Question

capacitance of a single and concentric sphere?
please no unnecessary answers​

Answer 1

Answer:

Let the concentric spheres have radius r1 and r2 then we can get the value of the electric field E for the charge sphere will be E = q/4πεr.

So, the value of the potential difference V for the concentric spheres of radius r1 and r2 will be integration of -E.ds from r2 to r1 where ds will be r.

Thus, on solving we will get the value of V to be  q(r2 - r1)/4πεr1r2.

We know that the value of Capacitance is q/V so we get the value of capacitance for concentric spheres to be (r2 - r1)/4πεr1r2.

Answer 2

Let us assume there is a charge of + q that has been distributed uniformly Over the surface of a sphere that has radius R1 Now this sphere share the same center with another sphere of radius R2 where are 2 is greater than R1.The electric field intensity of the sphere with radius R1 is given by :-

$\mathrm{E}=\frac{q}{4 \pi \varepsilon_{o} r^{2}} \mathrm{i}_{\mathrm{r}}$

The difference of potential between the sphere with radius R1 and the sphere with radius R2

$V_{12}=\frac{q}{4 \pi \varepsilon_{o}} \frac{r_{2}-r_{1}}{r_{1} r_{2}}$

The capacitance of the sphere will be  

$C_{12}=\frac{Q}{V_{12}}=4 \pi \varepsilon_{o} \frac{r_{1} r_{2}}{r_{2}-r_{1}}$