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● 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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 (a) Sketch the electric field lines and the location of all charges.

(b) Graph the electric-field magnitude as a function of r

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
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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