$\mathrm{dq}=2 \pi \mathrm{rdr} \sigma$
Work to bring $\mathrm{dq}=\mathrm{dU}=\mathrm{Vdq}$
$\mathrm{dU}=4 \sigma r \cdot 2 \pi \mathrm{rdr} \sigma=8 \pi \sigma^{2} \mathrm{r}^{2} \mathrm{dr}$
$\mathrm{U}=8 \pi \sigma^{2} \int_{0}^{\mathrm{R}} \mathrm{r}^{2} \mathrm{dr}=\frac{8 \pi \sigma^{2}}{3} \mathrm{R}^{3}$


$V$. Energy stored in the above combination is $\mathrm{E}$. When these capacitors are connected in series to the same supply, the stored energy is $\frac{9}{\mathrm{x}} \mathrm{E}$. The value of $x$ is___________.

