Show that the function ∇ 1 R = − ∇ ′ 1 R = − R R 3 .

Question

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Answer 1

Question

Chapter 7, Problem 58P

To determine

Show that the function ∇1R=−∇′1R=−RR3.

Expert Solution & Answer

Explanation of Solution

Calculation:

The variable R is the distance vector from the line element at the source point (x′,y′,z′) to the field point (x,y,z) and it is expressed as,

R=|R|R=|r−r′|=[(x−x′)2+(y−y′)2+(z−z′)2]12

To find ∇1R:

∇1R=(∂∂xax+∂∂yay+∂∂zaz)(1[(x−x′)2+(y−y′)2+(z−z′)2]12)=(∂∂xax+∂∂yay+∂∂zaz)[(x−x′)2+(y−y′)2+(z−z′)2]−12=(∂∂x[(x−x′)2+(y−y′)2+(z−z′)2]−12ax+∂∂y[(x−x′)2+(y−y′)2+(z−z′)2]−12ay+∂∂z[(x−x′)2+(y−y′)2+(z−z′)2]−12az)=(−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(x−x′)(1−0)ax−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(y−y′)(1−0)ay−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(z−z′)(1−0)az)

Simplify the above equation.

∇1R=(−[(x−x′)2+(y−y′)2+(z−z′)2]−32(x−x′)ax−[(x−x′)2+(y−y′)2+(z−z′)2]−32(y−y′)ay−[(x−x′)2+(y−y′)2+(z−z′)2]−32(z−z′)az)=−((x−x′)[(x−x′)2+(y−y′)2+(z−z′)2]32ax+(y−y′)[(x−x′)2+(y−y′)2+(z−z′)2]32ay+(z−z′)[(x−x′)2+(y−y′)2+(z−z′)2]32az)=−1[(x−x′)2+(y−y′)2+(z−z′)2]32[(x−x′)ax+(y−y′)ay+(z−z′)az]=−(x−x′)ax+(y−y′)ay+(z−z′)az{[(x−x′)2+(y−y′)2+(z−z′)2]12}3

Simplify the above equation.

∇1R=−RR3 {∵R=[(x−x′)2+(y−y′)2+(z−z′)2]12}

To find −∇′1R:

−∇′1R=−(∂∂x′ax+∂∂y′ay+∂∂z′az)(1[(x−x′)2+(y−y′)2+(z−z′)2]12)=−(∂∂x′ax+∂∂y′ay+∂∂z′az)[(x−x′)2+(y−y′)2+(z−z′)2]−12=−(∂∂x′[(x−x′)2+(y−y′)2+(z−z′)2]−12ax+∂∂y′[(x−x′)2+(y−y′)2+(z−z′)2]−12ay+∂∂z′[(x−x′)2+(y−y′)2+(z−z′)2]−12az)=−(−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(x−x′)(0−1)ax−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(y−y′)(0−1)ay−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(z−z′)(0−1)az)

Simplify the above equation.

−∇′1R=−([(x−x′)2+(y−y′)2+(z−z′)2]−32(x−x′)ax+[(x−x′)2+(y−y′)2+(z−z′)2]−32(y−y′)ay+[(x−x′)2+(y−y′)2+(z−z′)2]−32(z−z′)az)=−((x−x′)[(x−x′)2+(y−y′)2+(z−z′)2]32ax+(y−y′)[(x−x′)2+(y−y′)2+(z−z′)2]32ay+(z−z′)[(x−x′)2+(y−y′)2+(z−z′)2]32az)=−1[(x−x′)2+(y−y′)2+(z−z′)2]32[(x−x′)ax+(y−y′)ay+(z−z′)az]=−(x−x′)ax+(y−y′)ay+(z−z′)az{[(x−x′)2+(y−y′)2+(z−z′)2]12}3

Simplify the above equation.

−∇′1R=−RR3 {∵R=[(x−x′)2+(y−y′)2+(z−z′)2]12}

Conclusion:

Thus, the function ∇1R=−∇′1R=−RR3 and it is shown.

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Answer 2

Question

Chapter 7, Problem 58P

To determine

Show that the function ∇1R=−∇′1R=−RR3.

Expert Solution & Answer

Explanation of Solution

Calculation:

The variable R is the distance vector from the line element at the source point (x′,y′,z′) to the field point (x,y,z) and it is expressed as,

R=|R|R=|r−r′|=[(x−x′)2+(y−y′)2+(z−z′)2]12

To find ∇1R:

∇1R=(∂∂xax+∂∂yay+∂∂zaz)(1[(x−x′)2+(y−y′)2+(z−z′)2]12)=(∂∂xax+∂∂yay+∂∂zaz)[(x−x′)2+(y−y′)2+(z−z′)2]−12=(∂∂x[(x−x′)2+(y−y′)2+(z−z′)2]−12ax+∂∂y[(x−x′)2+(y−y′)2+(z−z′)2]−12ay+∂∂z[(x−x′)2+(y−y′)2+(z−z′)2]−12az)=(−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(x−x′)(1−0)ax−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(y−y′)(1−0)ay−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(z−z′)(1−0)az)

Simplify the above equation.

∇1R=(−[(x−x′)2+(y−y′)2+(z−z′)2]−32(x−x′)ax−[(x−x′)2+(y−y′)2+(z−z′)2]−32(y−y′)ay−[(x−x′)2+(y−y′)2+(z−z′)2]−32(z−z′)az)=−((x−x′)[(x−x′)2+(y−y′)2+(z−z′)2]32ax+(y−y′)[(x−x′)2+(y−y′)2+(z−z′)2]32ay+(z−z′)[(x−x′)2+(y−y′)2+(z−z′)2]32az)=−1[(x−x′)2+(y−y′)2+(z−z′)2]32[(x−x′)ax+(y−y′)ay+(z−z′)az]=−(x−x′)ax+(y−y′)ay+(z−z′)az{[(x−x′)2+(y−y′)2+(z−z′)2]12}3

Simplify the above equation.

∇1R=−RR3 {∵R=[(x−x′)2+(y−y′)2+(z−z′)2]12}

To find −∇′1R:

−∇′1R=−(∂∂x′ax+∂∂y′ay+∂∂z′az)(1[(x−x′)2+(y−y′)2+(z−z′)2]12)=−(∂∂x′ax+∂∂y′ay+∂∂z′az)[(x−x′)2+(y−y′)2+(z−z′)2]−12=−(∂∂x′[(x−x′)2+(y−y′)2+(z−z′)2]−12ax+∂∂y′[(x−x′)2+(y−y′)2+(z−z′)2]−12ay+∂∂z′[(x−x′)2+(y−y′)2+(z−z′)2]−12az)=−(−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(x−x′)(0−1)ax−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(y−y′)(0−1)ay−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(z−z′)(0−1)az)

Simplify the above equation.

−∇′1R=−([(x−x′)2+(y−y′)2+(z−z′)2]−32(x−x′)ax+[(x−x′)2+(y−y′)2+(z−z′)2]−32(y−y′)ay+[(x−x′)2+(y−y′)2+(z−z′)2]−32(z−z′)az)=−((x−x′)[(x−x′)2+(y−y′)2+(z−z′)2]32ax+(y−y′)[(x−x′)2+(y−y′)2+(z−z′)2]32ay+(z−z′)[(x−x′)2+(y−y′)2+(z−z′)2]32az)=−1[(x−x′)2+(y−y′)2+(z−z′)2]32[(x−x′)ax+(y−y′)ay+(z−z′)az]=−(x−x′)ax+(y−y′)ay+(z−z′)az{[(x−x′)2+(y−y′)2+(z−z′)2]12}3

Simplify the above equation.

−∇′1R=−RR3 {∵R=[(x−x′)2+(y−y′)2+(z−z′)2]12}

Conclusion:

Thus, the function ∇1R=−∇′1R=−RR3 and it is shown.

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Answer 3

Question

Chapter 7, Problem 58P

To determine

Show that the function ∇1R=−∇′1R=−RR3.

Expert Solution & Answer

Explanation of Solution

Calculation:

The variable R is the distance vector from the line element at the source point (x′,y′,z′) to the field point (x,y,z) and it is expressed as,

R=|R|R=|r−r′|=[(x−x′)2+(y−y′)2+(z−z′)2]12

To find ∇1R:

∇1R=(∂∂xax+∂∂yay+∂∂zaz)(1[(x−x′)2+(y−y′)2+(z−z′)2]12)=(∂∂xax+∂∂yay+∂∂zaz)[(x−x′)2+(y−y′)2+(z−z′)2]−12=(∂∂x[(x−x′)2+(y−y′)2+(z−z′)2]−12ax+∂∂y[(x−x′)2+(y−y′)2+(z−z′)2]−12ay+∂∂z[(x−x′)2+(y−y′)2+(z−z′)2]−12az)=(−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(x−x′)(1−0)ax−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(y−y′)(1−0)ay−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(z−z′)(1−0)az)

Simplify the above equation.

∇1R=(−[(x−x′)2+(y−y′)2+(z−z′)2]−32(x−x′)ax−[(x−x′)2+(y−y′)2+(z−z′)2]−32(y−y′)ay−[(x−x′)2+(y−y′)2+(z−z′)2]−32(z−z′)az)=−((x−x′)[(x−x′)2+(y−y′)2+(z−z′)2]32ax+(y−y′)[(x−x′)2+(y−y′)2+(z−z′)2]32ay+(z−z′)[(x−x′)2+(y−y′)2+(z−z′)2]32az)=−1[(x−x′)2+(y−y′)2+(z−z′)2]32[(x−x′)ax+(y−y′)ay+(z−z′)az]=−(x−x′)ax+(y−y′)ay+(z−z′)az{[(x−x′)2+(y−y′)2+(z−z′)2]12}3

Simplify the above equation.

∇1R=−RR3 {∵R=[(x−x′)2+(y−y′)2+(z−z′)2]12}

To find −∇′1R:

−∇′1R=−(∂∂x′ax+∂∂y′ay+∂∂z′az)(1[(x−x′)2+(y−y′)2+(z−z′)2]12)=−(∂∂x′ax+∂∂y′ay+∂∂z′az)[(x−x′)2+(y−y′)2+(z−z′)2]−12=−(∂∂x′[(x−x′)2+(y−y′)2+(z−z′)2]−12ax+∂∂y′[(x−x′)2+(y−y′)2+(z−z′)2]−12ay+∂∂z′[(x−x′)2+(y−y′)2+(z−z′)2]−12az)=−(−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(x−x′)(0−1)ax−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(y−y′)(0−1)ay−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(z−z′)(0−1)az)

Simplify the above equation.

−∇′1R=−([(x−x′)2+(y−y′)2+(z−z′)2]−32(x−x′)ax+[(x−x′)2+(y−y′)2+(z−z′)2]−32(y−y′)ay+[(x−x′)2+(y−y′)2+(z−z′)2]−32(z−z′)az)=−((x−x′)[(x−x′)2+(y−y′)2+(z−z′)2]32ax+(y−y′)[(x−x′)2+(y−y′)2+(z−z′)2]32ay+(z−z′)[(x−x′)2+(y−y′)2+(z−z′)2]32az)=−1[(x−x′)2+(y−y′)2+(z−z′)2]32[(x−x′)ax+(y−y′)ay+(z−z′)az]=−(x−x′)ax+(y−y′)ay+(z−z′)az{[(x−x′)2+(y−y′)2+(z−z′)2]12}3

Simplify the above equation.

−∇′1R=−RR3 {∵R=[(x−x′)2+(y−y′)2+(z−z′)2]12}

Conclusion:

Thus, the function ∇1R=−∇′1R=−RR3 and it is shown.

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Answer 4

Question

Chapter 7, Problem 58P

To determine

Show that the function ∇1R=−∇′1R=−RR3.

Expert Solution & Answer

Explanation of Solution

Calculation:

The variable R is the distance vector from the line element at the source point (x′,y′,z′) to the field point (x,y,z) and it is expressed as,

R=|R|R=|r−r′|=[(x−x′)2+(y−y′)2+(z−z′)2]12

To find ∇1R:

∇1R=(∂∂xax+∂∂yay+∂∂zaz)(1[(x−x′)2+(y−y′)2+(z−z′)2]12)=(∂∂xax+∂∂yay+∂∂zaz)[(x−x′)2+(y−y′)2+(z−z′)2]−12=(∂∂x[(x−x′)2+(y−y′)2+(z−z′)2]−12ax+∂∂y[(x−x′)2+(y−y′)2+(z−z′)2]−12ay+∂∂z[(x−x′)2+(y−y′)2+(z−z′)2]−12az)=(−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(x−x′)(1−0)ax−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(y−y′)(1−0)ay−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(z−z′)(1−0)az)

Simplify the above equation.

∇1R=(−[(x−x′)2+(y−y′)2+(z−z′)2]−32(x−x′)ax−[(x−x′)2+(y−y′)2+(z−z′)2]−32(y−y′)ay−[(x−x′)2+(y−y′)2+(z−z′)2]−32(z−z′)az)=−((x−x′)[(x−x′)2+(y−y′)2+(z−z′)2]32ax+(y−y′)[(x−x′)2+(y−y′)2+(z−z′)2]32ay+(z−z′)[(x−x′)2+(y−y′)2+(z−z′)2]32az)=−1[(x−x′)2+(y−y′)2+(z−z′)2]32[(x−x′)ax+(y−y′)ay+(z−z′)az]=−(x−x′)ax+(y−y′)ay+(z−z′)az{[(x−x′)2+(y−y′)2+(z−z′)2]12}3

Simplify the above equation.

∇1R=−RR3 {∵R=[(x−x′)2+(y−y′)2+(z−z′)2]12}

To find −∇′1R:

−∇′1R=−(∂∂x′ax+∂∂y′ay+∂∂z′az)(1[(x−x′)2+(y−y′)2+(z−z′)2]12)=−(∂∂x′ax+∂∂y′ay+∂∂z′az)[(x−x′)2+(y−y′)2+(z−z′)2]−12=−(∂∂x′[(x−x′)2+(y−y′)2+(z−z′)2]−12ax+∂∂y′[(x−x′)2+(y−y′)2+(z−z′)2]−12ay+∂∂z′[(x−x′)2+(y−y′)2+(z−z′)2]−12az)=−(−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(x−x′)(0−1)ax−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(y−y′)(0−1)ay−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(z−z′)(0−1)az)

Simplify the above equation.

−∇′1R=−([(x−x′)2+(y−y′)2+(z−z′)2]−32(x−x′)ax+[(x−x′)2+(y−y′)2+(z−z′)2]−32(y−y′)ay+[(x−x′)2+(y−y′)2+(z−z′)2]−32(z−z′)az)=−((x−x′)[(x−x′)2+(y−y′)2+(z−z′)2]32ax+(y−y′)[(x−x′)2+(y−y′)2+(z−z′)2]32ay+(z−z′)[(x−x′)2+(y−y′)2+(z−z′)2]32az)=−1[(x−x′)2+(y−y′)2+(z−z′)2]32[(x−x′)ax+(y−y′)ay+(z−z′)az]=−(x−x′)ax+(y−y′)ay+(z−z′)az{[(x−x′)2+(y−y′)2+(z−z′)2]12}3

Simplify the above equation.

−∇′1R=−RR3 {∵R=[(x−x′)2+(y−y′)2+(z−z′)2]12}

Conclusion:

Thus, the function ∇1R=−∇′1R=−RR3 and it is shown.

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Answer 5

Chapter 7, Problem 58P

To determine

Show that the function ∇1R=−∇′1R=−RR3.

Expert Solution & Answer

Explanation of Solution

Calculation:

The variable R is the distance vector from the line element at the source point (x′,y′,z′) to the field point (x,y,z) and it is expressed as,

R=|R|R=|r−r′|=[(x−x′)2+(y−y′)2+(z−z′)2]12

To find ∇1R:

∇1R=(∂∂xax+∂∂yay+∂∂zaz)(1[(x−x′)2+(y−y′)2+(z−z′)2]12)=(∂∂xax+∂∂yay+∂∂zaz)[(x−x′)2+(y−y′)2+(z−z′)2]−12=(∂∂x[(x−x′)2+(y−y′)2+(z−z′)2]−12ax+∂∂y[(x−x′)2+(y−y′)2+(z−z′)2]−12ay+∂∂z[(x−x′)2+(y−y′)2+(z−z′)2]−12az)=(−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(x−x′)(1−0)ax−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(y−y′)(1−0)ay−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(z−z′)(1−0)az)

Simplify the above equation.

∇1R=(−[(x−x′)2+(y−y′)2+(z−z′)2]−32(x−x′)ax−[(x−x′)2+(y−y′)2+(z−z′)2]−32(y−y′)ay−[(x−x′)2+(y−y′)2+(z−z′)2]−32(z−z′)az)=−((x−x′)[(x−x′)2+(y−y′)2+(z−z′)2]32ax+(y−y′)[(x−x′)2+(y−y′)2+(z−z′)2]32ay+(z−z′)[(x−x′)2+(y−y′)2+(z−z′)2]32az)=−1[(x−x′)2+(y−y′)2+(z−z′)2]32[(x−x′)ax+(y−y′)ay+(z−z′)az]=−(x−x′)ax+(y−y′)ay+(z−z′)az{[(x−x′)2+(y−y′)2+(z−z′)2]12}3

Simplify the above equation.

∇1R=−RR3 {∵R=[(x−x′)2+(y−y′)2+(z−z′)2]12}

To find −∇′1R:

−∇′1R=−(∂∂x′ax+∂∂y′ay+∂∂z′az)(1[(x−x′)2+(y−y′)2+(z−z′)2]12)=−(∂∂x′ax+∂∂y′ay+∂∂z′az)[(x−x′)2+(y−y′)2+(z−z′)2]−12=−(∂∂x′[(x−x′)2+(y−y′)2+(z−z′)2]−12ax+∂∂y′[(x−x′)2+(y−y′)2+(z−z′)2]−12ay+∂∂z′[(x−x′)2+(y−y′)2+(z−z′)2]−12az)=−(−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(x−x′)(0−1)ax−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(y−y′)(0−1)ay−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(z−z′)(0−1)az)

Simplify the above equation.

−∇′1R=−([(x−x′)2+(y−y′)2+(z−z′)2]−32(x−x′)ax+[(x−x′)2+(y−y′)2+(z−z′)2]−32(y−y′)ay+[(x−x′)2+(y−y′)2+(z−z′)2]−32(z−z′)az)=−((x−x′)[(x−x′)2+(y−y′)2+(z−z′)2]32ax+(y−y′)[(x−x′)2+(y−y′)2+(z−z′)2]32ay+(z−z′)[(x−x′)2+(y−y′)2+(z−z′)2]32az)=−1[(x−x′)2+(y−y′)2+(z−z′)2]32[(x−x′)ax+(y−y′)ay+(z−z′)az]=−(x−x′)ax+(y−y′)ay+(z−z′)az{[(x−x′)2+(y−y′)2+(z−z′)2]12}3

Simplify the above equation.

−∇′1R=−RR3 {∵R=[(x−x′)2+(y−y′)2+(z−z′)2]12}

Conclusion:

Thus, the function ∇1R=−∇′1R=−RR3 and it is shown.

Answer 6

Chapter 7, Problem 58P

To determine

Show that the function ∇1R=−∇′1R=−RR3.

Expert Solution & Answer

Explanation of Solution

Calculation:

The variable R is the distance vector from the line element at the source point (x′,y′,z′) to the field point (x,y,z) and it is expressed as,

R=|R|R=|r−r′|=[(x−x′)2+(y−y′)2+(z−z′)2]12

To find ∇1R:

∇1R=(∂∂xax+∂∂yay+∂∂zaz)(1[(x−x′)2+(y−y′)2+(z−z′)2]12)=(∂∂xax+∂∂yay+∂∂zaz)[(x−x′)2+(y−y′)2+(z−z′)2]−12=(∂∂x[(x−x′)2+(y−y′)2+(z−z′)2]−12ax+∂∂y[(x−x′)2+(y−y′)2+(z−z′)2]−12ay+∂∂z[(x−x′)2+(y−y′)2+(z−z′)2]−12az)=(−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(x−x′)(1−0)ax−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(y−y′)(1−0)ay−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(z−z′)(1−0)az)

Simplify the above equation.

∇1R=(−[(x−x′)2+(y−y′)2+(z−z′)2]−32(x−x′)ax−[(x−x′)2+(y−y′)2+(z−z′)2]−32(y−y′)ay−[(x−x′)2+(y−y′)2+(z−z′)2]−32(z−z′)az)=−((x−x′)[(x−x′)2+(y−y′)2+(z−z′)2]32ax+(y−y′)[(x−x′)2+(y−y′)2+(z−z′)2]32ay+(z−z′)[(x−x′)2+(y−y′)2+(z−z′)2]32az)=−1[(x−x′)2+(y−y′)2+(z−z′)2]32[(x−x′)ax+(y−y′)ay+(z−z′)az]=−(x−x′)ax+(y−y′)ay+(z−z′)az{[(x−x′)2+(y−y′)2+(z−z′)2]12}3

Simplify the above equation.

∇1R=−RR3 {∵R=[(x−x′)2+(y−y′)2+(z−z′)2]12}

To find −∇′1R:

−∇′1R=−(∂∂x′ax+∂∂y′ay+∂∂z′az)(1[(x−x′)2+(y−y′)2+(z−z′)2]12)=−(∂∂x′ax+∂∂y′ay+∂∂z′az)[(x−x′)2+(y−y′)2+(z−z′)2]−12=−(∂∂x′[(x−x′)2+(y−y′)2+(z−z′)2]−12ax+∂∂y′[(x−x′)2+(y−y′)2+(z−z′)2]−12ay+∂∂z′[(x−x′)2+(y−y′)2+(z−z′)2]−12az)=−(−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(x−x′)(0−1)ax−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(y−y′)(0−1)ay−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(z−z′)(0−1)az)

Simplify the above equation.

−∇′1R=−([(x−x′)2+(y−y′)2+(z−z′)2]−32(x−x′)ax+[(x−x′)2+(y−y′)2+(z−z′)2]−32(y−y′)ay+[(x−x′)2+(y−y′)2+(z−z′)2]−32(z−z′)az)=−((x−x′)[(x−x′)2+(y−y′)2+(z−z′)2]32ax+(y−y′)[(x−x′)2+(y−y′)2+(z−z′)2]32ay+(z−z′)[(x−x′)2+(y−y′)2+(z−z′)2]32az)=−1[(x−x′)2+(y−y′)2+(z−z′)2]32[(x−x′)ax+(y−y′)ay+(z−z′)az]=−(x−x′)ax+(y−y′)ay+(z−z′)az{[(x−x′)2+(y−y′)2+(z−z′)2]12}3

Simplify the above equation.

−∇′1R=−RR3 {∵R=[(x−x′)2+(y−y′)2+(z−z′)2]12}

Conclusion:

Thus, the function ∇1R=−∇′1R=−RR3 and it is shown.

Answer 7

Chapter 7, Problem 58P

To determine

Show that the function ∇1R=−∇′1R=−RR3.

Answer 8

Expert Solution & Answer

Explanation of Solution

Calculation:

The variable R is the distance vector from the line element at the source point (x′,y′,z′) to the field point (x,y,z) and it is expressed as,

R=|R|R=|r−r′|=[(x−x′)2+(y−y′)2+(z−z′)2]12

To find ∇1R:

∇1R=(∂∂xax+∂∂yay+∂∂zaz)(1[(x−x′)2+(y−y′)2+(z−z′)2]12)=(∂∂xax+∂∂yay+∂∂zaz)[(x−x′)2+(y−y′)2+(z−z′)2]−12=(∂∂x[(x−x′)2+(y−y′)2+(z−z′)2]−12ax+∂∂y[(x−x′)2+(y−y′)2+(z−z′)2]−12ay+∂∂z[(x−x′)2+(y−y′)2+(z−z′)2]−12az)=(−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(x−x′)(1−0)ax−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(y−y′)(1−0)ay−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(z−z′)(1−0)az)

Simplify the above equation.

∇1R=(−[(x−x′)2+(y−y′)2+(z−z′)2]−32(x−x′)ax−[(x−x′)2+(y−y′)2+(z−z′)2]−32(y−y′)ay−[(x−x′)2+(y−y′)2+(z−z′)2]−32(z−z′)az)=−((x−x′)[(x−x′)2+(y−y′)2+(z−z′)2]32ax+(y−y′)[(x−x′)2+(y−y′)2+(z−z′)2]32ay+(z−z′)[(x−x′)2+(y−y′)2+(z−z′)2]32az)=−1[(x−x′)2+(y−y′)2+(z−z′)2]32[(x−x′)ax+(y−y′)ay+(z−z′)az]=−(x−x′)ax+(y−y′)ay+(z−z′)az{[(x−x′)2+(y−y′)2+(z−z′)2]12}3

Simplify the above equation.

∇1R=−RR3 {∵R=[(x−x′)2+(y−y′)2+(z−z′)2]12}

To find −∇′1R:

−∇′1R=−(∂∂x′ax+∂∂y′ay+∂∂z′az)(1[(x−x′)2+(y−y′)2+(z−z′)2]12)=−(∂∂x′ax+∂∂y′ay+∂∂z′az)[(x−x′)2+(y−y′)2+(z−z′)2]−12=−(∂∂x′[(x−x′)2+(y−y′)2+(z−z′)2]−12ax+∂∂y′[(x−x′)2+(y−y′)2+(z−z′)2]−12ay+∂∂z′[(x−x′)2+(y−y′)2+(z−z′)2]−12az)=−(−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(x−x′)(0−1)ax−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(y−y′)(0−1)ay−12[(x−x′)2+(y−y′)2+(z−z′)2]−12−1(2)(z−z′)(0−1)az)

Simplify the above equation.

−∇′1R=−([(x−x′)2+(y−y′)2+(z−z′)2]−32(x−x′)ax+[(x−x′)2+(y−y′)2+(z−z′)2]−32(y−y′)ay+[(x−x′)2+(y−y′)2+(z−z′)2]−32(z−z′)az)=−((x−x′)[(x−x′)2+(y−y′)2+(z−z′)2]32ax+(y−y′)[(x−x′)2+(y−y′)2+(z−z′)2]32ay+(z−z′)[(x−x′)2+(y−y′)2+(z−z′)2]32az)=−1[(x−x′)2+(y−y′)2+(z−z′)2]32[(x−x′)ax+(y−y′)ay+(z−z′)az]=−(x−x′)ax+(y−y′)ay+(z−z′)az{[(x−x′)2+(y−y′)2+(z−z′)2]12}3