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A conducting spherical shell of inner radius b and outer radius c is concentric with a small metal sphere of radius

a < b. The metal sphere has a positive charge Q. The total charge on the conducting spherical shell is –Q. What is the potential of the metal sphere?

I thought: The potential of the metal sphere is measured when a < r < b and V= V_{metal sphere}+ V_{inner radius of shell}+V_{outer radius of shell}

1) The metal sphere is to be considered as a point charge, therefore [tex]V_s_p_h_e_r_e= \frac{kQ}{r} [/tex]

2) The inner radius has charge -Q in order to compensate the charge of the metal sphere and if a< r < b then r < the inner radius of the metal sphere (b). Because the rule for a spherical shell is:

[tex] V= \frac{kQ}{r} [/tex] if r > R

[tex] V=\frac{kQ}{R} [/tex] if r < R

with R the radius of the spherical shell and r the distance from its center

V_{ inner radius of shell}= [tex]\frac{kQ}{b} [/tex]

3) The outer radius has charge 0 because a charge of -Q is induced on the inner sphere, a charge of +Q will be found on the outer spherical shell which already was -Q before the electrostatic equilibrium was established. Therefore Q_{ outer radius}= 0 and V_{ outer radius}=0

Summerizing:

V_{total}= [tex]\frac{kQ}{r} + \frac{kQ}{b} = kQ{\frac{1}{r} - \frac{1}{b}[/tex]

though my textbook tells me V_{total}= [tex]kQ{\frac{1}{a} - \frac{1}{b}[/tex]

So who is right and who is wrong?

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# Homework Help: Little question on the potential in a conducting spherical shell

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