$C=\frac{2 \pi \varepsilon_0 l}{\ln \left(\frac{b}{a}\right)}$
$C=\frac{2 \pi \varepsilon_0\left(\frac{20}{100}\right)}{\ln ^2}$
$\Delta V=\frac{10 \times 10^{-8} \ln 3}{2 \pi \varepsilon_0\left(\frac{20}{100}\right)}$
$\Delta V=\frac{10^{-4} \ln 2}{4 \pi \varepsilon_0}$
$\Delta V=\frac{0.1 \ln 2}{4 \pi \varepsilon_0} mV$


Reason : Charges in a conductor reside only at its surface
($1$) The value of $R$ is. . . . meter.
($2$) The value of $b$ is. . . . . .meter.

