Question

Suppose the space between the two inner shells of the previous problem is filled with a dielectric of dielectric constant K. Find the capacitance of the system between A and B.

Answer 1

Since the space between the two inner shells is filled with a dielectric, capacitance C_AB becomes $C_{\mathrm{AB}}=\frac{4\pi {\in }_{0}abK}{(b-a)}$ and capacitance C_BC becomes $C_{\mathrm{BC}}=\frac{4\pi {\in }_{0}bc}{(c-b)}$.
Now, as the capacitors are in series, the equivalent capacitance is given by

$$\begin{gathered} \frac{1}{C}=\frac{1}{C_{\mathrm{AB}}}+\frac{1}{C_{\mathrm{BC}}} \\ \Rightarrow C=\frac{C_{\mathrm{AB}}{C}_{\mathrm{BC}}}{C_{\mathrm{AB}}+C_{\mathrm{BC}}} \\ \Rightarrow C=\frac{{\displaystyle \frac{(4\pi {\in }_0{)}^{2}a{b}^{2}Kc}{(b-a)(c-b)}}}{{\displaystyle \frac{4\pi {\in }_{0}abk}{(b-a)}}+{\displaystyle \frac{4\pi {\in }_{0}bc}{(c-b)}}} \\ \Rightarrow C=\frac{{\displaystyle \frac{4\pi {\in }_{0}Ka{b}^{2}c}{(b-a)(c-b)}}}{{\displaystyle \left(\frac{abk(c-b)+bc(b-a)}{(b-a)(c-b)}\right)}} \\ \Rightarrow C=\frac{4\pi {\in }_{0}Ka{b}^{2}c}{\left[abk\right(c-b)+bc(b-a\left)\right]} \\ \Rightarrow C=\frac{4\pi {\in }_{0}Kabc}{\left[ak\right(c-b)+c(b-a\left)\right]} \end{gathered}$$