Question 25.48: Figure 25-47 shows a parallel-plate capacitor with a plate a......
Figure 25-47 shows a parallel-plate capacitor with a plate area A = 5.56 cm² and separation d = 5.56 mm. The left half of the gap is filled with material of dielectric constant \kappa_1 = 7.00; the right half is filled with material of dielectric constant \kappa_2 = 12.0.What is the capacitance?
The capacitor can be viewed as two capacitors C_1and C_2 in parallel, each with surface area A/2 and plate separation d, filled with dielectric materials with dielectric constants \kappa_1 and \kappa_2, respectively. Thus, (in SI units),
\begin{aligned} C & =C_1+C_2=\frac{\varepsilon_0(A / 2) \kappa_1}{d}+\frac{\varepsilon_0(A / 2) \kappa_2}{d}=\frac{\varepsilon_0 A}{d}\left(\frac{\kappa_1+\kappa_2}{2}\right) \\ & =\frac{\left(8.85 \times 10^{-12} \,C ^2 / N \cdot m ^2\right)\left(5.56 \times 10^{-4} \,m ^2\right)}{5.56 \times 10^{-3} \,m }\left(\frac{7.00+12.00}{2}\right)=8.41 \times 10^{-12} \,F . \end{aligned}