A length d of the charge lies on the Z-axis infree space. The charge density on the line is p L = + C / m ( 0 < z < d / 2 ) and p L = − p 0 C / m ( − d / 2 < z < 0 ) where P 0 is a possitive constant. (a) Find the electric field intensity E everywhere in the xy plane, expressing your result as a function of cylindrical radius p , (b) simplify your part a result for the case in radius p d , and express this result in terms of charge q = p 0 d/2.

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Answer 1

Textbook Question

Chapter 2, Problem 2.17P

A length d of the charge lies on the Z-axis infree space. The charge density on the line is $p L = + C / m ( 0 < z < d / 2 )$ and $p L = − p 0 C / m ( − d / 2 < z < 0 )$

where P₀ is a possitive constant. (a) Find the electric field intensity E everywhere in the xy plane, expressing your result as a function of cylindrical radius p, (b) simplify your part a result for the case in radius pd, and express this result in terms of charge q=p₀d/2.

Expert Solution

To determine

(a)

The electric field intensity in x-y plane, as a function of cylindrical radius $ρ$ .

Answer to Problem 2.17P

The required electric field intensity is:

$E→(ρ)=−ρ02πε0ρ(1−( 1+ d 2 4 ρ 2 )−1/2)a^z V/m$ .

Explanation of Solution

Given Information:

The line charge density is given as,

$ρL={+ ρ 0 C/m0<z<d/2− ρ 0 C/m−d/2<z<0$

Calculation:

Let $(x,y,0)$ be a point on x-y plane, and $(0,0,z)$ is a point on line charge. Then differential electric field,

$dE→=ρL(xa ^x+ya ^y−za ^z)4πε0( x 2 + y 2 + z 2 )3/2$

So, the electric field intensity:

$E → = ∫ dE → = ∫ −d/2 d/2 ρ L ( x a ^ x +y a ^ y −z a ^ z ) 4π ε 0 ( x 2 + y 2 + z 2 ) 3/2 dz$

$= ∫ −d/2 0 − ρ 0 ( x a ^ x +y a ^ y −z a ^ z ) 4π ε 0 ( x 2 + y 2 + z 2 ) 3/2 dz + ∫ 0 d/2 ρ 0 ( x a ^ x +y a ^ y −z a ^ z ) 4π ε 0 ( x 2 + y 2 + z 2 ) 3/2 dz$

$= ρ 0 4π ε 0 [ ∫ −d/2 0 z a ^ z ( x 2 + y 2 + z 2 ) 3/2 dz− ∫ 0 d/2 z a ^ z ( x 2 + y 2 + z 2 ) 3/2 dz ]$

$= ρ 0 4π ε 0 ( [ − 1 ( x 2 + y 2 + z 2 ) 1/2 ] −d/2 0 − [ − 1 ( x 2 + y 2 + z 2 ) 1/2 ] 0 d/2 ) a ^ z$

$=− ρ 0 4π ε 0 ( [ 1 ( x 2 + y 2 ) 1/2 − 1 ( x 2 + y 2 + ( d/2 ) 2 ) 1/2 ]−[ 1 ( x 2 + y 2 + ( d/2 ) 2 ) 1/2 − 1 ( x 2 + y 2 ) 1/2 ] ) a ^ z$

$=− 2 ρ 0 4π ε 0 ( 1 ( x 2 + y 2 ) 1/2 − 1 ( x 2 + y 2 + ( d/2 ) 2 ) 1/2 ) a ^ z$

So, in cylindrical coordinates,

$E→(ρ)=−2ρ04πε0(1ρ−1 ( ρ 2 + ( d/2 ) 2 ) 1/2 )a^z ∵x2+y2=ρ2 =−ρ02πε0ρ(1−1 ( 1+ ( d/2ρ ) 2 ) 1/2 )a^z =−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1/2)a^z V/m$

Conclusion:

The required electric field intensity is:

$E→(ρ)=−ρ02πε0ρ(1−( 1+ d 2 4 ρ 2 )−1/2)a^z V/m$ .

Expert Solution

To determine

(b)

The electric field intensity in x-y plane for $ρ≫d$ .

Answer to Problem 2.17P

The required electric field intensity is,

$E→(ρ≫d)=−qd4πε0ρ3a^z V/m$ .

Explanation of Solution

Given Information:

The line charge density is given as,

$ρL={+ ρ 0 C/m0<z<d/2− ρ 0 C/m−d/2<z<0$

$E→(ρ)=−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1/2)a^z V/mq=ρ0d/2$

Calculation:

The electric field intensity,

$E→(ρ≫d)=−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1/2)a^z =−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1)a^z ∵(1+ d 2 4 ρ 2 )1/2≈(1+ d 2 4 ρ 2 ) =−ρ02πε0ρ( d 2 4 ρ 2 + d 2 ) =−ρ02πε0ρ( d 2 4 ρ 2 ) ∵4ρ2+d2≈4ρ2 =−2q2πε0ρd( d 2 4 ρ 2 ) ∵q=ρ0d/2 =−qd4πε0ρ3a^z V/m$

Conclusion:

The required electric field intensity is,

$E→(ρ≫d)=−qd4πε0ρ3a^z V/m$ .

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Answer 2

Textbook Question

Chapter 2, Problem 2.17P

A length d of the charge lies on the Z-axis infree space. The charge density on the line is $p L = + C / m ( 0 < z < d / 2 )$ and $p L = − p 0 C / m ( − d / 2 < z < 0 )$

where P₀ is a possitive constant. (a) Find the electric field intensity E everywhere in the xy plane, expressing your result as a function of cylindrical radius p, (b) simplify your part a result for the case in radius pd, and express this result in terms of charge q=p₀d/2.

Expert Solution

To determine

(a)

The electric field intensity in x-y plane, as a function of cylindrical radius $ρ$ .

Answer to Problem 2.17P

The required electric field intensity is:

$E→(ρ)=−ρ02πε0ρ(1−( 1+ d 2 4 ρ 2 )−1/2)a^z V/m$ .

Explanation of Solution

Given Information:

The line charge density is given as,

$ρL={+ ρ 0 C/m0<z<d/2− ρ 0 C/m−d/2<z<0$

Calculation:

Let $(x,y,0)$ be a point on x-y plane, and $(0,0,z)$ is a point on line charge. Then differential electric field,

$dE→=ρL(xa ^x+ya ^y−za ^z)4πε0( x 2 + y 2 + z 2 )3/2$

So, the electric field intensity:

$E → = ∫ dE → = ∫ −d/2 d/2 ρ L ( x a ^ x +y a ^ y −z a ^ z ) 4π ε 0 ( x 2 + y 2 + z 2 ) 3/2 dz$

$= ∫ −d/2 0 − ρ 0 ( x a ^ x +y a ^ y −z a ^ z ) 4π ε 0 ( x 2 + y 2 + z 2 ) 3/2 dz + ∫ 0 d/2 ρ 0 ( x a ^ x +y a ^ y −z a ^ z ) 4π ε 0 ( x 2 + y 2 + z 2 ) 3/2 dz$

$= ρ 0 4π ε 0 [ ∫ −d/2 0 z a ^ z ( x 2 + y 2 + z 2 ) 3/2 dz− ∫ 0 d/2 z a ^ z ( x 2 + y 2 + z 2 ) 3/2 dz ]$

$= ρ 0 4π ε 0 ( [ − 1 ( x 2 + y 2 + z 2 ) 1/2 ] −d/2 0 − [ − 1 ( x 2 + y 2 + z 2 ) 1/2 ] 0 d/2 ) a ^ z$

$=− ρ 0 4π ε 0 ( [ 1 ( x 2 + y 2 ) 1/2 − 1 ( x 2 + y 2 + ( d/2 ) 2 ) 1/2 ]−[ 1 ( x 2 + y 2 + ( d/2 ) 2 ) 1/2 − 1 ( x 2 + y 2 ) 1/2 ] ) a ^ z$

$=− 2 ρ 0 4π ε 0 ( 1 ( x 2 + y 2 ) 1/2 − 1 ( x 2 + y 2 + ( d/2 ) 2 ) 1/2 ) a ^ z$

So, in cylindrical coordinates,

$E→(ρ)=−2ρ04πε0(1ρ−1 ( ρ 2 + ( d/2 ) 2 ) 1/2 )a^z ∵x2+y2=ρ2 =−ρ02πε0ρ(1−1 ( 1+ ( d/2ρ ) 2 ) 1/2 )a^z =−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1/2)a^z V/m$

Conclusion:

The required electric field intensity is:

$E→(ρ)=−ρ02πε0ρ(1−( 1+ d 2 4 ρ 2 )−1/2)a^z V/m$ .

Expert Solution

To determine

(b)

The electric field intensity in x-y plane for $ρ≫d$ .

Answer to Problem 2.17P

The required electric field intensity is,

$E→(ρ≫d)=−qd4πε0ρ3a^z V/m$ .

Explanation of Solution

Given Information:

The line charge density is given as,

$ρL={+ ρ 0 C/m0<z<d/2− ρ 0 C/m−d/2<z<0$

$E→(ρ)=−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1/2)a^z V/mq=ρ0d/2$

Calculation:

The electric field intensity,

$E→(ρ≫d)=−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1/2)a^z =−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1)a^z ∵(1+ d 2 4 ρ 2 )1/2≈(1+ d 2 4 ρ 2 ) =−ρ02πε0ρ( d 2 4 ρ 2 + d 2 ) =−ρ02πε0ρ( d 2 4 ρ 2 ) ∵4ρ2+d2≈4ρ2 =−2q2πε0ρd( d 2 4 ρ 2 ) ∵q=ρ0d/2 =−qd4πε0ρ3a^z V/m$

Conclusion:

The required electric field intensity is,

$E→(ρ≫d)=−qd4πε0ρ3a^z V/m$ .

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Answer 3

Textbook Question

Chapter 2, Problem 2.17P

A length d of the charge lies on the Z-axis infree space. The charge density on the line is $p L = + C / m ( 0 < z < d / 2 )$ and $p L = − p 0 C / m ( − d / 2 < z < 0 )$

where P₀ is a possitive constant. (a) Find the electric field intensity E everywhere in the xy plane, expressing your result as a function of cylindrical radius p, (b) simplify your part a result for the case in radius pd, and express this result in terms of charge q=p₀d/2.

Expert Solution

To determine

(a)

The electric field intensity in x-y plane, as a function of cylindrical radius $ρ$ .

Answer to Problem 2.17P

The required electric field intensity is:

$E→(ρ)=−ρ02πε0ρ(1−( 1+ d 2 4 ρ 2 )−1/2)a^z V/m$ .

Explanation of Solution

Given Information:

The line charge density is given as,

$ρL={+ ρ 0 C/m0<z<d/2− ρ 0 C/m−d/2<z<0$

Calculation:

Let $(x,y,0)$ be a point on x-y plane, and $(0,0,z)$ is a point on line charge. Then differential electric field,

$dE→=ρL(xa ^x+ya ^y−za ^z)4πε0( x 2 + y 2 + z 2 )3/2$

So, the electric field intensity:

$E → = ∫ dE → = ∫ −d/2 d/2 ρ L ( x a ^ x +y a ^ y −z a ^ z ) 4π ε 0 ( x 2 + y 2 + z 2 ) 3/2 dz$

$= ∫ −d/2 0 − ρ 0 ( x a ^ x +y a ^ y −z a ^ z ) 4π ε 0 ( x 2 + y 2 + z 2 ) 3/2 dz + ∫ 0 d/2 ρ 0 ( x a ^ x +y a ^ y −z a ^ z ) 4π ε 0 ( x 2 + y 2 + z 2 ) 3/2 dz$

$= ρ 0 4π ε 0 [ ∫ −d/2 0 z a ^ z ( x 2 + y 2 + z 2 ) 3/2 dz− ∫ 0 d/2 z a ^ z ( x 2 + y 2 + z 2 ) 3/2 dz ]$

$= ρ 0 4π ε 0 ( [ − 1 ( x 2 + y 2 + z 2 ) 1/2 ] −d/2 0 − [ − 1 ( x 2 + y 2 + z 2 ) 1/2 ] 0 d/2 ) a ^ z$

$=− ρ 0 4π ε 0 ( [ 1 ( x 2 + y 2 ) 1/2 − 1 ( x 2 + y 2 + ( d/2 ) 2 ) 1/2 ]−[ 1 ( x 2 + y 2 + ( d/2 ) 2 ) 1/2 − 1 ( x 2 + y 2 ) 1/2 ] ) a ^ z$

$=− 2 ρ 0 4π ε 0 ( 1 ( x 2 + y 2 ) 1/2 − 1 ( x 2 + y 2 + ( d/2 ) 2 ) 1/2 ) a ^ z$

So, in cylindrical coordinates,

$E→(ρ)=−2ρ04πε0(1ρ−1 ( ρ 2 + ( d/2 ) 2 ) 1/2 )a^z ∵x2+y2=ρ2 =−ρ02πε0ρ(1−1 ( 1+ ( d/2ρ ) 2 ) 1/2 )a^z =−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1/2)a^z V/m$

Conclusion:

The required electric field intensity is:

$E→(ρ)=−ρ02πε0ρ(1−( 1+ d 2 4 ρ 2 )−1/2)a^z V/m$ .

Expert Solution

To determine

(b)

The electric field intensity in x-y plane for $ρ≫d$ .

Answer to Problem 2.17P

The required electric field intensity is,

$E→(ρ≫d)=−qd4πε0ρ3a^z V/m$ .

Explanation of Solution

Given Information:

The line charge density is given as,

$ρL={+ ρ 0 C/m0<z<d/2− ρ 0 C/m−d/2<z<0$

$E→(ρ)=−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1/2)a^z V/mq=ρ0d/2$

Calculation:

The electric field intensity,

$E→(ρ≫d)=−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1/2)a^z =−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1)a^z ∵(1+ d 2 4 ρ 2 )1/2≈(1+ d 2 4 ρ 2 ) =−ρ02πε0ρ( d 2 4 ρ 2 + d 2 ) =−ρ02πε0ρ( d 2 4 ρ 2 ) ∵4ρ2+d2≈4ρ2 =−2q2πε0ρd( d 2 4 ρ 2 ) ∵q=ρ0d/2 =−qd4πε0ρ3a^z V/m$

Conclusion:

The required electric field intensity is,

$E→(ρ≫d)=−qd4πε0ρ3a^z V/m$ .

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Answer 4

Textbook Question

Chapter 2, Problem 2.17P

A length d of the charge lies on the Z-axis infree space. The charge density on the line is $p L = + C / m ( 0 < z < d / 2 )$ and $p L = − p 0 C / m ( − d / 2 < z < 0 )$

where P₀ is a possitive constant. (a) Find the electric field intensity E everywhere in the xy plane, expressing your result as a function of cylindrical radius p, (b) simplify your part a result for the case in radius pd, and express this result in terms of charge q=p₀d/2.

Expert Solution

To determine

(a)

The electric field intensity in x-y plane, as a function of cylindrical radius $ρ$ .

Answer to Problem 2.17P

The required electric field intensity is:

$E→(ρ)=−ρ02πε0ρ(1−( 1+ d 2 4 ρ 2 )−1/2)a^z V/m$ .

Explanation of Solution

Given Information:

The line charge density is given as,

$ρL={+ ρ 0 C/m0<z<d/2− ρ 0 C/m−d/2<z<0$

Calculation:

Let $(x,y,0)$ be a point on x-y plane, and $(0,0,z)$ is a point on line charge. Then differential electric field,

$dE→=ρL(xa ^x+ya ^y−za ^z)4πε0( x 2 + y 2 + z 2 )3/2$

So, the electric field intensity:

$E → = ∫ dE → = ∫ −d/2 d/2 ρ L ( x a ^ x +y a ^ y −z a ^ z ) 4π ε 0 ( x 2 + y 2 + z 2 ) 3/2 dz$

$= ∫ −d/2 0 − ρ 0 ( x a ^ x +y a ^ y −z a ^ z ) 4π ε 0 ( x 2 + y 2 + z 2 ) 3/2 dz + ∫ 0 d/2 ρ 0 ( x a ^ x +y a ^ y −z a ^ z ) 4π ε 0 ( x 2 + y 2 + z 2 ) 3/2 dz$

$= ρ 0 4π ε 0 [ ∫ −d/2 0 z a ^ z ( x 2 + y 2 + z 2 ) 3/2 dz− ∫ 0 d/2 z a ^ z ( x 2 + y 2 + z 2 ) 3/2 dz ]$

$= ρ 0 4π ε 0 ( [ − 1 ( x 2 + y 2 + z 2 ) 1/2 ] −d/2 0 − [ − 1 ( x 2 + y 2 + z 2 ) 1/2 ] 0 d/2 ) a ^ z$

$=− ρ 0 4π ε 0 ( [ 1 ( x 2 + y 2 ) 1/2 − 1 ( x 2 + y 2 + ( d/2 ) 2 ) 1/2 ]−[ 1 ( x 2 + y 2 + ( d/2 ) 2 ) 1/2 − 1 ( x 2 + y 2 ) 1/2 ] ) a ^ z$

$=− 2 ρ 0 4π ε 0 ( 1 ( x 2 + y 2 ) 1/2 − 1 ( x 2 + y 2 + ( d/2 ) 2 ) 1/2 ) a ^ z$

So, in cylindrical coordinates,

$E→(ρ)=−2ρ04πε0(1ρ−1 ( ρ 2 + ( d/2 ) 2 ) 1/2 )a^z ∵x2+y2=ρ2 =−ρ02πε0ρ(1−1 ( 1+ ( d/2ρ ) 2 ) 1/2 )a^z =−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1/2)a^z V/m$

Conclusion:

The required electric field intensity is:

$E→(ρ)=−ρ02πε0ρ(1−( 1+ d 2 4 ρ 2 )−1/2)a^z V/m$ .

Expert Solution

To determine

(b)

The electric field intensity in x-y plane for $ρ≫d$ .

Answer to Problem 2.17P

The required electric field intensity is,

$E→(ρ≫d)=−qd4πε0ρ3a^z V/m$ .

Explanation of Solution

Given Information:

The line charge density is given as,

$ρL={+ ρ 0 C/m0<z<d/2− ρ 0 C/m−d/2<z<0$

$E→(ρ)=−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1/2)a^z V/mq=ρ0d/2$

Calculation:

The electric field intensity,

$E→(ρ≫d)=−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1/2)a^z =−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1)a^z ∵(1+ d 2 4 ρ 2 )1/2≈(1+ d 2 4 ρ 2 ) =−ρ02πε0ρ( d 2 4 ρ 2 + d 2 ) =−ρ02πε0ρ( d 2 4 ρ 2 ) ∵4ρ2+d2≈4ρ2 =−2q2πε0ρd( d 2 4 ρ 2 ) ∵q=ρ0d/2 =−qd4πε0ρ3a^z V/m$

Conclusion:

The required electric field intensity is,

$E→(ρ≫d)=−qd4πε0ρ3a^z V/m$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

To determine

(a)

The electric field intensity in x-y plane, as a function of cylindrical radius $ρ$ .

Answer to Problem 2.17P

The required electric field intensity is:

$E→(ρ)=−ρ02πε0ρ(1−( 1+ d 2 4 ρ 2 )−1/2)a^z V/m$ .

Explanation of Solution

Given Information:

The line charge density is given as,

$ρL={+ ρ 0 C/m0<z<d/2− ρ 0 C/m−d/2<z<0$

Calculation:

Let $(x,y,0)$ be a point on x-y plane, and $(0,0,z)$ is a point on line charge. Then differential electric field,

$dE→=ρL(xa ^x+ya ^y−za ^z)4πε0( x 2 + y 2 + z 2 )3/2$

So, the electric field intensity:

$E → = ∫ dE → = ∫ −d/2 d/2 ρ L ( x a ^ x +y a ^ y −z a ^ z ) 4π ε 0 ( x 2 + y 2 + z 2 ) 3/2 dz$

$= ∫ −d/2 0 − ρ 0 ( x a ^ x +y a ^ y −z a ^ z ) 4π ε 0 ( x 2 + y 2 + z 2 ) 3/2 dz + ∫ 0 d/2 ρ 0 ( x a ^ x +y a ^ y −z a ^ z ) 4π ε 0 ( x 2 + y 2 + z 2 ) 3/2 dz$

$= ρ 0 4π ε 0 [ ∫ −d/2 0 z a ^ z ( x 2 + y 2 + z 2 ) 3/2 dz− ∫ 0 d/2 z a ^ z ( x 2 + y 2 + z 2 ) 3/2 dz ]$

$= ρ 0 4π ε 0 ( [ − 1 ( x 2 + y 2 + z 2 ) 1/2 ] −d/2 0 − [ − 1 ( x 2 + y 2 + z 2 ) 1/2 ] 0 d/2 ) a ^ z$

$=− ρ 0 4π ε 0 ( [ 1 ( x 2 + y 2 ) 1/2 − 1 ( x 2 + y 2 + ( d/2 ) 2 ) 1/2 ]−[ 1 ( x 2 + y 2 + ( d/2 ) 2 ) 1/2 − 1 ( x 2 + y 2 ) 1/2 ] ) a ^ z$

$=− 2 ρ 0 4π ε 0 ( 1 ( x 2 + y 2 ) 1/2 − 1 ( x 2 + y 2 + ( d/2 ) 2 ) 1/2 ) a ^ z$

So, in cylindrical coordinates,

$E→(ρ)=−2ρ04πε0(1ρ−1 ( ρ 2 + ( d/2 ) 2 ) 1/2 )a^z ∵x2+y2=ρ2 =−ρ02πε0ρ(1−1 ( 1+ ( d/2ρ ) 2 ) 1/2 )a^z =−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1/2)a^z V/m$

Conclusion:

The required electric field intensity is:

$E→(ρ)=−ρ02πε0ρ(1−( 1+ d 2 4 ρ 2 )−1/2)a^z V/m$ .

Expert Solution

To determine

(b)

The electric field intensity in x-y plane for $ρ≫d$ .

Answer to Problem 2.17P

The required electric field intensity is,

$E→(ρ≫d)=−qd4πε0ρ3a^z V/m$ .

Explanation of Solution

Given Information:

The line charge density is given as,

$ρL={+ ρ 0 C/m0<z<d/2− ρ 0 C/m−d/2<z<0$

$E→(ρ)=−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1/2)a^z V/mq=ρ0d/2$

Calculation:

The electric field intensity,

$E→(ρ≫d)=−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1/2)a^z =−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1)a^z ∵(1+ d 2 4 ρ 2 )1/2≈(1+ d 2 4 ρ 2 ) =−ρ02πε0ρ( d 2 4 ρ 2 + d 2 ) =−ρ02πε0ρ( d 2 4 ρ 2 ) ∵4ρ2+d2≈4ρ2 =−2q2πε0ρd( d 2 4 ρ 2 ) ∵q=ρ0d/2 =−qd4πε0ρ3a^z V/m$

Conclusion:

The required electric field intensity is,

$E→(ρ≫d)=−qd4πε0ρ3a^z V/m$ .

Answer 8

Expert Solution

To determine

(a)

The electric field intensity in x-y plane, as a function of cylindrical radius $ρ$ .

Answer to Problem 2.17P

The required electric field intensity is:

$E→(ρ)=−ρ02πε0ρ(1−( 1+ d 2 4 ρ 2 )−1/2)a^z V/m$ .

Explanation of Solution

Given Information:

The line charge density is given as,

$ρL={+ ρ 0 C/m0<z<d/2− ρ 0 C/m−d/2<z<0$

Calculation:

Let $(x,y,0)$ be a point on x-y plane, and $(0,0,z)$ is a point on line charge. Then differential electric field,

$dE→=ρL(xa ^x+ya ^y−za ^z)4πε0( x 2 + y 2 + z 2 )3/2$

So, the electric field intensity:

$E → = ∫ dE → = ∫ −d/2 d/2 ρ L ( x a ^ x +y a ^ y −z a ^ z ) 4π ε 0 ( x 2 + y 2 + z 2 ) 3/2 dz$

$= ∫ −d/2 0 − ρ 0 ( x a ^ x +y a ^ y −z a ^ z ) 4π ε 0 ( x 2 + y 2 + z 2 ) 3/2 dz + ∫ 0 d/2 ρ 0 ( x a ^ x +y a ^ y −z a ^ z ) 4π ε 0 ( x 2 + y 2 + z 2 ) 3/2 dz$

$= ρ 0 4π ε 0 [ ∫ −d/2 0 z a ^ z ( x 2 + y 2 + z 2 ) 3/2 dz− ∫ 0 d/2 z a ^ z ( x 2 + y 2 + z 2 ) 3/2 dz ]$

$= ρ 0 4π ε 0 ( [ − 1 ( x 2 + y 2 + z 2 ) 1/2 ] −d/2 0 − [ − 1 ( x 2 + y 2 + z 2 ) 1/2 ] 0 d/2 ) a ^ z$

$=− ρ 0 4π ε 0 ( [ 1 ( x 2 + y 2 ) 1/2 − 1 ( x 2 + y 2 + ( d/2 ) 2 ) 1/2 ]−[ 1 ( x 2 + y 2 + ( d/2 ) 2 ) 1/2 − 1 ( x 2 + y 2 ) 1/2 ] ) a ^ z$

$=− 2 ρ 0 4π ε 0 ( 1 ( x 2 + y 2 ) 1/2 − 1 ( x 2 + y 2 + ( d/2 ) 2 ) 1/2 ) a ^ z$

So, in cylindrical coordinates,

$E→(ρ)=−2ρ04πε0(1ρ−1 ( ρ 2 + ( d/2 ) 2 ) 1/2 )a^z ∵x2+y2=ρ2 =−ρ02πε0ρ(1−1 ( 1+ ( d/2ρ ) 2 ) 1/2 )a^z =−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1/2)a^z V/m$

Conclusion:

The required electric field intensity is:

$E→(ρ)=−ρ02πε0ρ(1−( 1+ d 2 4 ρ 2 )−1/2)a^z V/m$ .

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

(a)

The electric field intensity in x-y plane, as a function of cylindrical radius $ρ$ .

Answer 13

Answer to Problem 2.17P

The required electric field intensity is:

$E→(ρ)=−ρ02πε0ρ(1−( 1+ d 2 4 ρ 2 )−1/2)a^z V/m$ .

Answer 14

Explanation of Solution

Given Information:

The line charge density is given as,

$ρL={+ ρ 0 C/m0<z<d/2− ρ 0 C/m−d/2<z<0$

Calculation:

Let $(x,y,0)$ be a point on x-y plane, and $(0,0,z)$ is a point on line charge. Then differential electric field,

$dE→=ρL(xa ^x+ya ^y−za ^z)4πε0( x 2 + y 2 + z 2 )3/2$

So, the electric field intensity:

$E → = ∫ dE → = ∫ −d/2 d/2 ρ L ( x a ^ x +y a ^ y −z a ^ z ) 4π ε 0 ( x 2 + y 2 + z 2 ) 3/2 dz$

$= ∫ −d/2 0 − ρ 0 ( x a ^ x +y a ^ y −z a ^ z ) 4π ε 0 ( x 2 + y 2 + z 2 ) 3/2 dz + ∫ 0 d/2 ρ 0 ( x a ^ x +y a ^ y −z a ^ z ) 4π ε 0 ( x 2 + y 2 + z 2 ) 3/2 dz$

$= ρ 0 4π ε 0 [ ∫ −d/2 0 z a ^ z ( x 2 + y 2 + z 2 ) 3/2 dz− ∫ 0 d/2 z a ^ z ( x 2 + y 2 + z 2 ) 3/2 dz ]$

$= ρ 0 4π ε 0 ( [ − 1 ( x 2 + y 2 + z 2 ) 1/2 ] −d/2 0 − [ − 1 ( x 2 + y 2 + z 2 ) 1/2 ] 0 d/2 ) a ^ z$

$=− ρ 0 4π ε 0 ( [ 1 ( x 2 + y 2 ) 1/2 − 1 ( x 2 + y 2 + ( d/2 ) 2 ) 1/2 ]−[ 1 ( x 2 + y 2 + ( d/2 ) 2 ) 1/2 − 1 ( x 2 + y 2 ) 1/2 ] ) a ^ z$

$=− 2 ρ 0 4π ε 0 ( 1 ( x 2 + y 2 ) 1/2 − 1 ( x 2 + y 2 + ( d/2 ) 2 ) 1/2 ) a ^ z$

So, in cylindrical coordinates,

$E→(ρ)=−2ρ04πε0(1ρ−1 ( ρ 2 + ( d/2 ) 2 ) 1/2 )a^z ∵x2+y2=ρ2 =−ρ02πε0ρ(1−1 ( 1+ ( d/2ρ ) 2 ) 1/2 )a^z =−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1/2)a^z V/m$

Conclusion:

The required electric field intensity is:

$E→(ρ)=−ρ02πε0ρ(1−( 1+ d 2 4 ρ 2 )−1/2)a^z V/m$ .

Answer 15

Expert Solution

To determine

(b)

The electric field intensity in x-y plane for $ρ≫d$ .

Answer to Problem 2.17P

The required electric field intensity is,

$E→(ρ≫d)=−qd4πε0ρ3a^z V/m$ .

Explanation of Solution

Given Information:

The line charge density is given as,

$ρL={+ ρ 0 C/m0<z<d/2− ρ 0 C/m−d/2<z<0$

$E→(ρ)=−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1/2)a^z V/mq=ρ0d/2$

Calculation:

The electric field intensity,

$E→(ρ≫d)=−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1/2)a^z =−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1)a^z ∵(1+ d 2 4 ρ 2 )1/2≈(1+ d 2 4 ρ 2 ) =−ρ02πε0ρ( d 2 4 ρ 2 + d 2 ) =−ρ02πε0ρ( d 2 4 ρ 2 ) ∵4ρ2+d2≈4ρ2 =−2q2πε0ρd( d 2 4 ρ 2 ) ∵q=ρ0d/2 =−qd4πε0ρ3a^z V/m$

Conclusion:

The required electric field intensity is,

$E→(ρ≫d)=−qd4πε0ρ3a^z V/m$ .

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

(b)

The electric field intensity in x-y plane for $ρ≫d$ .

Answer 20

Answer to Problem 2.17P

The required electric field intensity is,

$E→(ρ≫d)=−qd4πε0ρ3a^z V/m$ .

Answer 21

Explanation of Solution

Given Information:

The line charge density is given as,

$ρL={+ ρ 0 C/m0<z<d/2− ρ 0 C/m−d/2<z<0$

$E→(ρ)=−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1/2)a^z V/mq=ρ0d/2$

Calculation:

The electric field intensity,

$E→(ρ≫d)=−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1/2)a^z =−ρ02πε0ρ(1− ( 1+ d 2 4 ρ 2 ) −1)a^z ∵(1+ d 2 4 ρ 2 )1/2≈(1+ d 2 4 ρ 2 ) =−ρ02πε0ρ( d 2 4 ρ 2 + d 2 ) =−ρ02πε0ρ( d 2 4 ρ 2 ) ∵4ρ2+d2≈4ρ2 =−2q2πε0ρd( d 2 4 ρ 2 ) ∵q=ρ0d/2 =−qd4πε0ρ3a^z V/m$