Charge transfer from electrometer to metal ball

Question

I was solving a physics problem wherein an electrometer was charged to certain volts and then it was brought in contact to a neutral metal ball, then metal ball was taken away and earthed. This process was repeated $n$ times , where it was given that $m$ number of processes are required to reduce the potential of electrometer to half. Then how do we evaluate the potential of electrometer after the same process has been done $n$ times?
In the solution they have used half life method, that is, they claim since it takes $m$ number of processes to reduce the potential to half, so potential as a function of $n$ is given by $$V=dfrac{V_0}{2^{frac{n}{m}}}$$ where $V_0$ is the initial potential. But how do we know this will be the manner in which charge would get transferred? Couldn't it follow some other relation?

Philip Wood · Accepted Answer

Let the capacitance of the electrometer be $C_e$ and that of the ball be $C_b$. If the potential of the electrometer before the ball touches the electrometer for the $n^text{th}$ time is $V_{n-1}$ and the potential of the electrometer when the ball is in contact for the $n^text{th}$ time is $V_n$, then since charge is conserved and the ball, while in contact with the electrometer, is at the same potential:
$$C_e V_{n-1}= (C_e + C_b)V_n.$$
So
$$frac{V_n}{V_{n-1}}=frac{C_e}{C_e + C_b}= text{constant}$$
From which your result follows easily.

Charge transfer from electrometer to metal ball

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