● 22.46 A conducting spherical shell with inner radius a and outer radius b has a positive point charge Q located at its center. The total charge on the shell is -3Q, and it is insulated from its sur- roundings (Fig. P22.46). Figure P22.46 b -30
● 22.46 A conducting spherical shell with inner radius a and outer radius b has a positive point charge Q located at its center. The total charge on the shell is -3Q, and it is insulated from its sur- roundings (Fig. P22.46). Figure P22.46 b -30
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Question
(a) Sketch the electric field lines and the location of all charges.
(b) Graph the electric-field magnitude as a function of r

Transcribed Image Text:22.46 A conducting spherical shell with inner
radius a and outer radius b has a positive point
charge Q located at its center. The total charge on
the shell is -3Q, and it is insulated from its sur-
roundings (Fig. P22.46).
note that:
Part A
Derive the expression for the electric field magnitude in
terms of the distance r from the center for the region r < a.
We apply Gauss's law to the sphere of radius r shown with
the dotted line in the figure above.
* = f E-dà = E (4= r²) = 2
→ E
Q
4 to r
Part B
Derive the expression for the electric field magnitude in
terms of the distance r from the center for the region
a<r<b
The electric field in a conductor is zero.
has an r value in the conductor. Therefore
E=0.
Part C
Derive the expression for the electric field magnitude in
terms of the distance r from the center for the region r> b.
We apply Gauss's law to the sphere of radius r shown with
the dotted line in the figure above.
PE=
· = § Ë · dà = Е (4ñ r³²) = ²
Q-3Q
€0
Solving for the magnitude of E we get
2Q 1
Ancore=
E=
Figure
P22.46
Q1
2π co r²
a
b
-30
Part D
What is the surface charge density on the inner surface of
the conducting shell, Pin?
We apply Gauss's law to the sphere of radius r shown with
the dotted line in the figure above.
Þ£ = f Ë · dà = E (4ñ r²) =
Q+Pin (4π a²) = 0
Q+Pin (47a²)
The electric field, E, is zero inside the conductor so this
equation can be rewritten as
←
€0
Pin =
Q
4 a²
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