Question

Consider an assembly of three conducting concentric spherical shell of radii a, b and c as shown in figure (31-E28). Find the capacitance of the assembly between the points A and B.

Answer 1

Attached above hope you satisfied with my answer.

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Answer 2

The capacitance of the assembly between the points A and B is equal to $\frac{4 \pi \varepsilon_{0} \mathrm{ac}}{\mathrm{c}-\mathrm{a}}$

Explanation:

From the above figure we can consider three metallic hollow spheres form two spherical capacitors which are connected in series.

$C=\frac{4 \pi \varepsilon_{0} R_{2} R_{1}}{R_{2}-R_{2}}$

$Similarly for ( 2) and ( 3) C_{2}=\frac{4 \pi \varepsilon_{0} b c}{c-b}$

                                  $C_{\mathrm{eff}}=\frac{C_{1} C_{2}}{C_{1}+C_{2}}$ $\frac{\frac{\left(4 \pi \varepsilon_{0}\right)^{2} a b^{2} c}{(b-a)(c-a)}}{4 \pi \varepsilon_{0}\left[\frac{a b(c-b)+b c(b-a)}{(b-a)(c-b)}\right]}$

                                         $=\frac{4 \pi \varepsilon_{0} a b^{2} c}{a b c-a b^{2}+b^{2} c-a b c}$

                                         $=\frac{4 \pi \varepsilon_{0} a b^{2} c}{b^{2}(c-a)}=\frac{4 \pi \varepsilon_{0} a c}{c-a}$

From the above equation provided we can represent the variables as below,

Where C = Capacitance of the circuit

            A = Area of the plates of the capacitor

            d = distance between two capacitor plates

            k = dielectric constant of the material that used between the two plates of the capacitor. This dielectric constant is acts as a medium between the plates which helps to pass the conductivity between the plates.