Question

Spherical CapacitorIt consists of two concentric conducting spherical shells of radii $a$ and $b(a<b)$. The inner shell is given a charge $Q$ and outer shell is given a charge $Q$. The potential difference between the shells is given by
$V_i-V_0=-{\int }_b^{a}Edr$
$\Rightarrow V=-{\int }_b^a\frac{Q}{4\pi {\epsilon }_o{r}^2}dr$
$\Rightarrow V=\frac{Q(b-a)}{4\pi {\epsilon }_{o}ab}$
Now,$C=\frac{Q}{V}\Rightarrow C=\frac{4\pi {\epsilon }_o\left(ab\right)}{(b-a)}$

Answer 1

$V_i-V_0=-{\int }_b^{a}Edr$

V = - ${\int }_b^a\frac{Q}{4\pi {ϵ}_0{r}^2}$

V = $\frac{Q(b-a)}{4\pi {ϵ}_{0}ab}$

Now,

C = $\frac{Q}{V}$

C = $\frac{4\pi {ϵ}_0\left(ab\right)}{b-a)}$