Revision of $Capacitance$

Nice problem. Consider the geometry below:

This is the configuration of a spherical capacitor, the negative terminal of which is at infinity. The permittivities of each region is indicated in the diagram.

Let the charge on the innermost shereical conductor be $Q$ . At any point on the red spherical region, the radially outward electric field due to the conducting sphere is:

$E = \frac{Q}{4 \pi k \epsilon_o R^2}$

Let the electric potential at the surface of the conducting sphere be $V_i$ and that on the edge of the dielectrics be $V_o$ . To calculate the potential difference between these two surfaces:

$E = -\frac{dV}{dR} = \frac{Q}{4 \pi k \epsilon_o R^2}$

$dV = -\frac{Q}{4 \pi k \epsilon_o R^2} dR$ $\int_{V_i}^{V_o} dV = \int_{R_1}^{R_2} -\frac{Q}{4 \pi k \epsilon_o R^2} dR$ $\implies V_i - V_o = \frac{Q}{4 \pi k \epsilon_o} \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \dots (1)$

Performing the same steps to compute the potential difference between the edge of the dielectrics and the negative terminal of the capacitor:

$E = -\frac{dV}{dR} = \frac{Q}{4 \pi \epsilon_o R^2}$

$dV = -\frac{Q}{4 \pi \epsilon_o R^2} dR$ $\int_{V_o}^{V_{\infty}} dV = \int_{R_2}^{\infty} -\frac{Q}{4 \pi \epsilon_o R^2} dR$

$\implies V_o - V_{\infty} = \frac{Q}{4 \pi k \epsilon_o} \left(\frac{1}{R_2}\right) \dots (2)$

Adding (1) and (2) will give the potential difference between the terminals of the capacitor. After substituting and simplifying, that is:

$V_i - V_{\infty} =\frac{Q}{4 \pi k \epsilon_o R_1} \left(1- \frac{R_1(k-1)}{R_2}\right)$

Re-arranging and recognising that $Q = C (\Delta V)$ gives the required expression:

$C = \frac{4 \pi k \epsilon_o R_1}{1- \frac{R_1(k-1)}{R_2}}$

Therefore: $\boxed{a = 4 \ ; \ b = c = 1}$

The system is a series combination of two capacitors in the shapes of concentric spherical shells, one of inner radius $R_1$ , outer radius $R_2$ , electric permitivity $k \epsilon_0$ , and the other of inner radius $R_2$ , outer radius $\infty$ , electric permitivity $\epsilon_0$ . Their capacitances are $C_1=\dfrac{4πk \epsilon_0R_1R_2}{R_2-R_1}$ and $C_2=4π\epsilon_0R_2$ respectively. Their equivalent capacitance is $\dfrac{C_1C_2}{C_1+C_2}=\dfrac{4πk \epsilon_0R_1}{1+\dfrac{(k -1)R_1}{R_2}}$ . So $a=4, b=1, c=1$ and $a+b+c=\boxed 6$ .