Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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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
Check Mark

Explanation of Solution

Given:

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

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

  σsurface=2.5μC4π(60cm(1m100cm))2=0.55μC/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 2.5μC for Qenclosed and 90cm for R in the expression (2).

  σsurface=2.5μC4π(90cm(1m100cm))2=0.25μC/m2

Conclusion:

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

(b)

To determine

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

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

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

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.

  E4πr2=Qenclosedε0

Here r is the distance from the center.

Rearrange the above equation for E .

  E=Qenclosed4πr2ε0

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

  E=KQenclosedr2   ........ (3)

Calculation:

For r<60cm

Substitute 2.5μC for Qenclosed and 9×109Nm2/C2 for K in the expression (3).

  E=(9×109Nm2/C2)(2.5μC(1C106μC))r2=(2.25×104Nm2/C)1r2

For 60cm<r<90cm

Substitute 0μC for Qenclosed and 9×109Nm2/C2 for K in the expression (3).

  E=(9×109Nm2/C2)(0μC)r2=0

For r>90cm

Substitute 2.5μC for Qenclosed and 9×109Nm2/C2 for K in the expression (3).

  E=(9×109Nm2/C2)(2.5μC(1C106μC))r2=(2.25×104Nm2/C)1r2

Conclusion:

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

(c)

To determine

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

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

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

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.

  E4πr2=Qenclosedε0

Here r is the distance from the center.

Rearrange the above equation for E .

  E=Qenclosed4πr2ε0

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

  E=KQenclosedr2

Calculation:

For inner surface E=0 inside a conductor.

Substitute for 2.5μC

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

  σsurface=2.5μC4π(60cm(1m100cm))2=0.55μC/m2

For outer surface of the spherical shell

Write the expression for surface charge.

  qouter=3.50μC-qinner

Substitute 2.5μC for qinner

  qouter=6.0μC

For outer surface.

Substitute for 6.0μC

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

  σsurface=6.0μC4π(90cm(1m100cm))2=0.59μC/m2

For r<60cm

Substitute 2.5μC for Qenclosed and 9×109Nm2/C2 for K in the expression (3).

  E=(9×109Nm2/C2)(2.5μC(1C106μC))r2=(2.25×104Nm2/C)1r2

For 60cm<r<90cm

Substitute 0μC for Qenclosed and 9×109Nm2/C2 for K in the expression (3).

  E=(9×109Nm2/C2)(0μC)r2=0

For r>90cm

Substitute 6.0μC for Qenclosed and 9×109Nm2/C2 for K in the expression (3).

  E=(9×109Nm2/C2)(6.0μC(1C106μC))r2=(5.4×104Nm2/C)1r2

Conclusion:

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

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