$=\frac{Q}{\epsilon_0}$
$=\frac{C V}{\epsilon_0}$
If $D > > d,$ the potential energy of the system is best given $b$
(Take $V =0$ at infinity $)$

($1$) The value of $R$ is. . . . meter.
($2$) The value of $b$ is. . . . . .meter.
$(A)$ $\frac{E_1}{E_2}=1$ $(B)$ $\frac{E_1}{E_2}=\frac{1}{K}$ $(C)$ $\frac{Q_1}{Q_2}=\frac{3}{K}$ $(D)$ $\frac{ C }{ C _1}=\frac{2+ K }{ K }$