$\frac{k q}{1}-\frac{k q}{2}+\frac{k q}{4}-\frac{k q}{8}+\ldots \ldots$
$k q\left[1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\ldots \ldots\right]$
$\frac{k q \cdot 1}{1-\left(\frac{-1}{2}\right)}=V$
$V=\frac{2 k q}{3}$



$K(x) = K_0 + \lambda x$ ( $\lambda =$ constant)
The capacitance $C,$ of the capacitor, would be related to its vacuum capacitance $C_0$ for the relation


Charge $- q $ is distributed on the surfaces as