The magnitude and the direction of the electric field for a non- conducting spherical shell concentric with a solid sphere.

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Answer 1

Question

Chapter 22, Problem 76P

(a)

To determine

The magnitude and the direction of the electric field for a non-conducting spherical shell concentric with a solid sphere.

(a)

Expert Solution

Explanation of Solution

Given:

The diameter of the sphere is $1.20 m$ .

The volume charge density of the sphere is $+5.00 μC/m3$ .

The diameter of the shell is $2.40 m$ .

The surface charge density of the sphere is $−1.50 μC/m2$ .

Formula used:

Write the expression of the electric field at any point for a non-conducting sphere.

$E→sphere=k Qr2r^$ …… (1)

Here, $E→sphere$ is the electric field, $k$ is Coulomb’s constant, $Q$ is the total charge, $r$ is the distance of any point and $r^$ is the unit vector along $r$ .

Write the expression for charge for a sphere.

$Q = ρV$

Here, $ρ$ is the volume charge density and $V$ is the volume of the sphere.

Substitute $4π3a3$ for $V$ in the above expression.

$Q = ρ(4π3a3)$

Here, $a$ is the radius of the sphere.

Substitute $ρ(4π3a3)$ for $Q$ in equation (1) and rearrange.

$E→sphere =4π3k ρa3r2r^$ …… (2)

Write the above expression when $(r ≤ a)$ .

$E→sphere =4π3kρr r^$ …… (3)

Simplify the above equation.

$E→sphere =4π3kρrr^$

Write the expression for the electric field at any point due to a spherical shell.

$E→shell=k qr2r^$ …… (4)

Here, $E→shell$ is the electric field, $k$ is Coulomb’s constant, $q$ is the total charge, $r$ is the distance of any point and $r^$ is the unit vector along $r$ .

Write the expression charge of a spherical shell.

$q = ρA$

Here, $σ$ is the surface charge density and $A$ is the surface area of the sphere.

Substitute $4πa2$ for $A$ in the above equation.

$q = ρ(4πa2)$

Here, $a$ is the radius of the sphere.

Substitute $ρ(4πa2)$ for $q$ in equation (4).

$E→shell=k ρ(4πa2)r2r^$ …… (5)

Write the expression for the resultant electric field at any point in space due to a spherical shell and a solid sphere.

$E→=E→shell + E→sphere$ …… (6)

Calculation:

The electric field at point $x = 4.50 m , y = 0$ for the sphere is calculated below.

Substitute $8.988×109 N⋅m2 /C2$ for $k$ , $+5.00 μC/m3$ for $ρ$ , $(4.50 m −4.00 m)$ for $r$ and $i^$ for $r^$ in equation (3).

$E→sphere =4π3(8.988× 109 N⋅m 2 /C2)(+5.00 μC/m3( 10 −6 C 1μC ))(4.50 m −4.00 m)i^=94122.11 N/C i^≈ 94 kN/C i^$

The direction of $E→sphere$ is calculated below.

$θ =0°$

$(4.50 m , 0)$ is inside the spherical shell.

The electric field at point $x = 4.50 m , y = 0$ for the spherical shell is calculated below.

$E→shell=0$

The electric field at point $(4.50 m , 0)$ is calculated below.

Substitute $94 kN/C i^$ for $E→sphere$ and $0$ for $E→shell$ in equation (6).

$E→=94 kN/C i^$

Conclusion:

Thus, the electric magnitude and the direction of electric field at point $(4.50 m , 0)$ is $94 kN/C$ and $0°$ respectively.

(b)

To determine

The magnitude and the direction of the electric field for a non-conducting spherical shell concentric with a solid sphere.

(b)

Expert Solution

Explanation of Solution

Given:

The diameter of the sphere is $1.20 m$ .

The volume charge density of the sphere is $+5.00 μC/m3$ .

The diameter of the shell is $2.40 m$ .

The surface charge density of the sphere is $−1.50 μC/m2$ .

Formula used:

Write the expression of the electric field at any point for a non-conducting sphere.

$E→sphere=k Qr2r^$

Here, $E→sphere$ is the electric field, $k$ is Coulomb’s constant, $Q$ is the total charge, $r$ is the distance of any point and $r^$ is the unit vector along $r$ .

Write the expression for charge for a sphere.

$Q = ρV$

Here, $ρ$ is the volume charge density and $V$ is the volume of the sphere.

Substitute $4π3a3$ for $V$ in the above expression.

$Q = ρ(4π3a3)$

Here, $a$ is the radius of the sphere.

Substitute $ρ(4π3a3)$ for $Q$ in equation (1) and rearrange.

$E→sphere =4π3k ρa3r2r^$

Write the above expression when $(r ≤ a)$ .

$E→sphere =4π3kρr r^$

Simplify the above equation.

$E→sphere =4π3kρrr^$

Write the expression for the electric field at any point due to a spherical shell.

$E→shell=k qr2r^$

Here, $E→shell$ is the electric field, $k$ is Coulomb’s constant, $q$ is the total charge, $r$ is the distance of any point and $r^$ is the unit vector along $r$ .

Write the expression charge of a spherical shell.

$q = ρA$

Here, $σ$ is the surface charge density and $A$ is the surface area of the sphere.

Substitute $4πa2$ for $A$ in the above equation.

$q = ρ(4πa2)$

Here, $a$ is the radius of the sphere.

Substitute $ρ(4πa2)$ for $q$ in equation (4).

$E→shell=k ρ(4πa2)r2r^$

Resultant electric field at any point in space due to a spherical shell and a solid sphere.

$E→=E→shell + E→sphere$

Calculation:

The electric field at point $x = 4.0 m , y = 1.10 m$ for the sphere is calculated below.

Substitute $8.988×109 N⋅m2 /C2$ for $k$ , $+5.00 μC/m3$ for $ρ$ , $0.600 m$ for $a$ , $1.10 m$ for $r$ , $0.600 m$ for $a$ and $j^$ for $r^$ in equation (2).

$E→sphere =4π3(8.988× 109 N⋅m 2 /C2) ( +5.00 μC/m 3 ( 10 −6 C 1μC )) ( 0.600 )3 ( 1.10 m )2j^= 33603.92 N/C j^= 33.6 kN/C j^$

The direction of $E→sphere$ at point $(4.0 m , 1.10 m)$ is calculated below.

$θ = tan−1( 1.10 4.0 −4.0)= tan−1(0)= 90°$

$(4.0 m , 1.10 m)$ is inside the spherical shell.

The electric field at point $x = 4.0 m , y = 1.10 m$ for the spherical shell is calculated below.

$E→shell=0$

The electric field at point $(4.50 m , 0)$ is calculated below.

Substitute $33.6 kN/C j^$ for $E→sphere$ and $0$ for $E→shell$ in equation (6).

$E→=33.6 kN/C j^$

Conclusion:

Thus, the electric magnitude and the direction of electric field at point $(4.0 m , 1.10 m)$ is $33.6 kN/C$ and $90°$ respectively.

(c)

To determine

The magnitude and the direction of the electric field for a non-conducting spherical shell concentric with a solid sphere.

(c)

Expert Solution

Explanation of Solution

Given:

The diameter of the sphere is $1.20 m$ .

The volume charge density of the sphere is $+5.00 μC/m3$ .

The diameter of the shell is $2.40 m$ .

The surface charge density of the sphere is $−1.50 μC/m2$ .

Formula used:

Write the expression of the electric field at any point for a non-conducting sphere.

$E→sphere=k Qr2r^$

Here, $E→sphere$ is the electric field, $k$ is Coulomb’s constant, $Q$ is the total charge, $r$ is the distance of any point and $r^$ is the unit vector along $r$ .

Write the expression for charge for a sphere.

$Q = ρV$

Here, $ρ$ is the volume charge density and $V$ is the volume of the sphere.

Substitute $4π3a3$ for $V$ in the above expression.

$Q = ρ(4π3a3)$

Here, $a$ is the radius of the sphere.

Substitute $ρ(4π3a3)$ for $Q$ in equation (1) and rearrange.

$E→sphere =4π3k ρa3r2r^$

Write the above expression when $(r ≤ a)$ .

$E→sphere =4π3kρr r^$

Simplify the above equation.

$E→sphere =4π3kρrr^$

Write the expression for the electric field at any point due to a spherical shell.

$E→shell=k qr2r^$

Here, $E→shell$ is the electric field, $k$ is Coulomb’s constant, $q$ is the total charge, $r$ is the distance of any point and $r^$ is the unit vector along $r$ .

Write the expression charge of a spherical shell.

$q = ρA$

Here, $σ$ is the surface charge density and $A$ is the surface area of the sphere.

Substitute $4πa2$ for $A$ in the above equation.

$q = ρ(4πa2)$

Here, $a$ is the radius of the sphere.

Substitute $ρ(4πa2)$ for $q$ in equation (4).

$E→shell=k ρ(4πa2)r2r^$

Resultant electric field at any point in space due to a spherical shell and a solid sphere.

$E→=E→shell + E→sphere$

Calculation:

Write the expression for distance between the points $(4.0 m , 0)$ and $(2.0 m , 3.0 m)$ .

$r = ( 4.0 m−2.0m )2+ ( 0 m−3.0 m )2=3.606 m$

Write the direction for $r$ .

$θ = tan−1( 0 −3.0 4.0 −2.0)≈ −56.3°$

When the above value is subtracted from $360°$ .

$θ=304°$

Write the expression for unit vector along $r$ .

$r^ = cosθ i^ +sinθ j^$

Substitute $123.7°$ for $θ$ in the above equation.

$r = cos(123.7°) i^ +sin(123.7°) j^≈ −0.554 i^+ 0.832 j^$

The electric field at point $x = 2.0 m , y =3. 0 m$ for the sphere is calculated below.

Substitute $8.988×109 N⋅m2 /C2$ for $k$ , $+5.00 μC/m3$ for $ρ$ , $0.600 m$ for $a$ , $3.606 m$ for $r$ and $(−0.554 i^+ 0.832 j^)$ for $r^$ in equation (3).

$E→sphere =4π3(8.988× 109 N⋅m 2 /C2) ( +5.00 μC/m 3 ( 10 −6 C 1μC )) ( 0.600 )3 ( 3.606 m )2(−0.554 i^+ 0.832 j^)= 3.127 kN/C (−0.554 i^+ 0.832 j^)≈ −1.732 kN/C i^+ 2.601 kN/C j^$

The electric field at point $x = 2.0 m , y =3. 0 m$ for the spherical shell is calculated below.

Substitute $8.988×109 N⋅m2 /C2$ for $k$ , $+5.00 μC/m3$ for $ρ$ , $1.20 m$ for $a$ , $3.606 m$ for $r$ and $(−0.554 i^+ 0.832 j^)$ for $r^$ in equation (5).

$E→sphere =4π(8.988× 109 N⋅m 2 /C2) ( −1.50 μC/m 2 ( 10 −6 C 1μC )) ( 1.20 )2 ( 3.606 m )2(−0.554 i^+ 0.832 j^)= −18.77 kN/C (−0.554 i^+ 0.832 j^)≈ 10.40 kN/C i^−15.61 kN/C j^$

The electric field at point $(2.0 m , 3.0 m)$ is calculated below.

Substitute $(−1.732 kN/C i^+ 2.601 kN/C j^)$ for $E→sphere$ and $(10.40 kN/C i^−15.61 kN/C j^)$ for $E→shell$ in equation (6).

$E→=(−1.732 kN/C i^+ 2.601 kN/C j^)+(10.40 kN/C i^−15.61 kN/C j^)= 8.668 kN/C i^−13.01 kN/C j^$

The magnitude of electric field at $(2.0 m , 3.0 m)$ is calculated below.

$E = ( −13.01 kN/C )2+ ( 8.668 kN/C )2≈ 15.62 kN/C$

Conclusion:

Thus, the electric magnitude and the direction of electric field at point $(2.0 m , 3.0 m)$ is $15.62 kN/C$ and $304°$ respectively.

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