surface charge densities are

  σ1=3.0μnC/m2

  σ2=2.0μnC/m2

  σ3=3.0μC/m3

Formula Used:

Electric field

  E=σ2εor^

E is the electric field.

  σ is the surface charge density.

  r^ is the unit vector in the direction normal to the charged plane.

  εo is the permittivity of free space.

The resultant electric field at point is E=E1+E2+Eshpere

Calculations:

The resultant electric field at point is E=E1+E2+Eshpere

  E=σ2εor^

Electric field at point 1 due to sphere.

  E =(kQspherer2)r^ where Q is the charge in the sphere.

Where r^ is a unit vector pointing from (1.0m,-0.6m) to (2.50m,0)

  Qsphere=σAsphere

  =4πσR2

  Qsphere=4π(3.0μC/m2)(1.0m)2

  Qsphere=37.30μC

  r^=0.9285 i ^+0.3714j ^

  Esphere=(8.988×109N.m2/C2)(37.70μC)(1.616m)2r^

  Esphere=(129.8kN/C)(0.9285 i ^+0.3714j ^)

  Esphere=(120.5kN/C) i ^+(-48.22kN/C)j ^

Electric field at point 1 due to plane 1

Substituting values in the formula E=σ2εor^

  E1=3.0μnC/m22(8.85×1012C2/N.m2)j ^

  E1=(169.4kN/C)j ^

The electric field at point 1 due to plane 2.

  E2=-2.0μnC/m22(8.85×1012C2/N.m2)( i ^)

  E2=(112.9kN/C) i ^

Substituting in the equation

The resultant electric field at point is E=E1+E2+Eshpere

Substituting

  E=(112.9kN/C) i ^ +(169.4kN/C)j ^ +(120.5kN/C) i ^+(-48.22kN/C)j ^

  E=(233.5kN/C) i ^ +(121.2kN/C)j ^

The magnitude of the electric field is

  E=(x2+y2)

  E=(233.5kN/C)2+(121.2kN/C)2

  E=263kN/C

Direction:

  θ=tan1(yx)

  θ=tan1(121.2kN/C-233.5kN/C)=153°

Conclusion:

The electric field E=263kN/C pointing at θ=153° from the x axis.

b.

Expert Solution
Check Mark

Answer to Problem 77P

The electric field E=263kN/C pointing at θ=153° from the x axis.

Explanation of Solution

Given:

  Physics for Scientists and Engineers, Chapter 22, Problem 77P , additional homework tip  5

The charges are placed as shown in the figure. The first plane at y=0.6m . The second plane at x=1.0m . The spherical shell centered at the intersection of the planes at point (1.0,-0.6) in the x-y plane.

The surface charge densities are

  σ1=3.0μnC/m2

  σ2=2.0μnC/m2

  σ3=3.0μC/m3

Formula Used:

Electric field

  E=σ2εor^

E is the electric field.

  σ is the surface charge density.

  r^ is the unit vector in the direction normal to the charged plane.

  εo is the permittivity of free space.

The resultant electric field at point is E=E1+E2+Eshpere

Calculations:

The resultant electric field at point is E=E1+E2+Eshpere

  E=σ2εor^

Electric field at point 1 due to sphere.

  E =(kQspherer2)r^ where Q is the charge in the sphere.

Where r^ is a unit vector pointing from (1.0m,-0.6m) to (2.50m,0)

  Qsphere=σAsphere

  =4πσR2

  Qsphere=4π(3.0μC/m2)(1.0m)2

  Qsphere=37.30μC

  r^=0.9285 i ^+0.3714j ^

  Esphere=(8.988×109N.m2/C2)(37.70μC)(1.616m)2r^

  Esphere=(129.8kN/C)(0.9285 i ^+0.3714j ^)

  Esphere=(120.5kN/C) i ^+(-48.22kN/C)j ^

Electric field at point 1 due to plane 1

Substituting values in the formula E=σ2εor^

  E1=3.0μnC/m22(8.85×1012C2/N.m2)j ^

  E1=(169.4kN/C)j ^

The electric field at point 1 due to plane 2.

  E2=-2.0μnC/m22(8.85×1012C2/N.m2)( i ^)

  E2=(112.9kN/C) i ^

Substituting in the equation

The resultant electric field at point is E=E1+E2+Eshpere

Substituting

  E=(112.9kN/C) i ^ +(169.4kN/C)j ^ +(120.5kN/C) i ^+(-48.22kN/C)j ^

  E=(233.5kN/C) i ^ +(121.2kN/C)j ^

The magnitude of the electric field is

  E=(x2+y2)

  E=(233.5kN/C)2+(121.2kN/C)2

  E=263kN/C

Direction:

  θ=tan1(yx)

  θ=tan1(121.2kN/C-233.5kN/C)=153°

Conclusion:

The electric field E=263kN/C pointing at θ=153° from the x axis.

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Chapter 22 Solutions

Physics for Scientists and Engineers

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