General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Chapter 16, Problem 94E

(a)

To determine

The expression for the electric field for r>b.

(a)

Expert Solution
Check Mark

Answer to Problem 94E

The expression for the electric field for r>b is Q4πr2ε0_.

Explanation of Solution

The below Figure 1 shows the metallic sphere of outer radius b and inner radius a.

General Physics, 2nd Edition, Chapter 16, Problem 94E

Write the expression for the Gauss’s law.

  EdS=Qε0        (I)

Here, EdA is the area integral of the electric field E over a closed surface dS, Q is the amount of charge encompassed by that surface, and ε0 is the permittivity of the free space.

Write the expression for the surface area of a hollow sphere.

  dS=4πr2        (II)

Conclusion:

Substitute Equation (II) in (I) to find the electric field for r>b.

  E(4πr2)=Qε0E=Q4πr2ε0

Therefore, the expression for the electric field for r>b is Q4πr2ε0_.

(b)

To determine

The expression for the electric field for a<r<b.

(b)

Expert Solution
Check Mark

Answer to Problem 94E

The expression for the electric field for a<r<b is Q(r3a3)4πr2ε0(b3a3)_.

Explanation of Solution

From part (a), the expression for the Gauss’s law is,

  EdS=Qε0

And the expression for the surface area of a hollow sphere is,

  dS=4πr2

Use Equation (II) in (I).

E(4πr2)=Qε0E=Q4πr2ε0        (III)

The expression for the volume charge density of a hollow sphere is,

  ρ=Q43π(b3a3)        (IV)

Here, ρ is the volume charge density of a hollow sphere.

The volume of the insulating hollow sphere for a small element is,

  dV=4πr2dr        (V)

Here, dV is the volume of the smallest element, r is the radius of the sphere, and dr is the radius of the smallest element.

The charge enclosed by the smallest element in the hollow sphere between a<b is,

Q=arρ(r)dV        (VI)

Use Equation (V) in (VI).

Q=arρ(r)(4πr2dr)        (VII)

Use Equation (VII) in Equation (III).

E=arρ(r)(4πr2dr)4πr2ε0=4πρ4πr2ε0arr2(dr)=ρr2ε0[r33]ar=ρ3r2ε0(r3a3)        (VIII)

Conclusion:

Substitute Equation (IV) in (VIII) to find the electric field for a<r<b.

  E=Q(r3a3)43π(b3a3)(3ε0r2)=Q(r3a3)4πε0r2(b3a3)

Therefore, the expression for the electric field for a<r<b is Q(r3a3)4πr2ε0(b3a3)_.

(c)

To determine

The expression for the electric field of r>b.

(c)

Expert Solution
Check Mark

Answer to Problem 94E

The expression for the electric field for r>b is  zero.

Explanation of Solution

From part (a), the expression for the Gauss’s law is,

  EdS=Qε0

The charge q induces an equal and opposite charge on the inner surface of the spherical shell and will have equal charge on the outer surface of the sphere.

Any charge placed inside hollow spherical conductor attracts opposite charge from sphere.

Since sphere is neutral, an equal and opposite charge appears on inner surface of sphere. The charge inside a conducting shell is zero.

Conclusion:

Therefore, expression for the electric field for r>b is zero.

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Chapter 16 Solutions

General Physics, 2nd Edition

Ch. 16 - Prob. 11RQCh. 16 - Prob. 12RQCh. 16 - Prob. 13RQCh. 16 - Prob. 1ECh. 16 - Prob. 2ECh. 16 - Prob. 3ECh. 16 - Prob. 4ECh. 16 - Prob. 5ECh. 16 - Prob. 6ECh. 16 - Prob. 7ECh. 16 - Prob. 8ECh. 16 - Prob. 9ECh. 16 - Prob. 10ECh. 16 - Prob. 11ECh. 16 - Prob. 12ECh. 16 - Prob. 13ECh. 16 - Prob. 14ECh. 16 - Prob. 15ECh. 16 - Prob. 16ECh. 16 - Prob. 17ECh. 16 - Prob. 18ECh. 16 - Prob. 19ECh. 16 - Prob. 20ECh. 16 - Prob. 21ECh. 16 - Prob. 22ECh. 16 - Prob. 23ECh. 16 - Prob. 24ECh. 16 - Prob. 25ECh. 16 - Prob. 26ECh. 16 - Prob. 27ECh. 16 - Prob. 28ECh. 16 - Prob. 29ECh. 16 - Prob. 30ECh. 16 - Prob. 31ECh. 16 - Prob. 32ECh. 16 - Prob. 33ECh. 16 - Prob. 34ECh. 16 - Prob. 35ECh. 16 - Prob. 36ECh. 16 - Prob. 37ECh. 16 - Prob. 38ECh. 16 - Prob. 39ECh. 16 - Prob. 40ECh. 16 - Prob. 41ECh. 16 - Prob. 42ECh. 16 - Prob. 43ECh. 16 - Prob. 44ECh. 16 - Prob. 45ECh. 16 - Prob. 46ECh. 16 - Prob. 47ECh. 16 - Prob. 48ECh. 16 - Prob. 49ECh. 16 - Prob. 50ECh. 16 - Prob. 51ECh. 16 - Prob. 52ECh. 16 - Prob. 53ECh. 16 - Prob. 54ECh. 16 - Prob. 55ECh. 16 - Prob. 56ECh. 16 - Prob. 57ECh. 16 - Prob. 58ECh. 16 - Prob. 59ECh. 16 - Prob. 60ECh. 16 - Prob. 61ECh. 16 - Prob. 62ECh. 16 - Prob. 63ECh. 16 - Prob. 64ECh. 16 - Prob. 65ECh. 16 - Prob. 66ECh. 16 - Prob. 67ECh. 16 - Prob. 68ECh. 16 - Prob. 69ECh. 16 - Prob. 70ECh. 16 - Prob. 72ECh. 16 - Prob. 73ECh. 16 - Prob. 74ECh. 16 - Prob. 75ECh. 16 - Prob. 76ECh. 16 - Prob. 78ECh. 16 - Prob. 81ECh. 16 - Prob. 82ECh. 16 - Prob. 83ECh. 16 - Prob. 84ECh. 16 - Prob. 85ECh. 16 - Prob. 86ECh. 16 - Prob. 87ECh. 16 - Prob. 88ECh. 16 - Prob. 89ECh. 16 - Prob. 90ECh. 16 - Prob. 91ECh. 16 - Prob. 92ECh. 16 - Prob. 93ECh. 16 - Prob. 94ECh. 16 - Prob. 95ECh. 16 - Prob. 96ECh. 16 - Prob. 97ECh. 16 - Prob. 98ECh. 16 - Prob. 99ECh. 16 - Prob. 100ECh. 16 - Prob. 101ECh. 16 - Prob. 102ECh. 16 - Prob. 103ECh. 16 - Prob. 104ECh. 16 - Prob. 105ECh. 16 - Prob. 106ECh. 16 - Prob. 107ECh. 16 - Prob. 108E
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