Physics for Scientists and Engineers, Vol. 1
Physics for Scientists and Engineers, Vol. 1
6th Edition
ISBN: 9781429201322
Author: Paul A. Tipler, Gene Mosca
Publisher: Macmillan Higher Education
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Chapter 22, Problem 78P
To determine

The magnitude and the direction of the electric field in the z plane at x=1.50m, y=0.50m .

The electric field at point P is E=241kN/C pointing at θ=220° .

Given:

  Physics for Scientists and Engineers, Vol. 1, Chapter 22, Problem 78P , additional homework tip  1

The charges are placed as shown in the figure. The first plane at x=2.0m . The line charge passing through the origin at an angle of 45° with the x axis in the x-y plane. The spherical shell centered at point (1.5m,0.5m) in the x-y plane.

The charge densities are

  σ=2μnC/m2

  λ=4μnC/m2

  ρ=6.0μC/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+Eline+Eshpere

Calculations:

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

  E=σ2εor^

Electric field at point P due to sphere.

  Esphere=4π3kr'ρr^'

  r' Represents the distance from the center of the sphere to P. constructing a Gaussian sphere of radius r' centered at (1,0)

From the figure

  r'=0.5 i ^ -0.5j ^

  r'=(0.5)2+(0.5)2=0.707m

  r^'=0.707 i ^ -0.707j ^

Substituting the values

  Esphere=4π3(8.988×109N.m2/C2)(0.707m)(-6.00μC/m3)[0.707 i ^ +0.707j ^]

  Esphere=(112.9kN/C) i ^+(112.9kN/C)j ^

  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 P due to plane 1

Substituting values in the formula E=σ2εor^

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

  Eplane=(112.9kN/C) i ^

The electric field at point P due to line charge.

  Eline=2kλrr^

From the figure

  r'=0.5 i ^ -0.5j ^r'=(0.5)2+(0.5)2=0.707mr ^ '=0.707 i ^ -0.707j ^

Substituting the values

  Eline=2(8.988×109N.m2/C2)(4.0μC/)0.707m(0.707 i ^ -0.707j ^)

  Eline=(71.9kN/C) i ^+(71.9kN/C)j ^

Substituting in the equation

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

  E=(71.9kN/C) i ^+(71.9kN/C)j ^ +(112.9kN/C) i ^+(112.9kN/C)j ^ +(112.9kN/C) i ^

  E=(154kN/C) i ^ +(184.9kN/C)j ^

The magnitude of the electric field is

  vector magnitude E=(x2+y2)

  E=(154.0kN/C)2+(184.9kN/C)2

  E=241kN/C

  vector direction θ=tan1(xy)

  θ=tan1(-154kN/C-184.9kN/C)=220°

Conclusion:

The electric field E=241kN/C pointing at θ=220° .

Expert Solution & Answer
Check Mark

Answer to Problem 78P

The electric field at point P is E=241kN/C pointing at θ=220° .

Explanation of Solution

Given:

  Physics for Scientists and Engineers, Vol. 1, Chapter 22, Problem 78P , additional homework tip  2

The charges are placed as shown in the figure. The first plane at x=2.0m . The line charge passing through the origin at an angle of 45° with the x axis in the x-y plane. The spherical shell centered at point (1.5m,0.5m) in the x-y plane.

The charge densities are

  σ=2μnC/m2

  λ=4μnC/m2

  ρ=6.0μC/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+Eline+Eshpere

Calculations:

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

  E=σ2εor^

Electric field at point P due to sphere.

  Esphere=4π3kr'ρr^'

  r' Represents the distance from the center of the sphere to P. constructing a Gaussian sphere of radius r' centered at (1,0)

From the figure

  r'=0.5 i ^ -0.5j ^

  r'=(0.5)2+(0.5)2=0.707m

  r^'=0.707 i ^ -0.707j ^

Substituting the values

  Esphere=4π3(8.988×109N.m2/C2)(0.707m)(-6.00μC/m3)[0.707 i ^ +0.707j ^]

  Esphere=(112.9kN/C) i ^+(112.9kN/C)j ^

  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 P due to plane 1

Substituting values in the formula E=σ2εor^

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

  Eplane=(112.9kN/C) i ^

The electric field at point P due to line charge.

  Eline=2kλrr^

From the figure

  r'=0.5 i ^ -0.5j ^r'=(0.5)2+(0.5)2=0.707mr ^ '=0.707 i ^ -0.707j ^

Substituting the values

  Eline=2(8.988×109N.m2/C2)(4.0μC/)0.707m(0.707 i ^ -0.707j ^)

  Eline=(71.9kN/C) i ^+(71.9kN/C)j ^

Substituting in the equation

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

  E=(71.9kN/C) i ^+(71.9kN/C)j ^ +(112.9kN/C) i ^+(112.9kN/C)j ^ +(112.9kN/C) i ^

  E=(154kN/C) i ^ +(184.9kN/C)j ^

The magnitude of the electric field is

  vector magnitude E=(x2+y2)

  E=(154.0kN/C)2+(184.9kN/C)2

  E=241kN/C

  vector direction θ=tan1(xy)

  θ=tan1(-154kN/C-184.9kN/C)=220°

Conclusion:

The electric field E=241kN/C pointing at θ=220° .

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Chapter 22 Solutions

Physics for Scientists and Engineers, Vol. 1

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