The total and individual charge on the inner and outer surface of the spherical shell.

Question

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Answer 1

Question

Chapter 22, Problem 63P

(a)

To determine

The total and individual charge on the inner and outer surface of the spherical shell.

(a)

Expert Solution

Explanation of Solution

Given:

A positive point charge of $2.5 μC$ is placed at the center of a conducting sphere. The radius of the inner surface is $60 cm$ and the radius of the outer surface is $90 cm$ .

Formula used:

The positive charge at the center of the spherical sphere induces a charge of same magnitude but different in polarity on the inner surface. The negative charge on the inner surface induces a positive charge of same magnitude on the outer surface.

Write the expression for the surface charge density

$σsurface=QenclosedA$ ........ (1)

Here, $σsurface$ is the surface charge density, $Qenclosed$ is the total charge enclosed by the surface of the spherical shell and $A$ is the surface area of the spherical shell.

Write the expression for the surface area of the square sheet.

$A=4πR2$

Here $R$ is the radius of the spherical shell.

Substitute $4πR2$ for $A$ in expression (1).

$σsurface=Qenclosed4πR2$ ........ (2)

Forinner surface of the spherical shell.

The induced charge on the inner surface has same magnitude of the point charge .The polarity of the charge is opposite to the point charge.

Calculation:

Substitute $−2.5 μC$ for $Qenclosed$ and $60 cm$ for $R$ in the expression (2).

$σsurface=−2.5 μC4π(60 cm(1 m100 cm))2=−0.55 μC/m2$

For outer surface of the spherical shell

The induced charge on the outer surface has same magnitude of the point charge .The polarity of the charge is same to the point charge

Substitute $2.5 μC$ for $Qenclosed$ and $90 cm$ for $R$ in the expression (2).

$σsurface=2.5 μC4π(90 cm(1 m100 cm))2=0.25 μC/m2$

Conclusion:

Thus, the total charge on the inner surface is $−2.5 μC$ and the surface charge density is $−0.55 μC/m2$ .The total charge on the outer surface is $2.5 μC$ and the surface charge density is $0.25 μC/m2$ .

(b)

To determine

The electric field due to a positive point charge located at the spherical shell.

(b)

Expert Solution

Explanation of Solution

Given:

A positive point charge of $2.5 μC$ is placed at the center of a conducting sphere. The radius of the inner surface is $60 cm$ and the radius of the outer surface is $90 cm$ .

Formula used:

Write the expression for Gauss’s law.

$∫E→.ds→=Qenclosedε0$

Here, $E→$ is the electric field, $ds→$ is the elementary surface area, $Qenclosed$ is the total charge enclosed by the spherical shell and $ε0$ is the permittivity of free space.

Substitute $4πr2$ for $ds→$ in the above expression.

$∫E⋅4πr2=Qenclosedε0$

Here $r$ is the distance from the center.

Rearrange the above equation for $E$ .

$E=Qenclosed4πr2⋅ε0$

Substitute $K$ for $1/4πε0$ in the above equation.

$E=KQenclosedr2$ ........ (3)

Calculation:

For $r<60 cm$

Substitute $2.5 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(2.5 μC(1 C106 μC))r2=(2.25×104 Nm2/C)1r2$

For $60 cm<r<90 cm$

Substitute $0 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(0 μC)r2=0$

For $r>90 cm$

Substitute $2.5 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(2.5 μC(1 C106 μC))r2=(2.25×104 Nm2/C)1r2$

Conclusion:

Thus, the electric field of the spherical shell for $r<60 cm$ is $(2.5×104 Nm2/C)1r2$ , For $60 cm<r<90 cm$ is $0$ and For $r > 90 cm$ is $(2.5×104 Nm2/C)1r2$ .

(c)

To determine

The electric field and surface charge density of the spherical shell.

(c)

Expert Solution

Explanation of Solution

Given:

A positive point charge of $3.5 μC$ is placed at the center of a conducting sphere. The radius of the inner surface is $60 cm$ and the radius of the outer surface is $90 cm$ .

Formula used:

The positive charge at the center of the spherical sphere induces a charge of same magnitude but different in polarity on the inner surface. The negative charge on the inner surface induces a positive charge of same magnitude on the outer surface.

Write the expression for the surface charge density

$σsurface=QenclosedA$

Here, $σsurface$ is the surface charge density, $Qenclosed$ is the total charge enclosed by the surface of the spherical shell and $A$ is the surface area of the spherical shell.

Write the expression for the surface area of the square sheet.

$A=4πR2$

Here $R$ is the radius of the spherical shell.

Substitute $4πR2$ for $A$ in expression (1).

$σsurface=Qenclosed4πR2$

For inner surface of the spherical shell.

The induced charge on the inner surface has same magnitude of the point charge .The polarity of the charge is opposite to the point charge

Write the expression for Gauss’s law.

$∫E→.ds→=Qenclosedε0$

Here, $E→$ is the electric field, $ds→$ is the elementary surface area, $Qenclosed$ is the total charge enclosed by the spherical shell and $ε0$ is the permittivity of free space.

Substitute $4πr2$ for $ds→$ in the above expression.

$∫E⋅4πr2=Qenclosedε0$

Here $r$ is the distance from the center.

Rearrange the above equation for $E$ .

$E=Qenclosed4πr2⋅ε0$

Substitute $K$ for $1/4πε0$ in the above equation.

$E=KQenclosedr2$

Calculation:

For inner surface $E=0$ inside a conductor.

Substitute for $−2.5 μC$

$Qenclosed$ and $60 cm$ for $R$ in the expression (2).

$σsurface=−2.5 μC4π(60 cm(1 m100 cm))2=−0.55 μC/m2$

For outer surface of the spherical shell

Write the expression for surface charge.

$qouter=3.50 μC -qinner$

Substitute $−2.5 μC$ for $qinner$

$qouter=6.0 μC$

For outer surface.

Substitute for $6.0 μC$

$Qenclosed$ and $90 cm$ for $R$ in the expression (2).

$σsurface=6.0 μC4π(90 cm(1 m100 cm))2=0.59 μC/m2$

For $r<60 cm$

Substitute $2.5 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(2.5 μC(1 C106 μC))r2=(2.25×104 Nm2/C)1r2$

For $60 cm<r<90 cm$

Substitute $0 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(0 μC)r2=0$

For $r>90 cm$

Substitute $6.0 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(6.0 μC(1 C106 μC))r2=(5.4×104 Nm2/C)1r2$

Conclusion:

Thus, the total charge on the inner surface is $−2.5 μC$ and the surface charge density is $−0.55 μC/m2$ .The total charge on the outer surface is $6.0μC$ and the surface charge density is $0.59 μC/m2$ . Thus, the electric field of the spherical shell for $r<60 cm$ is $(2.25×104 Nm2/C)1r2$ , for $60 cm<r<90 cm$ is $0$ and for $r>90 cm$ is $(5.4×104 Nm2/C)1r2$ .

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Answer 2

Question

Chapter 22, Problem 63P

(a)

To determine

The total and individual charge on the inner and outer surface of the spherical shell.

(a)

Expert Solution

Explanation of Solution

Given:

A positive point charge of $2.5 μC$ is placed at the center of a conducting sphere. The radius of the inner surface is $60 cm$ and the radius of the outer surface is $90 cm$ .

Formula used:

The positive charge at the center of the spherical sphere induces a charge of same magnitude but different in polarity on the inner surface. The negative charge on the inner surface induces a positive charge of same magnitude on the outer surface.

Write the expression for the surface charge density

$σsurface=QenclosedA$ ........ (1)

Here, $σsurface$ is the surface charge density, $Qenclosed$ is the total charge enclosed by the surface of the spherical shell and $A$ is the surface area of the spherical shell.

Write the expression for the surface area of the square sheet.

$A=4πR2$

Here $R$ is the radius of the spherical shell.

Substitute $4πR2$ for $A$ in expression (1).

$σsurface=Qenclosed4πR2$ ........ (2)

Forinner surface of the spherical shell.

The induced charge on the inner surface has same magnitude of the point charge .The polarity of the charge is opposite to the point charge.

Calculation:

Substitute $−2.5 μC$ for $Qenclosed$ and $60 cm$ for $R$ in the expression (2).

$σsurface=−2.5 μC4π(60 cm(1 m100 cm))2=−0.55 μC/m2$

For outer surface of the spherical shell

The induced charge on the outer surface has same magnitude of the point charge .The polarity of the charge is same to the point charge

Substitute $2.5 μC$ for $Qenclosed$ and $90 cm$ for $R$ in the expression (2).

$σsurface=2.5 μC4π(90 cm(1 m100 cm))2=0.25 μC/m2$

Conclusion:

Thus, the total charge on the inner surface is $−2.5 μC$ and the surface charge density is $−0.55 μC/m2$ .The total charge on the outer surface is $2.5 μC$ and the surface charge density is $0.25 μC/m2$ .

(b)

To determine

The electric field due to a positive point charge located at the spherical shell.

(b)

Expert Solution

Explanation of Solution

Given:

A positive point charge of $2.5 μC$ is placed at the center of a conducting sphere. The radius of the inner surface is $60 cm$ and the radius of the outer surface is $90 cm$ .

Formula used:

Write the expression for Gauss’s law.

$∫E→.ds→=Qenclosedε0$

Here, $E→$ is the electric field, $ds→$ is the elementary surface area, $Qenclosed$ is the total charge enclosed by the spherical shell and $ε0$ is the permittivity of free space.

Substitute $4πr2$ for $ds→$ in the above expression.

$∫E⋅4πr2=Qenclosedε0$

Here $r$ is the distance from the center.

Rearrange the above equation for $E$ .

$E=Qenclosed4πr2⋅ε0$

Substitute $K$ for $1/4πε0$ in the above equation.

$E=KQenclosedr2$ ........ (3)

Calculation:

For $r<60 cm$

Substitute $2.5 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(2.5 μC(1 C106 μC))r2=(2.25×104 Nm2/C)1r2$

For $60 cm<r<90 cm$

Substitute $0 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(0 μC)r2=0$

For $r>90 cm$

Substitute $2.5 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(2.5 μC(1 C106 μC))r2=(2.25×104 Nm2/C)1r2$

Conclusion:

Thus, the electric field of the spherical shell for $r<60 cm$ is $(2.5×104 Nm2/C)1r2$ , For $60 cm<r<90 cm$ is $0$ and For $r > 90 cm$ is $(2.5×104 Nm2/C)1r2$ .

(c)

To determine

The electric field and surface charge density of the spherical shell.

(c)

Expert Solution

Explanation of Solution

Given:

A positive point charge of $3.5 μC$ is placed at the center of a conducting sphere. The radius of the inner surface is $60 cm$ and the radius of the outer surface is $90 cm$ .

Formula used:

The positive charge at the center of the spherical sphere induces a charge of same magnitude but different in polarity on the inner surface. The negative charge on the inner surface induces a positive charge of same magnitude on the outer surface.

Write the expression for the surface charge density

$σsurface=QenclosedA$

Here, $σsurface$ is the surface charge density, $Qenclosed$ is the total charge enclosed by the surface of the spherical shell and $A$ is the surface area of the spherical shell.

Write the expression for the surface area of the square sheet.

$A=4πR2$

Here $R$ is the radius of the spherical shell.

Substitute $4πR2$ for $A$ in expression (1).

$σsurface=Qenclosed4πR2$

For inner surface of the spherical shell.

The induced charge on the inner surface has same magnitude of the point charge .The polarity of the charge is opposite to the point charge

Write the expression for Gauss’s law.

$∫E→.ds→=Qenclosedε0$

Here, $E→$ is the electric field, $ds→$ is the elementary surface area, $Qenclosed$ is the total charge enclosed by the spherical shell and $ε0$ is the permittivity of free space.

Substitute $4πr2$ for $ds→$ in the above expression.

$∫E⋅4πr2=Qenclosedε0$

Here $r$ is the distance from the center.

Rearrange the above equation for $E$ .

$E=Qenclosed4πr2⋅ε0$

Substitute $K$ for $1/4πε0$ in the above equation.

$E=KQenclosedr2$

Calculation:

For inner surface $E=0$ inside a conductor.

Substitute for $−2.5 μC$

$Qenclosed$ and $60 cm$ for $R$ in the expression (2).

$σsurface=−2.5 μC4π(60 cm(1 m100 cm))2=−0.55 μC/m2$

For outer surface of the spherical shell

Write the expression for surface charge.

$qouter=3.50 μC -qinner$

Substitute $−2.5 μC$ for $qinner$

$qouter=6.0 μC$

For outer surface.

Substitute for $6.0 μC$

$Qenclosed$ and $90 cm$ for $R$ in the expression (2).

$σsurface=6.0 μC4π(90 cm(1 m100 cm))2=0.59 μC/m2$

For $r<60 cm$

Substitute $2.5 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(2.5 μC(1 C106 μC))r2=(2.25×104 Nm2/C)1r2$

For $60 cm<r<90 cm$

Substitute $0 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(0 μC)r2=0$

For $r>90 cm$

Substitute $6.0 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(6.0 μC(1 C106 μC))r2=(5.4×104 Nm2/C)1r2$

Conclusion:

Thus, the total charge on the inner surface is $−2.5 μC$ and the surface charge density is $−0.55 μC/m2$ .The total charge on the outer surface is $6.0μC$ and the surface charge density is $0.59 μC/m2$ . Thus, the electric field of the spherical shell for $r<60 cm$ is $(2.25×104 Nm2/C)1r2$ , for $60 cm<r<90 cm$ is $0$ and for $r>90 cm$ is $(5.4×104 Nm2/C)1r2$ .

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Answer 3

Question

Chapter 22, Problem 63P

(a)

To determine

The total and individual charge on the inner and outer surface of the spherical shell.

(a)

Expert Solution

Explanation of Solution

Given:

A positive point charge of $2.5 μC$ is placed at the center of a conducting sphere. The radius of the inner surface is $60 cm$ and the radius of the outer surface is $90 cm$ .

Formula used:

The positive charge at the center of the spherical sphere induces a charge of same magnitude but different in polarity on the inner surface. The negative charge on the inner surface induces a positive charge of same magnitude on the outer surface.

Write the expression for the surface charge density

$σsurface=QenclosedA$ ........ (1)

Here, $σsurface$ is the surface charge density, $Qenclosed$ is the total charge enclosed by the surface of the spherical shell and $A$ is the surface area of the spherical shell.

Write the expression for the surface area of the square sheet.

$A=4πR2$

Here $R$ is the radius of the spherical shell.

Substitute $4πR2$ for $A$ in expression (1).

$σsurface=Qenclosed4πR2$ ........ (2)

Forinner surface of the spherical shell.

The induced charge on the inner surface has same magnitude of the point charge .The polarity of the charge is opposite to the point charge.

Calculation:

Substitute $−2.5 μC$ for $Qenclosed$ and $60 cm$ for $R$ in the expression (2).

$σsurface=−2.5 μC4π(60 cm(1 m100 cm))2=−0.55 μC/m2$

For outer surface of the spherical shell

The induced charge on the outer surface has same magnitude of the point charge .The polarity of the charge is same to the point charge

Substitute $2.5 μC$ for $Qenclosed$ and $90 cm$ for $R$ in the expression (2).

$σsurface=2.5 μC4π(90 cm(1 m100 cm))2=0.25 μC/m2$

Conclusion:

Thus, the total charge on the inner surface is $−2.5 μC$ and the surface charge density is $−0.55 μC/m2$ .The total charge on the outer surface is $2.5 μC$ and the surface charge density is $0.25 μC/m2$ .

(b)

To determine

The electric field due to a positive point charge located at the spherical shell.

(b)

Expert Solution

Explanation of Solution

Given:

A positive point charge of $2.5 μC$ is placed at the center of a conducting sphere. The radius of the inner surface is $60 cm$ and the radius of the outer surface is $90 cm$ .

Formula used:

Write the expression for Gauss’s law.

$∫E→.ds→=Qenclosedε0$

Here, $E→$ is the electric field, $ds→$ is the elementary surface area, $Qenclosed$ is the total charge enclosed by the spherical shell and $ε0$ is the permittivity of free space.

Substitute $4πr2$ for $ds→$ in the above expression.

$∫E⋅4πr2=Qenclosedε0$

Here $r$ is the distance from the center.

Rearrange the above equation for $E$ .

$E=Qenclosed4πr2⋅ε0$

Substitute $K$ for $1/4πε0$ in the above equation.

$E=KQenclosedr2$ ........ (3)

Calculation:

For $r<60 cm$

Substitute $2.5 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(2.5 μC(1 C106 μC))r2=(2.25×104 Nm2/C)1r2$

For $60 cm<r<90 cm$

Substitute $0 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(0 μC)r2=0$

For $r>90 cm$

Substitute $2.5 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(2.5 μC(1 C106 μC))r2=(2.25×104 Nm2/C)1r2$

Conclusion:

Thus, the electric field of the spherical shell for $r<60 cm$ is $(2.5×104 Nm2/C)1r2$ , For $60 cm<r<90 cm$ is $0$ and For $r > 90 cm$ is $(2.5×104 Nm2/C)1r2$ .

(c)

To determine

The electric field and surface charge density of the spherical shell.

(c)

Expert Solution

Explanation of Solution

Given:

A positive point charge of $3.5 μC$ is placed at the center of a conducting sphere. The radius of the inner surface is $60 cm$ and the radius of the outer surface is $90 cm$ .

Formula used:

The positive charge at the center of the spherical sphere induces a charge of same magnitude but different in polarity on the inner surface. The negative charge on the inner surface induces a positive charge of same magnitude on the outer surface.

Write the expression for the surface charge density

$σsurface=QenclosedA$

Here, $σsurface$ is the surface charge density, $Qenclosed$ is the total charge enclosed by the surface of the spherical shell and $A$ is the surface area of the spherical shell.

Write the expression for the surface area of the square sheet.

$A=4πR2$

Here $R$ is the radius of the spherical shell.

Substitute $4πR2$ for $A$ in expression (1).

$σsurface=Qenclosed4πR2$

For inner surface of the spherical shell.

The induced charge on the inner surface has same magnitude of the point charge .The polarity of the charge is opposite to the point charge

Write the expression for Gauss’s law.

$∫E→.ds→=Qenclosedε0$

Here, $E→$ is the electric field, $ds→$ is the elementary surface area, $Qenclosed$ is the total charge enclosed by the spherical shell and $ε0$ is the permittivity of free space.

Substitute $4πr2$ for $ds→$ in the above expression.

$∫E⋅4πr2=Qenclosedε0$

Here $r$ is the distance from the center.

Rearrange the above equation for $E$ .

$E=Qenclosed4πr2⋅ε0$

Substitute $K$ for $1/4πε0$ in the above equation.

$E=KQenclosedr2$

Calculation:

For inner surface $E=0$ inside a conductor.

Substitute for $−2.5 μC$

$Qenclosed$ and $60 cm$ for $R$ in the expression (2).

$σsurface=−2.5 μC4π(60 cm(1 m100 cm))2=−0.55 μC/m2$

For outer surface of the spherical shell

Write the expression for surface charge.

$qouter=3.50 μC -qinner$

Substitute $−2.5 μC$ for $qinner$

$qouter=6.0 μC$

For outer surface.

Substitute for $6.0 μC$

$Qenclosed$ and $90 cm$ for $R$ in the expression (2).

$σsurface=6.0 μC4π(90 cm(1 m100 cm))2=0.59 μC/m2$

For $r<60 cm$

Substitute $2.5 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(2.5 μC(1 C106 μC))r2=(2.25×104 Nm2/C)1r2$

For $60 cm<r<90 cm$

Substitute $0 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(0 μC)r2=0$

For $r>90 cm$

Substitute $6.0 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(6.0 μC(1 C106 μC))r2=(5.4×104 Nm2/C)1r2$

Conclusion:

Thus, the total charge on the inner surface is $−2.5 μC$ and the surface charge density is $−0.55 μC/m2$ .The total charge on the outer surface is $6.0μC$ and the surface charge density is $0.59 μC/m2$ . Thus, the electric field of the spherical shell for $r<60 cm$ is $(2.25×104 Nm2/C)1r2$ , for $60 cm<r<90 cm$ is $0$ and for $r>90 cm$ is $(5.4×104 Nm2/C)1r2$ .

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Answer 4

Question

Chapter 22, Problem 63P

(a)

To determine

The total and individual charge on the inner and outer surface of the spherical shell.

(a)

Expert Solution

Explanation of Solution

Given:

A positive point charge of $2.5 μC$ is placed at the center of a conducting sphere. The radius of the inner surface is $60 cm$ and the radius of the outer surface is $90 cm$ .

Formula used:

The positive charge at the center of the spherical sphere induces a charge of same magnitude but different in polarity on the inner surface. The negative charge on the inner surface induces a positive charge of same magnitude on the outer surface.

Write the expression for the surface charge density

$σsurface=QenclosedA$ ........ (1)

Here, $σsurface$ is the surface charge density, $Qenclosed$ is the total charge enclosed by the surface of the spherical shell and $A$ is the surface area of the spherical shell.

Write the expression for the surface area of the square sheet.

$A=4πR2$

Here $R$ is the radius of the spherical shell.

Substitute $4πR2$ for $A$ in expression (1).

$σsurface=Qenclosed4πR2$ ........ (2)

Forinner surface of the spherical shell.

The induced charge on the inner surface has same magnitude of the point charge .The polarity of the charge is opposite to the point charge.

Calculation:

Substitute $−2.5 μC$ for $Qenclosed$ and $60 cm$ for $R$ in the expression (2).

$σsurface=−2.5 μC4π(60 cm(1 m100 cm))2=−0.55 μC/m2$

For outer surface of the spherical shell

The induced charge on the outer surface has same magnitude of the point charge .The polarity of the charge is same to the point charge

Substitute $2.5 μC$ for $Qenclosed$ and $90 cm$ for $R$ in the expression (2).

$σsurface=2.5 μC4π(90 cm(1 m100 cm))2=0.25 μC/m2$

Conclusion:

Thus, the total charge on the inner surface is $−2.5 μC$ and the surface charge density is $−0.55 μC/m2$ .The total charge on the outer surface is $2.5 μC$ and the surface charge density is $0.25 μC/m2$ .

(b)

To determine

The electric field due to a positive point charge located at the spherical shell.

(b)

Expert Solution

Explanation of Solution

Given:

A positive point charge of $2.5 μC$ is placed at the center of a conducting sphere. The radius of the inner surface is $60 cm$ and the radius of the outer surface is $90 cm$ .

Formula used:

Write the expression for Gauss’s law.

$∫E→.ds→=Qenclosedε0$

Here, $E→$ is the electric field, $ds→$ is the elementary surface area, $Qenclosed$ is the total charge enclosed by the spherical shell and $ε0$ is the permittivity of free space.

Substitute $4πr2$ for $ds→$ in the above expression.

$∫E⋅4πr2=Qenclosedε0$

Here $r$ is the distance from the center.

Rearrange the above equation for $E$ .

$E=Qenclosed4πr2⋅ε0$

Substitute $K$ for $1/4πε0$ in the above equation.

$E=KQenclosedr2$ ........ (3)

Calculation:

For $r<60 cm$

Substitute $2.5 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(2.5 μC(1 C106 μC))r2=(2.25×104 Nm2/C)1r2$

For $60 cm<r<90 cm$

Substitute $0 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(0 μC)r2=0$

For $r>90 cm$

Substitute $2.5 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(2.5 μC(1 C106 μC))r2=(2.25×104 Nm2/C)1r2$

Conclusion:

Thus, the electric field of the spherical shell for $r<60 cm$ is $(2.5×104 Nm2/C)1r2$ , For $60 cm<r<90 cm$ is $0$ and For $r > 90 cm$ is $(2.5×104 Nm2/C)1r2$ .

(c)

To determine

The electric field and surface charge density of the spherical shell.

(c)

Expert Solution

Explanation of Solution

Given:

A positive point charge of $3.5 μC$ is placed at the center of a conducting sphere. The radius of the inner surface is $60 cm$ and the radius of the outer surface is $90 cm$ .

Formula used:

The positive charge at the center of the spherical sphere induces a charge of same magnitude but different in polarity on the inner surface. The negative charge on the inner surface induces a positive charge of same magnitude on the outer surface.

Write the expression for the surface charge density

$σsurface=QenclosedA$

Here, $σsurface$ is the surface charge density, $Qenclosed$ is the total charge enclosed by the surface of the spherical shell and $A$ is the surface area of the spherical shell.

Write the expression for the surface area of the square sheet.

$A=4πR2$

Here $R$ is the radius of the spherical shell.

Substitute $4πR2$ for $A$ in expression (1).

$σsurface=Qenclosed4πR2$

For inner surface of the spherical shell.

The induced charge on the inner surface has same magnitude of the point charge .The polarity of the charge is opposite to the point charge

Write the expression for Gauss’s law.

$∫E→.ds→=Qenclosedε0$

Here, $E→$ is the electric field, $ds→$ is the elementary surface area, $Qenclosed$ is the total charge enclosed by the spherical shell and $ε0$ is the permittivity of free space.

Substitute $4πr2$ for $ds→$ in the above expression.

$∫E⋅4πr2=Qenclosedε0$

Here $r$ is the distance from the center.

Rearrange the above equation for $E$ .

$E=Qenclosed4πr2⋅ε0$

Substitute $K$ for $1/4πε0$ in the above equation.

$E=KQenclosedr2$

Calculation:

For inner surface $E=0$ inside a conductor.

Substitute for $−2.5 μC$

$Qenclosed$ and $60 cm$ for $R$ in the expression (2).

$σsurface=−2.5 μC4π(60 cm(1 m100 cm))2=−0.55 μC/m2$

For outer surface of the spherical shell

Write the expression for surface charge.

$qouter=3.50 μC -qinner$

Substitute $−2.5 μC$ for $qinner$

$qouter=6.0 μC$

For outer surface.

Substitute for $6.0 μC$

$Qenclosed$ and $90 cm$ for $R$ in the expression (2).

$σsurface=6.0 μC4π(90 cm(1 m100 cm))2=0.59 μC/m2$

For $r<60 cm$

Substitute $2.5 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(2.5 μC(1 C106 μC))r2=(2.25×104 Nm2/C)1r2$

For $60 cm<r<90 cm$

Substitute $0 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(0 μC)r2=0$

For $r>90 cm$

Substitute $6.0 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(6.0 μC(1 C106 μC))r2=(5.4×104 Nm2/C)1r2$

Conclusion:

Thus, the total charge on the inner surface is $−2.5 μC$ and the surface charge density is $−0.55 μC/m2$ .The total charge on the outer surface is $6.0μC$ and the surface charge density is $0.59 μC/m2$ . Thus, the electric field of the spherical shell for $r<60 cm$ is $(2.25×104 Nm2/C)1r2$ , for $60 cm<r<90 cm$ is $0$ and for $r>90 cm$ is $(5.4×104 Nm2/C)1r2$ .

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Answer 5

Chapter 22, Problem 63P

(a)

To determine

The total and individual charge on the inner and outer surface of the spherical shell.

(a)

Expert Solution

Explanation of Solution

Given:

A positive point charge of $2.5 μC$ is placed at the center of a conducting sphere. The radius of the inner surface is $60 cm$ and the radius of the outer surface is $90 cm$ .

Formula used:

The positive charge at the center of the spherical sphere induces a charge of same magnitude but different in polarity on the inner surface. The negative charge on the inner surface induces a positive charge of same magnitude on the outer surface.

Write the expression for the surface charge density

$σsurface=QenclosedA$ ........ (1)

Here, $σsurface$ is the surface charge density, $Qenclosed$ is the total charge enclosed by the surface of the spherical shell and $A$ is the surface area of the spherical shell.

Write the expression for the surface area of the square sheet.

$A=4πR2$

Here $R$ is the radius of the spherical shell.

Substitute $4πR2$ for $A$ in expression (1).

$σsurface=Qenclosed4πR2$ ........ (2)

Forinner surface of the spherical shell.

The induced charge on the inner surface has same magnitude of the point charge .The polarity of the charge is opposite to the point charge.

Calculation:

Substitute $−2.5 μC$ for $Qenclosed$ and $60 cm$ for $R$ in the expression (2).

$σsurface=−2.5 μC4π(60 cm(1 m100 cm))2=−0.55 μC/m2$

For outer surface of the spherical shell

The induced charge on the outer surface has same magnitude of the point charge .The polarity of the charge is same to the point charge

Substitute $2.5 μC$ for $Qenclosed$ and $90 cm$ for $R$ in the expression (2).

$σsurface=2.5 μC4π(90 cm(1 m100 cm))2=0.25 μC/m2$

Conclusion:

Thus, the total charge on the inner surface is $−2.5 μC$ and the surface charge density is $−0.55 μC/m2$ .The total charge on the outer surface is $2.5 μC$ and the surface charge density is $0.25 μC/m2$ .

(b)

To determine

The electric field due to a positive point charge located at the spherical shell.

(b)

Expert Solution

Explanation of Solution

Given:

A positive point charge of $2.5 μC$ is placed at the center of a conducting sphere. The radius of the inner surface is $60 cm$ and the radius of the outer surface is $90 cm$ .

Formula used:

Write the expression for Gauss’s law.

$∫E→.ds→=Qenclosedε0$

Here, $E→$ is the electric field, $ds→$ is the elementary surface area, $Qenclosed$ is the total charge enclosed by the spherical shell and $ε0$ is the permittivity of free space.

Substitute $4πr2$ for $ds→$ in the above expression.

$∫E⋅4πr2=Qenclosedε0$

Here $r$ is the distance from the center.

Rearrange the above equation for $E$ .

$E=Qenclosed4πr2⋅ε0$

Substitute $K$ for $1/4πε0$ in the above equation.

$E=KQenclosedr2$ ........ (3)

Calculation:

For $r<60 cm$

Substitute $2.5 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(2.5 μC(1 C106 μC))r2=(2.25×104 Nm2/C)1r2$

For $60 cm<r<90 cm$

Substitute $0 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(0 μC)r2=0$

For $r>90 cm$

Substitute $2.5 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(2.5 μC(1 C106 μC))r2=(2.25×104 Nm2/C)1r2$

Conclusion:

Thus, the electric field of the spherical shell for $r<60 cm$ is $(2.5×104 Nm2/C)1r2$ , For $60 cm<r<90 cm$ is $0$ and For $r > 90 cm$ is $(2.5×104 Nm2/C)1r2$ .

(c)

To determine

The electric field and surface charge density of the spherical shell.

(c)

Expert Solution

Explanation of Solution

Given:

A positive point charge of $3.5 μC$ is placed at the center of a conducting sphere. The radius of the inner surface is $60 cm$ and the radius of the outer surface is $90 cm$ .

Formula used:

The positive charge at the center of the spherical sphere induces a charge of same magnitude but different in polarity on the inner surface. The negative charge on the inner surface induces a positive charge of same magnitude on the outer surface.

Write the expression for the surface charge density

$σsurface=QenclosedA$

Here, $σsurface$ is the surface charge density, $Qenclosed$ is the total charge enclosed by the surface of the spherical shell and $A$ is the surface area of the spherical shell.

Write the expression for the surface area of the square sheet.

$A=4πR2$

Here $R$ is the radius of the spherical shell.

Substitute $4πR2$ for $A$ in expression (1).

$σsurface=Qenclosed4πR2$

For inner surface of the spherical shell.

The induced charge on the inner surface has same magnitude of the point charge .The polarity of the charge is opposite to the point charge

Write the expression for Gauss’s law.

$∫E→.ds→=Qenclosedε0$

Here, $E→$ is the electric field, $ds→$ is the elementary surface area, $Qenclosed$ is the total charge enclosed by the spherical shell and $ε0$ is the permittivity of free space.

Substitute $4πr2$ for $ds→$ in the above expression.

$∫E⋅4πr2=Qenclosedε0$

Here $r$ is the distance from the center.

Rearrange the above equation for $E$ .

$E=Qenclosed4πr2⋅ε0$

Substitute $K$ for $1/4πε0$ in the above equation.

$E=KQenclosedr2$

Calculation:

For inner surface $E=0$ inside a conductor.

Substitute for $−2.5 μC$

$Qenclosed$ and $60 cm$ for $R$ in the expression (2).

$σsurface=−2.5 μC4π(60 cm(1 m100 cm))2=−0.55 μC/m2$

For outer surface of the spherical shell

Write the expression for surface charge.

$qouter=3.50 μC -qinner$

Substitute $−2.5 μC$ for $qinner$

$qouter=6.0 μC$

For outer surface.

Substitute for $6.0 μC$

$Qenclosed$ and $90 cm$ for $R$ in the expression (2).

$σsurface=6.0 μC4π(90 cm(1 m100 cm))2=0.59 μC/m2$

For $r<60 cm$

Substitute $2.5 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(2.5 μC(1 C106 μC))r2=(2.25×104 Nm2/C)1r2$

For $60 cm<r<90 cm$

Substitute $0 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(0 μC)r2=0$

For $r>90 cm$

Substitute $6.0 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(6.0 μC(1 C106 μC))r2=(5.4×104 Nm2/C)1r2$

Conclusion:

Thus, the total charge on the inner surface is $−2.5 μC$ and the surface charge density is $−0.55 μC/m2$ .The total charge on the outer surface is $6.0μC$ and the surface charge density is $0.59 μC/m2$ . Thus, the electric field of the spherical shell for $r<60 cm$ is $(2.25×104 Nm2/C)1r2$ , for $60 cm<r<90 cm$ is $0$ and for $r>90 cm$ is $(5.4×104 Nm2/C)1r2$ .

Answer 6

Chapter 22, Problem 63P

(a)

To determine

The total and individual charge on the inner and outer surface of the spherical shell.

(a)

Expert Solution

Explanation of Solution

Given:

A positive point charge of $2.5 μC$ is placed at the center of a conducting sphere. The radius of the inner surface is $60 cm$ and the radius of the outer surface is $90 cm$ .

Formula used:

The positive charge at the center of the spherical sphere induces a charge of same magnitude but different in polarity on the inner surface. The negative charge on the inner surface induces a positive charge of same magnitude on the outer surface.

Write the expression for the surface charge density

$σsurface=QenclosedA$ ........ (1)

Here, $σsurface$ is the surface charge density, $Qenclosed$ is the total charge enclosed by the surface of the spherical shell and $A$ is the surface area of the spherical shell.

Write the expression for the surface area of the square sheet.

$A=4πR2$

Here $R$ is the radius of the spherical shell.

Substitute $4πR2$ for $A$ in expression (1).

$σsurface=Qenclosed4πR2$ ........ (2)

Forinner surface of the spherical shell.

The induced charge on the inner surface has same magnitude of the point charge .The polarity of the charge is opposite to the point charge.

Calculation:

Substitute $−2.5 μC$ for $Qenclosed$ and $60 cm$ for $R$ in the expression (2).

$σsurface=−2.5 μC4π(60 cm(1 m100 cm))2=−0.55 μC/m2$

For outer surface of the spherical shell

The induced charge on the outer surface has same magnitude of the point charge .The polarity of the charge is same to the point charge

Substitute $2.5 μC$ for $Qenclosed$ and $90 cm$ for $R$ in the expression (2).

$σsurface=2.5 μC4π(90 cm(1 m100 cm))2=0.25 μC/m2$

Conclusion:

Thus, the total charge on the inner surface is $−2.5 μC$ and the surface charge density is $−0.55 μC/m2$ .The total charge on the outer surface is $2.5 μC$ and the surface charge density is $0.25 μC/m2$ .

Answer 7

Chapter 22, Problem 63P

(a)

To determine

The total and individual charge on the inner and outer surface of the spherical shell.

Answer 8

(a)

Expert Solution

Explanation of Solution

Given:

A positive point charge of $2.5 μC$ is placed at the center of a conducting sphere. The radius of the inner surface is $60 cm$ and the radius of the outer surface is $90 cm$ .

Formula used:

The positive charge at the center of the spherical sphere induces a charge of same magnitude but different in polarity on the inner surface. The negative charge on the inner surface induces a positive charge of same magnitude on the outer surface.

Write the expression for the surface charge density

$σsurface=QenclosedA$ ........ (1)

Here, $σsurface$ is the surface charge density, $Qenclosed$ is the total charge enclosed by the surface of the spherical shell and $A$ is the surface area of the spherical shell.

Write the expression for the surface area of the square sheet.

$A=4πR2$

Here $R$ is the radius of the spherical shell.

Substitute $4πR2$ for $A$ in expression (1).

$σsurface=Qenclosed4πR2$ ........ (2)

Forinner surface of the spherical shell.

The induced charge on the inner surface has same magnitude of the point charge .The polarity of the charge is opposite to the point charge.

Calculation:

Substitute $−2.5 μC$ for $Qenclosed$ and $60 cm$ for $R$ in the expression (2).

$σsurface=−2.5 μC4π(60 cm(1 m100 cm))2=−0.55 μC/m2$

For outer surface of the spherical shell

The induced charge on the outer surface has same magnitude of the point charge .The polarity of the charge is same to the point charge

Substitute $2.5 μC$ for $Qenclosed$ and $90 cm$ for $R$ in the expression (2).

$σsurface=2.5 μC4π(90 cm(1 m100 cm))2=0.25 μC/m2$

Conclusion:

Thus, the total charge on the inner surface is $−2.5 μC$ and the surface charge density is $−0.55 μC/m2$ .The total charge on the outer surface is $2.5 μC$ and the surface charge density is $0.25 μC/m2$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

The electric field due to a positive point charge located at the spherical shell.

(b)

Expert Solution

Explanation of Solution

Given:

A positive point charge of $2.5 μC$ is placed at the center of a conducting sphere. The radius of the inner surface is $60 cm$ and the radius of the outer surface is $90 cm$ .

Formula used:

Write the expression for Gauss’s law.

$∫E→.ds→=Qenclosedε0$

Here, $E→$ is the electric field, $ds→$ is the elementary surface area, $Qenclosed$ is the total charge enclosed by the spherical shell and $ε0$ is the permittivity of free space.

Substitute $4πr2$ for $ds→$ in the above expression.

$∫E⋅4πr2=Qenclosedε0$

Here $r$ is the distance from the center.

Rearrange the above equation for $E$ .

$E=Qenclosed4πr2⋅ε0$

Substitute $K$ for $1/4πε0$ in the above equation.

$E=KQenclosedr2$ ........ (3)

Calculation:

For $r<60 cm$

Substitute $2.5 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(2.5 μC(1 C106 μC))r2=(2.25×104 Nm2/C)1r2$

For $60 cm<r<90 cm$

Substitute $0 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(0 μC)r2=0$

For $r>90 cm$

Substitute $2.5 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(2.5 μC(1 C106 μC))r2=(2.25×104 Nm2/C)1r2$

Conclusion:

Thus, the electric field of the spherical shell for $r<60 cm$ is $(2.5×104 Nm2/C)1r2$ , For $60 cm<r<90 cm$ is $0$ and For $r > 90 cm$ is $(2.5×104 Nm2/C)1r2$ .

Answer 13

(b)

To determine

The electric field due to a positive point charge located at the spherical shell.

Answer 14

(b)

Expert Solution

Explanation of Solution

Given:

A positive point charge of $2.5 μC$ is placed at the center of a conducting sphere. The radius of the inner surface is $60 cm$ and the radius of the outer surface is $90 cm$ .

Formula used:

Write the expression for Gauss’s law.

$∫E→.ds→=Qenclosedε0$

Here, $E→$ is the electric field, $ds→$ is the elementary surface area, $Qenclosed$ is the total charge enclosed by the spherical shell and $ε0$ is the permittivity of free space.

Substitute $4πr2$ for $ds→$ in the above expression.

$∫E⋅4πr2=Qenclosedε0$

Here $r$ is the distance from the center.

Rearrange the above equation for $E$ .

$E=Qenclosed4πr2⋅ε0$

Substitute $K$ for $1/4πε0$ in the above equation.

$E=KQenclosedr2$ ........ (3)

Calculation:

For $r<60 cm$

Substitute $2.5 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(2.5 μC(1 C106 μC))r2=(2.25×104 Nm2/C)1r2$

For $60 cm<r<90 cm$

Substitute $0 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(0 μC)r2=0$

For $r>90 cm$

Substitute $2.5 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(2.5 μC(1 C106 μC))r2=(2.25×104 Nm2/C)1r2$

Conclusion:

Thus, the electric field of the spherical shell for $r<60 cm$ is $(2.5×104 Nm2/C)1r2$ , For $60 cm<r<90 cm$ is $0$ and For $r > 90 cm$ is $(2.5×104 Nm2/C)1r2$ .

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

To determine

The electric field and surface charge density of the spherical shell.

(c)

Expert Solution

Explanation of Solution

Given:

A positive point charge of $3.5 μC$ is placed at the center of a conducting sphere. The radius of the inner surface is $60 cm$ and the radius of the outer surface is $90 cm$ .

Formula used:

The positive charge at the center of the spherical sphere induces a charge of same magnitude but different in polarity on the inner surface. The negative charge on the inner surface induces a positive charge of same magnitude on the outer surface.

Write the expression for the surface charge density

$σsurface=QenclosedA$

Here, $σsurface$ is the surface charge density, $Qenclosed$ is the total charge enclosed by the surface of the spherical shell and $A$ is the surface area of the spherical shell.

Write the expression for the surface area of the square sheet.

$A=4πR2$

Here $R$ is the radius of the spherical shell.

Substitute $4πR2$ for $A$ in expression (1).

$σsurface=Qenclosed4πR2$

For inner surface of the spherical shell.

The induced charge on the inner surface has same magnitude of the point charge .The polarity of the charge is opposite to the point charge

Write the expression for Gauss’s law.

$∫E→.ds→=Qenclosedε0$

Here, $E→$ is the electric field, $ds→$ is the elementary surface area, $Qenclosed$ is the total charge enclosed by the spherical shell and $ε0$ is the permittivity of free space.

Substitute $4πr2$ for $ds→$ in the above expression.

$∫E⋅4πr2=Qenclosedε0$

Here $r$ is the distance from the center.

Rearrange the above equation for $E$ .

$E=Qenclosed4πr2⋅ε0$

Substitute $K$ for $1/4πε0$ in the above equation.

$E=KQenclosedr2$

Calculation:

For inner surface $E=0$ inside a conductor.

Substitute for $−2.5 μC$

$Qenclosed$ and $60 cm$ for $R$ in the expression (2).

$σsurface=−2.5 μC4π(60 cm(1 m100 cm))2=−0.55 μC/m2$

For outer surface of the spherical shell

Write the expression for surface charge.

$qouter=3.50 μC -qinner$

Substitute $−2.5 μC$ for $qinner$

$qouter=6.0 μC$

For outer surface.

Substitute for $6.0 μC$

$Qenclosed$ and $90 cm$ for $R$ in the expression (2).

$σsurface=6.0 μC4π(90 cm(1 m100 cm))2=0.59 μC/m2$

For $r<60 cm$

Substitute $2.5 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(2.5 μC(1 C106 μC))r2=(2.25×104 Nm2/C)1r2$

For $60 cm<r<90 cm$

Substitute $0 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(0 μC)r2=0$

For $r>90 cm$

Substitute $6.0 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(6.0 μC(1 C106 μC))r2=(5.4×104 Nm2/C)1r2$

Conclusion:

Thus, the total charge on the inner surface is $−2.5 μC$ and the surface charge density is $−0.55 μC/m2$ .The total charge on the outer surface is $6.0μC$ and the surface charge density is $0.59 μC/m2$ . Thus, the electric field of the spherical shell for $r<60 cm$ is $(2.25×104 Nm2/C)1r2$ , for $60 cm<r<90 cm$ is $0$ and for $r>90 cm$ is $(5.4×104 Nm2/C)1r2$ .

Answer 19

(c)

To determine

The electric field and surface charge density of the spherical shell.

Answer 20

(c)

Expert Solution

Explanation of Solution

Given:

A positive point charge of $3.5 μC$ is placed at the center of a conducting sphere. The radius of the inner surface is $60 cm$ and the radius of the outer surface is $90 cm$ .

Formula used:

The positive charge at the center of the spherical sphere induces a charge of same magnitude but different in polarity on the inner surface. The negative charge on the inner surface induces a positive charge of same magnitude on the outer surface.

Write the expression for the surface charge density

$σsurface=QenclosedA$

Here, $σsurface$ is the surface charge density, $Qenclosed$ is the total charge enclosed by the surface of the spherical shell and $A$ is the surface area of the spherical shell.

Write the expression for the surface area of the square sheet.

$A=4πR2$

Here $R$ is the radius of the spherical shell.

Substitute $4πR2$ for $A$ in expression (1).

$σsurface=Qenclosed4πR2$

For inner surface of the spherical shell.

The induced charge on the inner surface has same magnitude of the point charge .The polarity of the charge is opposite to the point charge

Write the expression for Gauss’s law.

$∫E→.ds→=Qenclosedε0$

Here, $E→$ is the electric field, $ds→$ is the elementary surface area, $Qenclosed$ is the total charge enclosed by the spherical shell and $ε0$ is the permittivity of free space.

Substitute $4πr2$ for $ds→$ in the above expression.

$∫E⋅4πr2=Qenclosedε0$

Here $r$ is the distance from the center.

Rearrange the above equation for $E$ .

$E=Qenclosed4πr2⋅ε0$

Substitute $K$ for $1/4πε0$ in the above equation.

$E=KQenclosedr2$

Calculation:

For inner surface $E=0$ inside a conductor.

Substitute for $−2.5 μC$

$Qenclosed$ and $60 cm$ for $R$ in the expression (2).

$σsurface=−2.5 μC4π(60 cm(1 m100 cm))2=−0.55 μC/m2$

For outer surface of the spherical shell

Write the expression for surface charge.

$qouter=3.50 μC -qinner$

Substitute $−2.5 μC$ for $qinner$

$qouter=6.0 μC$

For outer surface.

Substitute for $6.0 μC$

$Qenclosed$ and $90 cm$ for $R$ in the expression (2).

$σsurface=6.0 μC4π(90 cm(1 m100 cm))2=0.59 μC/m2$

For $r<60 cm$

Substitute $2.5 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(2.5 μC(1 C106 μC))r2=(2.25×104 Nm2/C)1r2$

For $60 cm<r<90 cm$

Substitute $0 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(0 μC)r2=0$

For $r>90 cm$

Substitute $6.0 μC$ for $Qenclosed$ and $9×109N⋅m2/C2$ for $K$ in the expression (3).

$E=(9×109N⋅m2/C2)(6.0 μC(1 C106 μC))r2=(5.4×104 Nm2/C)1r2$

Conclusion:

Thus, the total charge on the inner surface is $−2.5 μC$ and the surface charge density is $−0.55 μC/m2$ .The total charge on the outer surface is $6.0μC$ and the surface charge density is $0.59 μC/m2$ . Thus, the electric field of the spherical shell for $r<60 cm$ is $(2.25×104 Nm2/C)1r2$ , for $60 cm<r<90 cm$ is $0$ and for $r>90 cm$ is $(5.4×104 Nm2/C)1r2$ .