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Question
Show that if we connect the smaller and the outer sphere by a wire, the charge q on the former will always flow to the latter, independent of how large the charge Q is.
Solution
To calculate the electric field intensity at any point P', where point P' lies outside of the spherical shell, imagine a Gaussian surface with centre O and radius x', as shown in the figure given above.
According to Gauss's theorem,
` E' xx 4πx^'2 = (q+Q)/ε_0 `
⇒`E = (q+Q)/(4πx'^ 2 )`
As the charge always resides only on the outer surface of a conduction shell, the charge flows essentially from the sphere to the shell when they are connected by a wire. It does not depend on the magnitude and sign of charge Q.
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