Physics for Scientists and Engineers with Modern Physics, Technology Update
Physics for Scientists and Engineers with Modern Physics, Technology Update
9th Edition
ISBN: 9781305401969
Author: SERWAY, Raymond A.; Jewett, John W.
Publisher: Cengage Learning
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Chapter 25, Problem 71CP

(a)

To determine

The proof for V=pkecosθr2 at the point P.

(a)

Expert Solution
Check Mark

Answer to Problem 71CP

The electric potential at the point P is V=pkecosθr2.

Explanation of Solution

The following figure shows the diagram of the dipoles and the point P.

Physics for Scientists and Engineers with Modern Physics, Technology Update, Chapter 25, Problem 71CP

Figure-(1)

The dipole moment’s magnitude is given as.

  p=2qa

Here, p is the magnitude of the dipole moment, q is the charge and 2a is the separating distance between the charges, r1 is the distance between the positive charge and the point P and r2 is the distance between the point P and the negative charge.

Write the equation for the electric potential at the point P due to the dipole moment.

  V=ke[qr1qr2]=keq[1r11r2]                                                                                                       (I)

Here, V is the potential, ke is the Coulomb’s constant,

Write the equation for the distance between the positive charge and the point P.

  r21=r2(12arcosθ)                                                                                              (II)

Write the equation for the distance between the negative charge and the point P.

  r22=r2(1+2arcosθ)                                                                                           (III)

Rearrange the equation (II) and (III) to calculate 1r1 and 1r2.

  1r1=1r(12arcosθ)12 and 1r2=1r(1+2arcosθ)12

Expand the above equation binomially and neglect the higher terms to calculate 1r1 and 1r2.

  1r1=1r(1(12)2arcosθ) and 1r2=1r(1+(12)2arcosθ)

  1r1=1r(1+arcosθ) and 1r2=1r(1arcosθ)

Substitute 1r(1+arcosθ) for 1r1 and 1r(1arcosθ) for 1r2 in the equation (I) to calculate the electric potential.

  V=keq[1r(1+arcosθ)1r(1arcosθ)]=keqr[2arcosθ]=ke2qar2cosθ

Substitute p for 2aq in the above equation to calculate the electric potential.

  V=pkecosθr2

Therefore, the electric potential at the point P is V=pkecosθr2.

(b)

To determine

The electric field’s radial component Er and the perpendicular component Eθ.

(b)

Expert Solution
Check Mark

Answer to Problem 71CP

The electric field’s radial component is Er=2pkecosθr3 and the perpendicular component is Eθ=pkesinθr3.

Explanation of Solution

Write the equation for the radial component of the electric field.

  Er=dVdr                                                                                                               (IV)

Here, Er is the radial component of the electric field.

Write the equation for the perpendicular component of the electric field.

  Eθ=(1r)dVdθ                                                                                                        (V)

Here, Eθ is the perpendicular component of the electric field.

Conclusion:

Substitute pkecosθr2 for V in the equation (IV) to calculate the radial component of the electric field.

  Er=d(pkecosθr2)dr=2pkecosθr3

Substitute pkecosθr2 for V in the equation (V) to calculate the perpendicular component of the electric field.

  Eθ=(1r)d(pkecosθr2)dθ=(1r)pke(sinθ)r3=pkesinθr3

Therefore, the electric field’s radial component is Er=2pkecosθr3 and the perpendicular component is Eθ=pkesinθr3.

(c)

To determine

Whether the results for Er and Eθ hold good when θ=0° and 90°.

(c)

Expert Solution
Check Mark

Answer to Problem 71CP

Yes, the results for Er and Eθ hold good when θ=0° and θ=90°.

Explanation of Solution

Write the equation for Er.

  Er=2pkecosθr3                                                                                                     (VI)

Write the equation for Eθ.

  Eθ=pkesinθr3                                                                                                       (VII)

Conclusion:

Substitute 0° for θ in the equation (VI) to calculate Er.

  Er=2pkecos0°r3=2pker3

Substitute 0° for θ in the equation (VII) to calculate Eθ.

  Eθ=pkesin0°r3=0

Substitute 90° for θ in the equation (VI) to calculate Er.

  Er=2pkecos90°r3=0

Substitute 90° for θ in the equation (VII) to calculate Eθ.

    Eθ=pkesin90°r3=pker3

Therefore, the results for Er and Eθ hold good when θ=0° and θ=90°.

(d)

To determine

Whether the results for Er and Eθ hold good when r=0.

(d)

Expert Solution
Check Mark

Answer to Problem 71CP

No, the results for Er and Eθ do not hold good when r=0.

Explanation of Solution

Write the equation for Er.

  Er=2pkecosθr3                                                                                                    (VIII)

Write the equation for Eθ.

  Eθ=pkesinθr3                                                                                                         (IX)

Conclusion:

Substitute 0 for r in the equation (VIII) and (IX) to calculate Er.

  Er=2pkecosθ03=

Substitute 0 for r in the equation (IX) to calculate Eθ.

    Eθ=pkesinθ03=

Therefore, the results for Er and Eθ do not hold good when r=0.

(e)

To determine

The expression for the electric potential in terms of the Cartesian coordinates.

(e)

Expert Solution
Check Mark

Answer to Problem 71CP

The expression for the electric potential in terms of the Cartesian coordinates is V=pkey(x2+y2)32.

Explanation of Solution

Write the equation for the electric potential.

  V=pkecosθr2

Conclusion:

Substitute x2+y2 for r2 and yx2+y2 for cosθ in the equation above to calculate the electric potential.

  V=pke(yx2+y2)(x2+y2)2=pkey(x2+y2)32

Therefore, the expression for the electric potential in terms of the Cartesian coordinates is V=pkey(x2+y2)32.

(f)

To determine

The x component and the y component of the electric field.

(f)

Expert Solution
Check Mark

Answer to Problem 71CP

The x component and the y component of the electric field are Ex=3pkexy(x2+y2)52 and Ey=pke×(2y2x2)(x2+y2)52.

Explanation of Solution

Write the equation for the x component of the electric field.

  Ex=dVdx                                                                                                                (X)

Here, Ex is the x component of the electric field.

Write the equation for the y component of the electric field.

  Ex=dVdy                                                                                                              (XI)

Here, Ey is the y component of the electric field.

Conclusion:

Substitute pkey(x2+y2)32 for V in the equation (X) to calculate the x component of the electric field.

  Ex=d[pkey(x2+y2)32]dx=3pkexy(x2+y2)52

Substitute pkey(x2+y2)32 for V in the equation (XI) to calculate the y component of the electric field.

  Ey=d[pkey(x2+y2)32]dy=pke(2y2x2)(x2+y2)52

Therefore, the x component and the y component of the electric field are Ex=3pkexy(x2+y2)52 and Ey=pke×(2y2x2)(x2+y2)52.

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Chapter 25 Solutions

Physics for Scientists and Engineers with Modern Physics, Technology Update

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