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Recently I came upon the following question $-$

A conducting sphere of radius $R = 1 m$ is charged to a potential $V = 1000$ volts. A thin metal disc of radius $r = 1 cm$ mounted on an insulating handle is touched with the sphere making contact with one of its flat faces and then separated. After separation the disc is earthed and the process is repeated until the potential of the sphere becomes $V_{r} = 999$ volts. Approximately how many times has this process been repeated?

The official solution given claims that charge removed from the sphere is proportional to surface area of the disc.

However when I equate the potential at the contact point approximating the potential of the disc at its center, I get that charge removed is proportional to radii of the sphere and disc.

Which one is correct? What am I missing?

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When you touch the charged conducting sphere with the small isolated metal disc, the disk practically becomes part of the surface of the sphere so that its charge per area is equal to the charge per area of the disk. When you separate and ground the disk, you remove this charge from the disk and the remaining smaller charge redistributes uniformly. The removed charge is thus given by the ratio of disk to sphere surface area times the respective charge on the sphere. Using the capacitance formula for the sphere and its initial voltage to obtain the sphere charge, you can easily calculate how often you have to repeat the process to reduce the voltage to 0.999 of the original voltage.

This assumes that the insulator on which the disk is mounted does not significantly change the field distribution of the sphere.

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  • $\begingroup$ Thanks for the answer! But why does equating the potentials of disc and sphere not work though? $\endgroup$
    – Eisenstein
    Commented Apr 28 at 20:19
  • $\begingroup$ @Eisenstein It is implicit in the described process that the potentials of disk and sphere equate upon contact. $\endgroup$
    – freecharly
    Commented Apr 28 at 20:25
  • $\begingroup$ Also is the approximation that the disc becomes part of the sphere valid at this scale? $\endgroup$
    – Eisenstein
    Commented Apr 28 at 20:36
  • $\begingroup$ @Eisenstein Yes, if you have a very small and thin disk and the isolator it is mounted on has a negligible influence on the symmetric field distribution of the sphere when you touch with the disk! The disk should also make in good approximation a flat contact with the sphere. $\endgroup$
    – freecharly
    Commented Apr 28 at 20:45
  • $\begingroup$ " The disk should also make in good approximation a flat contact with the sphere" But the disc can make contact with the sphere at only one point though $\endgroup$
    – Eisenstein
    Commented Apr 28 at 20:57

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