Capacitance of a spherical conductor

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choose two spherical conductor systems as shown in the figure with the inner conductor – $M_{2}$ -radius-a and outer conductor – $M_{1}$ -radius-b .

Now induced field is directed from $M_{2}$ to $M_{1}$ . then the potential difference between the two conductors is

$V= -\int_{1}^{2}\overrightarrow{E} .\overrightarrow{dl}$ .

by assuming a point P between the two conductors such that P is out of inner spherical conductor ( $M_{2}$ ) an inside the outer conductor ( $M_{1}$ ).

$\therefore \overrightarrow{E}_{at P} = \frac{Q}{4\pi \epsilon _{o}r^{2} }\overrightarrow{a}_{r}$ .

$\therefore V = -\int_{1}^{2}\overrightarrow{E}.\overrightarrow{dl}$ .

$V = -\int_{1}^{2} \frac{Q}{4\pi \epsilon _{o}r^{2} }\overrightarrow{a}_{r}.(dr \overrightarrow{a}_{r }+d\phi \overrightarrow{a}_{\phi }+d\theta \overrightarrow{a}_{\theta })$ .

$V = -\int_{r=b}^{a} \frac{Q}{4\pi \epsilon _{o}r^{2} }\overrightarrow{a}_{r}.(dr \overrightarrow{a}_{r }+d\phi \overrightarrow{a}_{\phi }+d\theta \overrightarrow{a}_{\theta })$ .