A solid metallic sphere of radius r is charged to a voltage V. It is enclosed by a thin spherical shell of radius 2r. If q is the charge on the sphere and the shell each, the potential difference between them will be $\frac{x V}{2}$ . What is x? ___

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Solution

Charge on the inner sphere is, $q = c V = 4 π ϵ_{0} r V$
The potential difference between the two spheres is,
$V^{^{'}} = V_{1} - V_{2} = \frac{q}{4 π ϵ_{0}} (\frac{1}{r} - \frac{1}{2 r}) = \frac{q}{4 π ϵ_{0}} \frac{1}{2 r} = \frac{V}{2}$

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A solid metallic sphere of radius r is charged to a voltage V. It is enclosed by a thin spherical shell of radius 2r. If q is the charge on the sphere and the shell each, the potential difference between them will be xV2. What is x? ___

Charge on the inner sphere is, q=cV=4πϵ0rV The potential difference between the two spheres is, V′=V1−V2=q4πϵ0(1r−12r)=q4πϵ012r=V2

A solid metallic sphere of radius r is charged to a voltage V. It is enclosed by a thin spherical shell of radius 2r. If q is the charge on the sphere and the shell each, the potential difference between them will be $\frac{x V}{2}$ . What is x? ___

Charge on the inner sphere is, $q = c V = 4 π ϵ_{0} r V$
The potential difference between the two spheres is,
$V^{^{'}} = V_{1} - V_{2} = \frac{q}{4 π ϵ_{0}} (\frac{1}{r} - \frac{1}{2 r}) = \frac{q}{4 π ϵ_{0}} \frac{1}{2 r} = \frac{V}{2}$